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Complex Numbers: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 158 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. Powers that go around . 13 points. Question 1 of 10.

This question spends the powers of $i$ twice: once on two single powers, and once on a long sum of them.
1. Part A.
  Evaluate $i^{95}$ and $i^{40}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Evaluate the sum $i + i^2 + i^3 + \cdots + i^{95}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Explain why any four consecutive powers of $i$ add to $0$ , in a way that covers every starting exponent rather than one particular case. Then give the value of $i + i^2 + \cdots + i^N$ for each of the four possibilities for $N$ , and say how you know the four possibilities cover every $N$ .
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

$i^{95} = -i$ and $i^{40} = 1$ .
- $i^{95} = -i$ may be written $0 - 1i$ ; what is not the same is $i$ , which is the value at remainder $1$
Part B

The sum is $-1$ .
- $-1$ may be written $-1 + 0i$ ; an answer of $0$ has discarded a block that was never complete
Part C

Four consecutive powers factor as $i^k\left(1 + i + i^2 + i^3\right)$ , and the bracket is $0$ , so the block vanishes whatever $k$ is. What survives is the leftover part of a block, so by the remainder of $N$ on division by $4$ the sum is $0$ , $i$ , $i - 1$ or $-1$ .
Worked solution

Part A

Divide each exponent by $4$ and keep the remainder, since $i^n = i^r$ whenever $n = 4q + r$ .

$95 = 4 \times 23 + 3 \ \Rightarrow\ i^{95} = i^3 = -i, \qquad 40 = 4 \times 10 + 0 \ \Rightarrow\ i^{40} = i^0 = 1$

A remainder of $0$ means the exponent is a multiple of $4$ , and $\left(i^4\right)^{10} = 1^{10} = 1$ .

Part B

Take the terms in consecutive blocks of four. Each block is a power of $i$ times $1 + i + i^2 + i^3$ , and that bracket is $0$ .

$1 + i + i^2 + i^3 = 1 + i - 1 - i = 0$

The terms $i^1$ through $i^{92}$ form $23$ complete blocks and contribute nothing, leaving three terms. Since $93 = 4 \times 23 + 1$ , those are $i$ , $-1$ and $-i$ :

$i^{93} + i^{94} + i^{95} = i - 1 - i = -1$

Part C

Why a block vanishes. Four consecutive powers starting at $i^{k+1}$ share the factor $i^k$ :

$i^{k+1} + i^{k+2} + i^{k+3} + i^{k+4} = i^k\left(i + i^2 + i^3 + i^4\right) = i^k(i - 1 - i + 1) = i^k \cdot 0 = 0.$

The argument never used the value of $k$ , so it covers every starting exponent.

The four values. Write $N = 4q + r$ with $r$ one of $0, 1, 2, 3$ . The first $4q$ terms fall into $q$ complete blocks and contribute $0$ , so the whole sum equals the total of the first $r$ terms of the next block, and there are only four of those totals:

$r = 0: \ 0, \qquad r = 1: \ i, \qquad r = 2: \ i - 1, \qquad r = 3: \ -1.$

So the running sum only ever takes one of those four values, whatever $N$ is. Nothing is left out because dividing by $4$ hands every positive integer exactly one of those four remainders.

In one line

$i^{95} = -i$ and $i^{40} = 1$ ; the sum $i + i^2 + \cdots + i^{95}$ is $-1$ . Any four consecutive powers factor as $i^k\left(1 + i + i^2 + i^3\right) = 0$ , so all that survives a running sum $i + \cdots + i^N$ is the leftover part of a block: by the remainder of $N$ on division by $4$ , the value is $0$ , $i$ , $i - 1$ or $-1$ , and those four remainders cover every $N$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Reduces each exponent by its remainder on division by $4$ rather than by writing out a table of powers. . Worth 2 points.

Reports both values with their signs, and treats remainder $0$ as landing on $1$ . . Worth 1 point.

Part B 5 points

Groups the terms into blocks of four consecutive powers and shows that a block totals $0$ . . Worth 2 points.

Identifies exactly which terms are left over after the complete blocks and evaluates each of them. . Worth 2 points.

Reports a single number in standard form. . Worth 1 point.

Part C 5 points

Factors a common power of $i$ out of four consecutive terms and identifies the remaining bracket as $0$ , with no dependence on where the block starts. . Worth 3 points. needs an explanation, not just an answer

Gives the value of the running sum for each of the four remainders, and says why those four cover every $N$ . . Worth 2 points. needs an explanation, not just an answer
2. One family, and the two slots that sort it . 14 points. Question 2 of 10.

For each real number $t$ the expression $(2 + ti)^2$ names a complex number, and the whole family is described by two real expressions in $t$ : one for the real part and one for the imaginary part. Which members are real and which are pure imaginary is settled by those two slots and nothing else.
1. Part A.
  Write $(2 + ti)^2$ in standard form $a + bi$ , with $a$ and $b$ given in terms of the real number $t$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Find every real $t$ for which $(2 + ti)^2$ is a real number, and every real $t$ for which it is pure imaginary. Give the value the square takes in each case.
  
  Carry your own answer forward Work from the two expressions you produced in part A, whatever they were. The credit here is for the reasoning you run on your own expressions, not for landing on particular values of $t$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate reads part B and concludes that the square of a non-real number is never real. Decide whether that general claim is true, and settle your decision with a specific number.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 5 points
The answer
Part A

$(2 + ti)^2 = \left(4 - t^2\right) + 4t\,i$ , so $a = 4 - t^2$ and $b = 4t$ .
- $4 - t^2 + 4ti$ is the same expression unbracketed; $4 + t^2 + 4ti$ is not, since $t^2i^2$ contributes with its sign flipped
Part B

Real only at $t = 0$ , where the square is $4$ . Pure imaginary at $t = 2$ and $t = -2$ , where the square is $8i$ and $-8i$ .
- $t = \pm 2$ names the same two values as $t = 2$ and $t = -2$ ; reporting $t = 0$ as pure imaginary is not the same, since $4$ has imaginary part $0$
Part C

The claim is false. The number $4i$ is not real, yet $(4i)^2 = 16i^2 = -16$ , which is real. Part B only ever ranged over the family $2 + ti$ , so it cannot settle a claim about all complex numbers.
Worked solution

Part A

Expand as an ordinary square and then exchange $i^2$ for $-1$ , which moves the last term into the real part with its sign flipped.

$(2 + ti)^2 = 4 + 4ti + t^2i^2 = \left(4 - t^2\right) + 4t\,i$

Both $4 - t^2$ and $4t$ are real numbers for every real $t$ , so this really is standard form.

Part B

A complex number is real when its imaginary part is $0$ , and pure imaginary when its real part is $0$ while the imaginary part is not.

$b = 4t = 0 \ \Longrightarrow\ t = 0, \qquad a = 4 - t^2 = 0 \ \Longrightarrow\ t = \pm 2$

At $t = 0$ the square is $4$ , a real number. At $t = 2$ it is $8i$ and at $t = -2$ it is $-8i$ , and in both cases the imaginary part $4t$ is nonzero, so both really are pure imaginary.

Part C

The claim is false, and one number refutes it. Take $z = 4i$ , whose imaginary part is $4 \ne 0$ , so $z$ is not real:

$(4i)^2 = 4^2 i^2 = 16(-1) = -16.$

The square is a real number, so a non-real number can certainly have a real square.

What went wrong in the reasoning is the range of the evidence. Part B searched one family, the numbers $2 + ti$ with real part fixed at $2$ , and inside that family the only real square did come from $t = 0$ . A statement about every complex number cannot be established by a family that leaves most of them out, and $4i$ is one of the numbers left out.

In one line

$(2 + ti)^2 = \left(4 - t^2\right) + 4t\,i$ . It is real only at $t = 0$ , where it equals $4$ , and pure imaginary at $t = \pm 2$ , where it equals $\pm 8i$ . The general claim that a non-real number never has a real square is false: $4i$ is not real and $(4i)^2 = -16$ is.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Expands the square completely, keeping the middle term. . Worth 1 point.

Replaces $t^2i^2$ by $-t^2$ and collects it into the real part. . Worth 2 points.

Names which expression is the real part and which is the imaginary part. . Worth 1 point.

Part B 5 points

Sets the imaginary part to $0$ for the real case and the real part to $0$ for the pure imaginary case, rather than the other way round. . Worth 2 points.

Solves both equations and keeps both values from the one that is quadratic. . Worth 2 points.

Checks that the other slot is nonzero in the pure imaginary case, and reports the value of the square each time. . Worth 1 point.

Part C 5 points

Gives a specific non-real number and computes its square, showing the square is real. . Worth 3 points.

Says why part B could not have settled the general claim, naming the restriction that the family carried. . Worth 2 points. needs an explanation, not just an answer
3. Clearing an i out of the basement . 15 points. Question 3 of 10.

Division is the one operation of this chapter that needed a theorem before it could even be attempted. This question runs it twice and then asks what the theorem was for.
1. Part A.
  Write $\dfrac{13 - 13i}{3 + 2i}$ in standard form $a + bi$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Write $\dfrac{1}{3 + 2i}$ in standard form, then verify your answer by multiplying it by $3 + 2i$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Explain why the multiplier you used in parts A and B is the right one, and why the procedure never breaks down for any divisor except $0$ . Your account should say what kind of number the new denominator is and why it cannot be zero when the divisor is not.
  
  Explain why it works A sentence or two. Reasons, not steps. 6 points
The answer
Part A

$1 - 5i$ .
- $\frac{13 - 65i}{13}$ is the same value before dividing each part by $13$ , so it is unfinished rather than wrong
Part B

$\dfrac{1}{3 + 2i} = \dfrac{3}{13} - \dfrac{2}{13}i$ , and multiplying it by $3 + 2i$ gives $1$ .
- $\frac{3 - 2i}{13}$ is the same number with the division left undone across the parts
Part C

Multiplying by $\frac{c - di}{c - di}$ is multiplying by $1$ , so the value is unchanged, and the new denominator $(c + di)(c - di) = c^2 + d^2$ is real, which is what makes the parts separable. A sum of two real squares is $0$ only when both are $0$ , so the denominator vanishes only for the divisor $0$ .
Worked solution

Part A

Multiply numerator and denominator by the conjugate of the DENOMINATOR, which is multiplying by $1$ . The denominator becomes $3^2 + 2^2 = 13$ , a real number.

$\frac{13 - 13i}{3 + 2i} = \frac{(13 - 13i)(3 - 2i)}{13} = \frac{39 - 26i - 39i + 26i^2}{13} = \frac{13 - 65i}{13} = 1 - 5i$

Check by multiplying back: $(1 - 5i)(3 + 2i) = 3 + 2i - 15i + 10 = 13 - 13i$ .

Part B

The same move with $1$ on top gives the reciprocal directly.

$\frac{1}{3 + 2i} = \frac{3 - 2i}{(3 + 2i)(3 - 2i)} = \frac{3 - 2i}{13} = \frac{3}{13} - \frac{2}{13}i$

The verification is one product, and the conjugate product does the work again:

$(3 + 2i)\cdot\frac{3 - 2i}{13} = \frac{3^2 + 2^2}{13} = \frac{13}{13} = 1.$

Part C

Why the conjugate. The fraction $\frac{c - di}{c - di}$ is $1$ for any nonzero $c + di$ , so multiplying by it changes the costume and not the value. What it buys is that the new denominator is a conjugate product, and the cross terms cancel while $i^2$ turns the subtracted square into an added one:

$(c + di)(c - di) = c^2 - d^2i^2 = c^2 + d^2.$

That is a real number, and dividing a complex number by a real number is done part by part, which is exactly what standard form requires.

Why it never fails. Both $c^2$ and $d^2$ are squares of real numbers, so neither is negative and their sum is $0$ only when both are $0$ , that is only when $c = d = 0$ . So the only divisor this procedure cannot handle is $0$ itself, the same single exception the real numbers always carried.

In one line

$\dfrac{13 - 13i}{3 + 2i} = 1 - 5i$ and $\dfrac{1}{3 + 2i} = \dfrac{3}{13} - \dfrac{2}{13}i$ , which multiplies back to $1$ . The denominator's conjugate works because $\frac{c - di}{c - di}$ is $1$ and $(c + di)(c - di) = c^2 + d^2$ is real, and that sum of two real squares is zero only when the divisor is $0$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Multiplies numerator and denominator by the denominator's conjugate, not by the numerator's and not by the denominator itself. . Worth 2 points.

Expands the numerator with every $i^2$ replaced by $-1$ , and evaluates the denominator as a real number. . Worth 2 points.

Divides BOTH parts by the real denominator and reports one real part and one imaginary part. . Worth 1 point.

Part B 4 points

Produces a reciprocal in standard form, with the real denominator split across both parts. . Worth 2 points.

Carries out the check and reaches $1$ exactly, rather than asserting that it works. . Worth 2 points.

Part C 6 points

Identifies the multiplier as a form of $1$ , so that the value of the quotient is unchanged. . Worth 2 points. needs an explanation, not just an answer

Computes the new denominator as $c^2 + d^2$ and names it as a real number, with the sign flip from $i^2$ accounted for. . Worth 2 points.

Argues from a sum of two real squares that the denominator is zero only for the divisor $0$ . . Worth 2 points. needs an explanation, not just an answer
4. Four corners built from two numbers . 16 points. Question 4 of 10.

Let $z = 13 + 4i$ and $w = 11 + 8i$ . Plot $0$ , $z$ , $w$ and $z + w$ . Because adding $w$ and then $z$ lands in the same place as adding $z$ and then $w$ , those four points are the corners of a parallelogram.
1. Part A.
  Compute $z + w$ and $z - w$ , then compute $|z|$ and $|w|$ and say how the two moduli compare.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  One diagonal of the parallelogram runs from $0$ to $z + w$ ; the other joins the corners at $w$ and at $z$ . Give the length of each diagonal, and decide which one is longer without estimating either square root.
  
  Carry your own answer forward Use the sum and the difference you found in part A, whatever they were. The credit here is for measuring the correct two segments and for settling the comparison exactly.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate argues from your part A moduli that all four sides of this parallelogram are the same length, and therefore that its two diagonals must be the same length as each other. Decide whether each half of that argument holds, and explain what side lengths do and do not fix.
  
  Carry your own answer forward Judge the claim against the two diagonal lengths you computed in part B, even if they were not the expected ones, and say honestly what your own numbers show. The credit is for the account of what a modulus does and does not record.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$z + w = 24 + 12i$ and $z - w = 2 - 4i$ ; $|z| = \sqrt{185}$ and $|w| = \sqrt{185}$ , so the two moduli are equal and the two arrows are the same length.
- $\sqrt{185}$ may be left exactly as it is; replacing it by a rounded decimal loses the fact that the two moduli are exactly equal
Part B

The diagonal from $0$ has length $|z + w| = \sqrt{720}$ ; the diagonal joining $w$ and $z$ has length $|z - w| = \sqrt{20}$ . The first is longer, because $720 > 20$ and both lengths are non-negative.
- $\sqrt{720} = 12\sqrt{5}$ and $\sqrt{20} = 2\sqrt{5}$ are the same two lengths simplified; comparing $720$ with $20$ is the same comparison as comparing the roots, since both are non-negative
Part C

The first half holds: opposite sides are copies of the arrows for $z$ and $w$ , so equal moduli make all four sides equal. The second half fails, and part B refutes it, $\sqrt{720}$ against $\sqrt{20}$ . Equal sides fix how far each arrow reaches, not how the two are placed, and the diagonals depend on the placement.
Worked solution

Part A

Combine the parts separately for the sum and the difference, distributing the minus sign through both parts of $w$ .

$z + w = (13 + 11) + (4 + 8)i = 24 + 12i, \qquad z - w = (13 - 11) + (4 - 8)i = 2 - 4i$

Each modulus is a Pythagorean computation, and the two happen to agree:

$|z| = \sqrt{169 + 16} = \sqrt{185}, \qquad |w| = \sqrt{121 + 64} = \sqrt{185}.$

Part B

The gap between two complex numbers is the modulus of their difference, so the second diagonal has length $|z - w|$ , while the first runs from $0$ and has length $|z + w|$ .

$|z + w| = \sqrt{24^2 + 12^2} = \sqrt{720} = 12\sqrt{5}, \qquad |z - w| = \sqrt{2^2 + (-4)^2} = \sqrt{20} = 2\sqrt{5}$

No decimals are needed for the comparison: both numbers are non-negative, so the one with the larger square is the larger number, and $720 > 20$ .

Part C

The first half is sound. The four sides are translated copies of the arrows for $z$ and for $w$ , two of each, and translation does not change length. So the sides have lengths $|z|$ , $|w|$ , $|z|$ , $|w|$ , and $|z| = |w| = \sqrt{185}$ makes all four equal.

The second half is false, and part B already refutes it:

$|z + w| = \sqrt{720} \ne \sqrt{20} = |z - w|.$

What the moduli record is how far each arrow reaches from $0$ , and nothing about the direction it reaches in. The diagonals are built from the sum and the difference, and those depend on how the two arrows are placed relative to each other. Two arrows of equal length can be nearly aligned, making $|z + w|$ nearly $2|z|$ and $|z - w|$ nearly $0$ , or nearly opposite, which reverses the two. Here they are nearly aligned: the triangle inequality caps the long diagonal at $|z| + |w| = 2\sqrt{185} = \sqrt{740}$ , and $\sqrt{720}$ sits just under it.

In one line

$z + w = 24 + 12i$ , $z - w = 2 - 4i$ , and $|z| = |w| = \sqrt{185}$ , so all four sides are equal. The diagonals are $|z + w| = \sqrt{720} = 12\sqrt{5}$ and $|z - w| = \sqrt{20} = 2\sqrt{5}$ , and the first is longer because $720 > 20$ . Equal sides therefore do not force equal diagonals: a modulus records how far an arrow reaches, not how the two arrows are placed relative to each other.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Adds and subtracts part by part, with the minus sign reaching both parts of the second number. . Worth 2 points.

Squares each part, adds, then takes the square root, leaving the two moduli exact. . Worth 2 points.

Compares the two moduli and says what the comparison means for the lengths of the two arrows. . Worth 1 point.

Part B 5 points

Measures the second diagonal as the modulus of a difference rather than as a difference of moduli. . Worth 2 points.

Evaluates both lengths exactly. . Worth 2 points.

Settles which is longer by comparing the quantities under the roots, with the reason that both lengths are non-negative. . Worth 1 point.

Part C 6 points

Accepts the first half with a reason tied to the sides being translated copies of the two arrows. . Worth 2 points. needs an explanation, not just an answer

Rejects the second half by pointing at two unequal diagonal lengths. . Worth 2 points.

Explains that a modulus records reach and not placement, so equal sides leave the diagonals free. . Worth 2 points. needs an explanation, not just an answer
5. Two roots, located twice . 17 points. Question 5 of 10.

The equation $2x^2 + 36x + 194 = 0$ has real coefficients. This question locates its two roots in the complex plane twice over: once by solving the equation, and once from the coefficients alone, without solving anything.
1. Part A.
  Solve the equation, reporting both roots in the form $p + qi$ with $p$ and $q$ real.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Now use the coefficients only. Give the distance from $0$ to each root, and the vertical line that both roots lie on, without using your answer to part A. Then check both against part A.
  
  Carry your own answer forward Compute this part from the coefficients, then compare it with whatever pair you produced in part A. If the two accounts disagree, say so and say which one you would trust; the credit is for the two routes and the comparison, not for agreement.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Explain why the two roots of ANY real quadratic with a negative discriminant must be the same distance from $0$ , and why the two loci in part B always meet in exactly two points. Then describe what becomes of that circle and that line as the discriminant of a real quadratic rises to $0$ and then past it.
  
  Explain why it works A sentence or two. Reasons, not steps. 7 points
The answer
Part A

$x = -9 + 4i$ and $x = -9 - 4i$ .
- $x = -9 \pm 4i$ names the same pair on one line; reporting only $-9 + 4i$ does not, because the square root carries both signs
Part B

Each root is $\sqrt{97}$ from $0$ , since the product $\tfrac{c}{a} = 97$ is the squared modulus, and both sit on the line of real part $-\tfrac{b}{2a} = -9$ . Part A agrees: $81 + 16 = 97$ .
- $\sqrt{97}$ is the distance and $97$ is its square; giving $97$ as the distance reports the product of the roots instead
Part C

A conjugate pair $p \pm qi$ has both moduli $\sqrt{p^2 + q^2}$ , since squaring erases the sign of $q$ , and the line meets that circle exactly where the imaginary part is $\pm q$ . At $\Delta = 0$ the two points merge into a single touch; past $0$ the roots are real and the line keeps only their midpoint.
Worked solution

Part A

Every coefficient is even, so divide through by $2$ first, which changes neither root. Then the discriminant is small enough to read.

$x^2 + 18x + 97 = 0, \qquad \Delta = 18^2 - 4(1)(97) = 324 - 388 = -64$

A negative discriminant no longer stops the work: $\sqrt{-64} = 8i$ , and the whole numerator is divided by $2a = 2$ .

$x = \frac{-18 \pm 8i}{2} = -9 \pm 4i$

The shared real part $-9$ is the axis of symmetry $-\tfrac{b}{2a}$ , computed from the original equation as $-\tfrac{36}{4}$ .

Part B

For a non-real pair $p \pm qi$ the product of the roots is a conjugate product, so it is the squared modulus of either root, and the product is also $\tfrac{c}{a}$ .

$r_1 r_2 = \frac{c}{a} = \frac{194}{2} = 97 = p^2 + q^2 = |z|^2 \quad\Longrightarrow\quad |z| = \sqrt{97}$

The sum of the roots is $-\tfrac{b}{a} = -18$ , and the pair shares one real part, so that real part is half the sum:

$p = -\frac{b}{2a} = -\frac{36}{4} = -9.$

So both roots sit on the circle of radius $\sqrt{97}$ about $0$ and on the vertical line of numbers with real part $-9$ . The roots from part A satisfy both: $(-9)^2 + (\pm 4)^2 = 97$ .

Part C

Equal distances. With real coefficients, a negative discriminant returns $x = -\tfrac{b}{2a} \pm \tfrac{\sqrt{4ac - b^2}}{2a}\,i$ , a conjugate pair $p \pm qi$ . The two moduli are computed from the same two squares, since squaring erases the sign of the imaginary part:

$|p + qi| = \sqrt{p^2 + q^2} = |p - qi|.$

Geometrically, conjugation reflects the plane across the real axis, and a reflection moves no point closer to or further from $0$ .

Exactly two meeting points. The circle has radius $\sqrt{p^2 + q^2}$ about $0$ and the line is the set of numbers with real part $p$ . A point $p + bi$ on that line lies on the circle when $p^2 + b^2 = p^2 + q^2$ , that is when $b = \pm q$ , and $q \ne 0$ because the discriminant is strictly negative. Two values of $b$ , two points, mirror images of each other.

As the discriminant rises. Keep $p = -\tfrac{b}{2a}$ , which is a real number whatever $\Delta$ does. While $\Delta < 0$ the offset $q$ is nonzero and the picture is the one just described. As $\Delta$ climbs toward $0$ , the radicand $4ac - b^2$ shrinks to $0$ , so $q$ shrinks to $0$ and the two meeting points slide together down the line:

$\Delta = 0 \quad\Longrightarrow\quad q = 0 \quad\Longrightarrow\quad \text{both roots equal } p.$

At that moment the product of the roots is $p^2$ , so the circle has shrunk to radius $|p|$ , and the line of real part $p$ meets it at exactly one point, the repeated root: the two crossings have become a single touch.

Past $0$ the two descriptions stop applying, and it is worth saying exactly how. With $\Delta > 0$ the roots are real numbers $p + d$ and $p - d$ with $d \ne 0$ . Their real parts are $p + d$ and $p - d$ , so the line of real part $p$ holds neither of them; it keeps only their midpoint. And $\tfrac{c}{a}$ is still their product, but the product of two real numbers is not a squared modulus: it is negative whenever the roots sit on opposite sides of $0$ , and then there is no circle of radius $\sqrt{c/a}$ to draw.

In one line

$2x^2 + 36x + 194 = 0$ has the roots $-9 \pm 4i$ . From the coefficients alone, the product $\tfrac{c}{a} = 97$ is the squared modulus, so each root is $\sqrt{97}$ from $0$ , and $-\tfrac{b}{2a} = -9$ is the shared real part: the roots are the two meeting points of that circle and that vertical line. As the discriminant of a real quadratic rises to $0$ those two points merge into one, where the line touches the circle; past $0$ the roots are real, the line keeps only their midpoint, and the product $\tfrac{c}{a}$ is no longer a squared modulus.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Computes the discriminant from correctly identified coefficients and finds it negative. . Worth 2 points.

Rewrites the root of the negative discriminant as $i$ times a real root and divides the WHOLE numerator by $2a$ . . Worth 2 points.

Reports both roots in the requested form, with a real part and an imaginary part. . Worth 1 point.

Part B 5 points

Uses the product of the roots as the squared modulus, dividing $c$ by $a$ rather than using $c$ alone. . Worth 2 points.

Obtains the shared real part as half the sum of the roots and states the line it determines. . Worth 2 points.

Checks both figures against the roots found in part A. . Worth 1 point.

Part C 7 points

Derives equal moduli from the roots being a conjugate pair, whether by the formula or by reflection across the real axis. . Worth 3 points. needs an explanation, not just an answer

Shows the line meets the circle exactly twice, using that the imaginary part is nonzero. . Worth 2 points. needs an explanation, not just an answer

Follows both loci through $\Delta = 0$ , naming the single touching point there, and says what stops holding once the roots are real. . Worth 2 points. needs an explanation, not just an answer
6. A root offered without any work . 16 points. Question 6 of 10.

A student claims that $1 + 5i$ is a root of $x^2 - 2x + 26$ and shows no working at all. The claim can be settled without solving the equation.
1. Part A.
  Evaluate $x^2 - 2x + 26$ at $x = 1 + 5i$ , showing each step, and state whether the claim holds.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Name the second root without solving the equation, state the hypothesis that entitles you to it, and confirm your pair against the product of the roots.
  
  Carry your own answer forward Pair up whichever root part A left you working with, even if the verification there did not come out as expected, and check your own pair against the coefficients. The credit is for naming the hypothesis and running the check, not for one particular pair.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  The student now says the two roots are "the same size, so neither of them is the bigger root". Decide which half of that sentence states something about these numbers and which half states nothing, and give the one number that the meaningful half reports.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

The value is $0$ , so the claim holds and $1 + 5i$ is a root.
- $0$ may be written $0 + 0i$ ; a nonzero answer means the substitution slipped, since both parts must vanish
Part B

The second root is $1 - 5i$ : every coefficient of $x^2 - 2x + 26$ is real, so a non-real root cannot appear without its conjugate. The product $(1 + 5i)(1 - 5i) = 26$ matches $\tfrac{c}{a}$ .
- $\overline{1 + 5i}$ names the same number as $1 - 5i$ ; $-1 - 5i$ does not, since conjugation leaves the real part alone
Part C

"The same size" is meaningful: it compares the moduli, which are real numbers, and both are $\sqrt{26}$ . "The bigger root" is not, because no ordering of the complex numbers survives the arithmetic rules, so there is no fact about which root is larger to report.
Worked solution

Part A

Square first, spending $i^2 = -1$ , then assemble the whole expression.

$(1 + 5i)^2 = 1 + 10i + 25i^2 = -24 + 10i$

$(-24 + 10i) - 2(1 + 5i) + 26 = (-24 - 2 + 26) + (10 - 10)i = 0$

The real part and the imaginary part vanish separately, which is what it means for a complex expression to be $0$ . So the claim holds.

Part B

The coefficients $1$ , $-2$ and $26$ are all real, so conjugating the whole equation carries a root to a root, and the second root is the conjugate.

$\overline{1 + 5i} = 1 - 5i$

The product of the two roots should be $\tfrac{c}{a} = 26$ , and a conjugate product is a sum of two squares:

$(1 + 5i)(1 - 5i) = 1^2 + 5^2 = 26. \ \checkmark$

The sum agrees too: $(1 + 5i) + (1 - 5i) = 2 = -\tfrac{b}{a}$ .

Part C

The meaningful half. Size, for a complex number, means the modulus, its distance from $0$ , and that is a real number that can be compared. Both roots give the same one:

$|1 + 5i| = \sqrt{1 + 25} = \sqrt{26} = |1 - 5i|,$

which also follows from the fact that conjugation reflects a point across the real axis and so cannot change its distance from $0$ .

The empty half. Asking which root is bigger presumes an ordering of the complex numbers that behaves like the one on the real line. There is none: if $i > 0$ , multiplying that inequality by the positive number $i$ gives $i^2 > 0$ , that is $-1 > 0$ ; if $i < 0$ then $-i > 0$ and the same move gives $(-i)^2 = -1 > 0$ again; and $i = 0$ fails since $0^2 \ne -1$ . Every placement collapses, so the phrase names nothing. Equal moduli say the two roots are the same distance from $0$ , and that is the whole of what "the same size" can report: it does not make them equal, since $1 + 5i \ne 1 - 5i$ .

In one line

Substituting gives $x^2 - 2x + 26 = 0$ at $x = 1 + 5i$ , so the claim holds; the second root is the conjugate $1 - 5i$ , licensed by the coefficients all being real and confirmed by the product $26 = \tfrac{c}{a}$ . Of the student's sentence, only the size half means anything: both roots have modulus $\sqrt{26}$ , while "the bigger root" names nothing, because no ordering of the complex numbers is consistent with the arithmetic.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Squares the number correctly, with $25i^2$ resolved into the real part. . Worth 2 points.

Distributes the $-2$ across both parts and collects real with real and imaginary with imaginary. . Worth 2 points.

Reads the result back as a verdict on the claim, noting that both parts came out $0$ . . Worth 1 point.

Part B 5 points

Gives the conjugate as the second root, flipping the sign of the imaginary part only. . Worth 1 point.

States the real-coefficient hypothesis as what licenses the pairing, rather than treating it as automatic. . Worth 2 points. needs an explanation, not just an answer

Checks the pair against the product of the roots, computed as a conjugate product. . Worth 2 points.

Part C 6 points

Identifies the size comparison as a statement about moduli and evaluates the common value. . Worth 2 points.

Rules out an ordering of the complex numbers by an argument, not by assertion. . Worth 3 points. needs an explanation, not just an answer

Notes that equal moduli do not make the two numbers equal. . Worth 1 point. needs an explanation, not just an answer
7. Two products that trade places . 16 points. Question 7 of 10.

For complex numbers $z$ and $w$ , the two products $z\overline{w}$ and $\overline{z}w$ are assembled from the same four real numbers and are not usually equal. This question finds the exact relationship between them, and then reads two consequences off it.
1. Part A.
  For $z = 6 + i$ and $w = 2 + 5i$ , compute $z\overline{w}$ and $\overline{z}w$ in standard form, and say how the two results are related.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Compute $z\overline{w} + \overline{z}w$ and $z\overline{w} - \overline{z}w$ for those same two numbers, and say what kind of number each result is.
  
  Carry your own answer forward Combine whichever two products you obtained in part A, whatever they were, and describe honestly what kind of numbers your own results are. The credit here is for the combination and the classification.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Using the conjugate test for realness, settle what kind of number $z\overline{w} + \overline{z}w$ is for EVERY pair of complex numbers, and prove it. Then do the same for $z\overline{w} - \overline{z}w$ , and say which real values that second combination is able to take.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 7 points
The answer
Part A

$z\overline{w} = 17 - 28i$ and $\overline{z}w = 17 + 28i$ . The two are conjugates of each other.
- $17 - 28i$ and $17 + 28i$ in either order is the same pair; a pair whose imaginary parts are not opposites of each other means a conjugate was taken on the wrong factor, or on both
Part B

The sum is $34$ , a real number. The difference is $-56i$ , a pure imaginary number.
- $34$ may be written $34 + 0i$ and $-56i$ as $0 - 56i$ ; reporting the difference as $56i$ reverses the order of subtraction
Part C

With $u = z\overline{w}$ , the rules that conjugation passes through products and undoes itself give $\overline{u} = \overline{z}w$ . The sum is $u + \overline{u}$ , which equals its own conjugate and so is real. The difference is $u - \overline{u}$ , the opposite of its own conjugate, so it is real only when it is $0$ .
Worked solution

Part A

Conjugate the correct factor each time, then expand and spend $i^2 = -1$ .

$z\overline{w} = (6 + i)(2 - 5i) = 12 - 30i + 2i + 5 = 17 - 28i$

$\overline{z}w = (6 - i)(2 + 5i) = 12 + 30i - 2i + 5 = 17 + 28i$

The real parts agree and the imaginary parts are opposites, so each result is the conjugate of the other.

Part B

Add and subtract the two results from part A part by part.

$(17 - 28i) + (17 + 28i) = 34, \qquad (17 - 28i) - (17 + 28i) = -56i$

In the sum the imaginary parts cancel and the real parts double; in the difference the real parts cancel and the imaginary parts double. So one is real and the other is pure imaginary.

Part C

The sum is always real. Put $u = z\overline{w}$ . Conjugation passes through a product and undoes itself, so

$\overline{u} = \overline{z\overline{w}} = \overline{z}\,\overline{\overline{w}} = \overline{z}w.$

The second product is therefore exactly $\overline{u}$ , and the sum in question is $u + \overline{u}$ . Conjugating it and using that conjugation passes through a sum,

$\overline{u + \overline{u}} = \overline{u} + u = u + \overline{u},$

so the sum equals its own conjugate and is real. (In parts, $u = p + qi$ gives $u + \overline{u} = 2p$ , twice the real part, which is where the $34$ in part B came from.)

The difference is real only at $0$ . The same two rules give

$\overline{u - \overline{u}} = \overline{u} - u = -\left(u - \overline{u}\right),$

so this number is the opposite of its own conjugate. If it were also real it would equal its own conjugate, so it would equal its own opposite: $v = -v$ forces $2v = 0$ and $v = 0$ . So the difference is never a nonzero real number. In parts it is $2qi$ , pure imaginary, and it is $0$ exactly when $z\overline{w}$ is real.

In one line

For $z = 6 + i$ and $w = 2 + 5i$ , $z\overline{w} = 17 - 28i$ and $\overline{z}w = 17 + 28i$ , so their sum is $34$ and their difference is $-56i$ . In general $\overline{z\overline{w}} = \overline{z}w$ , so the sum is $u + \overline{u}$ and is always real, while the difference is $u - \overline{u}$ , the opposite of its own conjugate, and is therefore real only when it is $0$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Forms each product with the conjugate on the correct factor, changing the sign of the imaginary part only. . Worth 2 points.

Expands both products and resolves every $i^2$ into the real part. . Worth 2 points.

Names the relationship between the two answers, that each is the conjugate of the other. . Worth 1 point.

Part B 4 points

Computes both combinations part by part, with the minus sign reaching both parts of the second product. . Worth 2 points.

Classifies each result, naming one as real and the other as pure imaginary. . Worth 2 points.

Part C 7 points

Identifies the second product as the conjugate of the first, citing that conjugation passes through products and that conjugating twice returns the number. . Worth 3 points. needs an explanation, not just an answer

Concludes that the sum is real from its equalling its own conjugate, rather than from the one worked example. . Worth 2 points. needs an explanation, not just an answer

Shows the difference is the opposite of its own conjugate and deduces that only $0$ can be real. . Worth 2 points. needs an explanation, not just an answer
8. Every number that cannot tell the two apart . 17 points. Question 8 of 10.

Consider the complex numbers $z$ that satisfy $|z| = |z - 6|$ , that is, the numbers whose distance from $0$ equals their distance from $6$ . Both sides are moduli, so both are distances.
1. Part A.
  Write $z = a + bi$ with $a$ and $b$ real, turn the equation into a statement about $a$ and $b$ , and simplify it as far as it will go.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Describe the solution set as a picture in the complex plane, and give two of its members, checking one of them in the original equation.
  
  Carry your own answer forward Describe and test the set your part A actually produced, even if it was not the expected one. The credit here is for turning your own condition into a picture and for checking a member of it honestly in the original equation.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
3. Part C.
  A classmate insists the answer must be a circle, since an equation in moduli describes a circle. Decide whether the objection has any force, and explain what feature of this particular equation produces the form of solution set you found in part B. Say what would have to change for the solution set to be a circle.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer
Part A

The equation becomes $a^2 + b^2 = (a - 6)^2 + b^2$ , which simplifies to $12a = 36$ , that is $a = 3$ . No condition on $b$ survives.
- $a = 3$ is the whole content; $a = 3$ together with any restriction on $b$ is not, since $b$ cancels completely
Part B

It is the vertical line of all numbers with real part $3$ , the mirror line midway between $0$ and $6$ . Two members are $3$ and $3 + 4i$ ; for the second, $|3 + 4i| = 5$ and $|3 + 4i - 6| = |-3 + 4i| = 5$ .
- The line may be named as the numbers $3 + bi$ with $b$ real, or as the perpendicular bisector of the segment from $0$ to $6$
Part C

The objection fails. A circle needs $|z - c| = r$ with $r$ a positive constant. Here both sides carry $z$ with coefficient $1$ , so squaring cancels $a^2$ and $b^2$ and leaves a linear equation, whose solutions form a line. Replacing one side by a positive constant, as in $|z - 6| = 4$ , restores the squared terms and with them the circle.
Worked solution

Part A

Both distances are moduli, so square both sides, which is safe because both are non-negative.

$|a + bi|^2 = |(a - 6) + bi|^2 \ \Longrightarrow\ a^2 + b^2 = (a - 6)^2 + b^2$

The $b^2$ terms are identical on the two sides and cancel, and so does $a^2$ once the right side is expanded:

$a^2 + b^2 = a^2 - 12a + 36 + b^2 \ \Longrightarrow\ 0 = -12a + 36 \ \Longrightarrow\ a = 3.$

Every squared term has gone, leaving one linear equation in $a$ and no condition at all on $b$ .

Part B

The condition $a = 3$ with $b$ free describes every number of the form $3 + bi$ : a vertical line in the plane, crossing the real axis at $3$ , which is the midpoint of $0$ and $6$ .

Take $z = 3$ and $z = 3 + 4i$ . The check on the second is two modulus computations:

$|3 + 4i| = \sqrt{9 + 16} = 5, \qquad |(3 + 4i) - 6| = |-3 + 4i| = \sqrt{9 + 16} = 5.$

The two distances agree, as the equation demands. The picture makes the answer unsurprising: the numbers equally far from two fixed points are exactly the numbers on the line that cuts the segment between them in half at a right angle.

Part C

The objection has no force as stated. What produces a circle is an equation of the form $|z - c| = r$ in which one side is a fixed POSITIVE number: squaring then gives $(a - p)^2 + (b - q)^2 = r^2$ , and the squared terms are what bend the picture.

Why this one comes out straight. Here $z$ sits inside the modulus on both sides, and on each side it carries coefficient $1$ , so both sides produce the same $a^2 + b^2$ and squaring destroys exactly those terms:

$a^2 + b^2 = (a - 6)^2 + b^2 \ \Longrightarrow\ 12a = 36.$

What is left is linear in $a$ and $b$ , and a linear equation in the two coordinates describes a line. The geometry says the same thing: the equation asks for the points that cannot tell $0$ from $6$ , and those are the points of the perpendicular bisector of the segment joining them.

What would restore a circle. Fix the right side as a positive constant. The equation $|z - 6| = 4$ keeps its squared terms, and its solution set is the circle of radius $4$ about $6$ . A modulus equation gives a circle when one side depends on $z$ and the other is a positive constant; put $0$ on the right and the solution set shrinks to a single point, and put a negative number there and nothing satisfies it at all.

In one line

Writing $z = a + bi$ turns $|z| = |z - 6|$ into $a^2 + b^2 = (a - 6)^2 + b^2$ , which collapses to $a = 3$ with $b$ free: the vertical line of numbers $3 + bi$ , the perpendicular bisector of the segment from $0$ to $6$ , containing $3$ and $3 + 4i$ . It is a line rather than a circle because both sides carry $z$ with coefficient $1$ , so squaring cancels $a^2$ and $b^2$ and leaves a linear equation; an equation such as $|z - 6| = 4$ , whose right side is a positive constant, keeps its squared terms and gives a circle.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Writes both distances with the modulus formula, subtracting $6$ from the real part only. . Worth 2 points.

Squares both sides and cancels the squared terms correctly. . Worth 2 points.

Reports the surviving condition and states that $b$ is unconstrained. . Worth 1 point.

Part B 5 points

Names the set as a vertical line and locates it, rather than describing it only in symbols. . Worth 2 points.

Gives two distinct members of the set. . Worth 1 point.

Verifies one member by computing both moduli in the original equation. . Worth 2 points.

Part C 7 points

Rejects the objection and identifies a positive constant on one side as what a circle equation requires. . Worth 2 points. needs an explanation, not just an answer

Traces the shape back to the cancellation of the squared terms between the two sides. . Worth 3 points. needs an explanation, not just an answer

Gives a modified equation whose solution set is a circle. . Worth 2 points.
9. A number recovered from two totals . 17 points. Question 9 of 10.

A complex number $z$ is never named, but two combinations of it and its conjugate are: $z + \overline{z} = 8$ and $z\,\overline{z} = 52$ . Notice that both totals are real numbers, which is a clue about what kind of information they can carry.
1. Part A.
  Find every complex number $z$ meeting both conditions.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Multiply out $(x - z)\left(x - \overline{z}\right)$ for your pair, report the quadratic, and say which of its coefficients could have been written down straight from the two given totals.
  
  Carry your own answer forward Build the quadratic from whichever number you found in part A and its conjugate, even if it was not the expected one, and compare its coefficients with the totals you were given. The credit here is for the expansion and for reading the coefficients off it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Explain why this construction returns real coefficients for EVERY complex $z$ , not just for this one. Then take a non-real $z$ and decide whether any second root other than $\overline{z}$ could give a monic quadratic with both coefficients real.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer
Part A

$z = 4 + 6i$ or $z = 4 - 6i$ .
- $4 \pm 6i$ names the same two numbers; giving only $4 + 6i$ omits the second, since the sign of the imaginary part is not determined
Part B

$(x - z)\left(x - \overline{z}\right) = x^2 - 8x + 52$ . Both come straight from the totals: the $x$ coefficient is the opposite of $z + \overline{z}$ , and the constant is $z\,\overline{z}$ .
- $x^2 - 8x + 52$ may be presented factored as $(x - 4 - 6i)(x - 4 + 6i)$ ; the expanded form is what shows the coefficients are real
Part C

For $z = a + bi$ the two totals are $z + \overline{z} = 2a$ and $z\overline{z} = a^2 + b^2$ , both real whatever $a$ and $b$ are, and the coefficients are $-(z + \overline{z})$ and $z\overline{z}$ . No other partner works: pairing $z$ with $w$ needs $z + w$ and $zw$ both real, and for non-real $z$ those two demands force $w = \overline{z}$ .
Worked solution

Part A

Write $z = a + bi$ with $a$ and $b$ real. The two combinations are the standard ones: the sum with the conjugate cancels the imaginary parts, and the product with the conjugate is a sum of squares.

$z + \overline{z} = 2a = 8 \ \Longrightarrow\ a = 4$

$z\,\overline{z} = a^2 + b^2 = 52 \ \Longrightarrow\ 16 + b^2 = 52 \ \Longrightarrow\ b^2 = 36 \ \Longrightarrow\ b = \pm 6$

Both signs are genuine solutions, so there are exactly two such numbers, $4 + 6i$ and $4 - 6i$ , and they are conjugates of each other.

Part B

Expand in letters first, so the pattern is visible before any numbers arrive.

$(x - z)\left(x - \overline{z}\right) = x^2 - \left(z + \overline{z}\right)x + z\,\overline{z}$

The two totals are exactly the two coefficients, so nothing needs to be computed:

$x^2 - 8x + 52.$

The check with the numbers from part A agrees: $(4 + 6i) + (4 - 6i) = 8$ and $(4 + 6i)(4 - 6i) = 16 + 36 = 52$ .

Part C

Why the coefficients are always real. With $z = a + bi$ the two totals are computed once and for all, and the quadratic's coefficients are the NEGATIVE of the first and the second as it stands:

$z + \overline{z} = (a + bi) + (a - bi) = 2a, \qquad z\,\overline{z} = a^2 + b^2.$

Both are built from the real numbers $a$ and $b$ by real arithmetic, so both are real for every choice of $z$ , and so is $-2a$ . The imaginary parts cancel in the sum and the $i^2$ turns the difference of squares into a sum in the product.

Why no other partner works. A monic quadratic with roots $z$ and $w$ is $x^2 - (z + w)x + zw$ , so both $z + w$ and $zw$ must be real. Take $z = a + bi$ with $b \ne 0$ , and write $w = c + di$ .

The sum $z + w = (a + c) + (b + d)i$ is real only if $d = -b$ , so $w = c - bi$ . With that, the product is

$zw = (a + bi)(c - bi) = \left(ac + b^2\right) + b(c - a)\,i,$

which is real only if $b(c - a) = 0$ . Since $b \ne 0$ , this forces $c = a$ , so $w = a - bi = \overline{z}$ . The conjugate is not merely one possible partner for a non-real root; it is the only one.

In one line

The conditions give $2a = 8$ and $a^2 + b^2 = 52$ , so $z = 4 + 6i$ or $z = 4 - 6i$ , and $(x - z)\left(x - \overline{z}\right) = x^2 - 8x + 52$ , whose coefficients are the two given totals. The construction is always real because $z + \overline{z} = 2a$ and $z\overline{z} = a^2 + b^2$ ; and for a non-real $z$ the conjugate is the ONLY partner that keeps both the sum and the product real.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Turns each condition into a real equation, using $2a$ for the sum and $a^2 + b^2$ for the product. . Worth 2 points.

Solves for both parts and keeps both signs of the imaginary part. . Worth 2 points.

Reports both numbers and notes that they are a conjugate pair. . Worth 1 point.

Part B 5 points

Expands the product in letters, reaching a middle coefficient that is the opposite of the sum and a constant that is the product. . Worth 2 points.

Reports the quadratic with real coefficients and no $i$ remaining. . Worth 2 points.

Identifies which coefficients were readable straight from the given totals. . Worth 1 point.

Part C 7 points

Computes the sum and the product in general letters and observes that both are real for every $z$ . . Worth 2 points. needs an explanation, not just an answer

Sets up the second question as the two demands that the sum and the product both be real. . Worth 2 points.

Derives from those two demands that the partner must be the conjugate, using that the imaginary part of $z$ is nonzero. . Worth 3 points. needs an explanation, not just an answer
10. Two unknown coefficients, and one equation that is really two . 17 points. Question 10 of 10.

Real numbers $b$ and $c$ are such that $12 + i$ satisfies $x^2 + bx + c = 0$ . There is one equation and two unknowns, which over the real numbers would leave the answer undetermined. Here it does not.
1. Part A.
  Substitute $12 + i$ into $x^2 + bx + c$ and write the result in standard form, with a real part and an imaginary part each expressed in terms of $b$ and $c$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Use part A to find $b$ and $c$ , then confirm your quadratic by checking that its other root is what the coefficients predict.
  
  Carry your own answer forward Work from whichever expression you produced in part A, even if it was not the expected one. The credit here is for what you do with your own expression and for running a check on the result.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Explain what entitled you to move from part A to part B at all, and say exactly where the assumption that $b$ and $c$ are real was spent. Then decide what happens to the question if $b$ and $c$ are allowed to be non-real: is the pair still determined?
  
  Explain why it works A sentence or two. Reasons, not steps. 7 points
The answer
Part A

$(143 + 12b + c) + (24 + b)\,i$ .
- $143 + 12b + c + (24 + b)i$ is the same expression unbracketed; a version with $145$ in the real part has read $i^2$ as $+1$
Part B

$b = -24$ and $c = 145$ , giving $x^2 - 24x + 145$ . Its other root is $12 - i$ : the sum $24$ matches $-b$ and the product $145$ matches $c$ .
- $x^2 - 24x + 145$ may be written with the coefficients listed separately; $b = 24$ has matched the parts without solving $24 + b = 0$
Part C

Two complex numbers are equal exactly when both parts match, which needs those parts to BE real; that is what the hypothesis supplies, and it is spent when the brackets of part A are called the real and imaginary parts. Non-real $b$ and $c$ carry four real unknowns against two real equations, so the pair is undetermined.
Worked solution

Part A

Square first, spending $i^2 = -1$ , then distribute $b$ and collect.

$(12 + i)^2 = 144 + 24i + i^2 = 143 + 24i$

$\left(143 + 24i\right) + b(12 + i) + c = (143 + 12b + c) + (24 + b)\,i$

Because $b$ and $c$ are real, the two brackets really are the real part and the imaginary part.

Part B

The expression must be $0$ , and a complex number is $0$ exactly when both of its parts are $0$ . So one complex equation becomes two real ones.

$24 + b = 0 \ \Longrightarrow\ b = -24, \qquad 143 + 12b + c = 0 \ \Longrightarrow\ 143 - 288 + c = 0 \ \Longrightarrow\ c = 145$

The quadratic is $x^2 - 24x + 145$ . Its coefficients are real, so the other root should be the conjugate $12 - i$ , and the two relations agree:

$(12 + i) + (12 - i) = 24 = -b, \qquad (12 + i)(12 - i) = 144 + 1 = 145 = c.$

Part C

What licensed the split. Equality of complex numbers is equality part by part: $a + bi = c + di$ exactly when $a = c$ and $b = d$ , PROVIDED all four of those are real numbers. Part A produced two brackets, and calling one of them the real part and the other the imaginary part is only correct because $b$ and $c$ are real; that is the single step where the hypothesis is spent. Without it, $143 + 12b + c$ could itself carry an imaginary contribution and there would be nothing to match.

Without the hypothesis. Let $b = b_1 + b_2i$ and $c = c_1 + c_2i$ with all four pieces real. Substituting and collecting still gives one complex equation, hence two real ones, but now there are FOUR real unknowns:

$143 + 12b_1 - b_2 + c_1 = 0, \qquad 24 + b_1 + 12b_2 + c_2 = 0.$

Two equations cannot pin down four numbers, so infinitely many pairs $(b, c)$ have $12 + i$ as a root. One is easy to name: $(x - (12 + i))(x - i) = x^2 - (12 + 2i)x + (-1 + 12i)$ has $12 + i$ as a root with non-real coefficients, and its second root is $i$ rather than $12 - i$ . So the determination in part B was bought entirely by the reality of the coefficients.

In one line

Substituting gives $(143 + 12b + c) + (24 + b)i = 0$ , so $b = -24$ and $c = 145$ , and $x^2 - 24x + 145$ has the second root $12 - i$ , matching the sum $24$ and the product $145$ . The split into two real equations is licensed by part-by-part equality, whose hypothesis is that the parts are real; if $b$ and $c$ may be non-real there are four real unknowns against two real equations and the coefficients are no longer determined.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Squares the number correctly, with the $i^2$ resolved into the real part. . Worth 2 points.

Distributes $b$ across both parts and gathers the expression into one real bracket and one imaginary bracket. . Worth 2 points.

States that the brackets are the two parts because $b$ and $c$ are real. . Worth 1 point.

Part B 5 points

Sets the real part and the imaginary part separately to zero, producing two real equations from one complex one. . Worth 2 points.

Solves the pair correctly, using the imaginary equation first because it carries one unknown. . Worth 2 points.

Checks the answer against the sum and the product of the roots. . Worth 1 point.

Part C 7 points

States part-by-part equality with its hypothesis that the parts are real, rather than as an unconditional rule. . Worth 2 points. needs an explanation, not just an answer

Points at the exact step where the reality of $b$ and $c$ is used. . Worth 2 points. needs an explanation, not just an answer

Argues that non-real coefficients leave four real unknowns against two real equations, so the pair is undetermined, and supports it with an example or a count. . Worth 3 points. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. Powers that go around 13 points

Part A 3

Reduces each exponent by its remainder on division by $4$ rather than by writing out a table of powers. 2
Reports both values with their signs, and treats remainder $0$ as landing on $1$ . 1

Part B 5

Groups the terms into blocks of four consecutive powers and shows that a block totals $0$ . 2
Identifies exactly which terms are left over after the complete blocks and evaluates each of them. 2
Reports a single number in standard form. 1

Part C 5

Factors a common power of $i$ out of four consecutive terms and identifies the remaining bracket as $0$ , with no dependence on where the block starts. 3
Gives the value of the running sum for each of the four remainders, and says why those four cover every $N$ . 2

2. One family, and the two slots that sort it 14 points

Part A 4

Expands the square completely, keeping the middle term. 1
Replaces $t^2i^2$ by $-t^2$ and collects it into the real part. 2
Names which expression is the real part and which is the imaginary part. 1

Part B 5

Sets the imaginary part to $0$ for the real case and the real part to $0$ for the pure imaginary case, rather than the other way round. 2
Solves both equations and keeps both values from the one that is quadratic. 2
Checks that the other slot is nonzero in the pure imaginary case, and reports the value of the square each time. 1

Part C 5

Gives a specific non-real number and computes its square, showing the square is real. 3
Says why part B could not have settled the general claim, naming the restriction that the family carried. 2

3. Clearing an i out of the basement 15 points

Part A 5

Multiplies numerator and denominator by the denominator's conjugate, not by the numerator's and not by the denominator itself. 2
Expands the numerator with every $i^2$ replaced by $-1$ , and evaluates the denominator as a real number. 2
Divides BOTH parts by the real denominator and reports one real part and one imaginary part. 1

Part B 4

Produces a reciprocal in standard form, with the real denominator split across both parts. 2
Carries out the check and reaches $1$ exactly, rather than asserting that it works. 2

Part C 6

Identifies the multiplier as a form of $1$ , so that the value of the quotient is unchanged. 2
Computes the new denominator as $c^2 + d^2$ and names it as a real number, with the sign flip from $i^2$ accounted for. 2
Argues from a sum of two real squares that the denominator is zero only for the divisor $0$ . 2

4. Four corners built from two numbers 16 points

Part A 5

Adds and subtracts part by part, with the minus sign reaching both parts of the second number. 2
Squares each part, adds, then takes the square root, leaving the two moduli exact. 2
Compares the two moduli and says what the comparison means for the lengths of the two arrows. 1

Part B 5

Measures the second diagonal as the modulus of a difference rather than as a difference of moduli. 2
Evaluates both lengths exactly. 2
Settles which is longer by comparing the quantities under the roots, with the reason that both lengths are non-negative. 1

Part C 6

Accepts the first half with a reason tied to the sides being translated copies of the two arrows. 2
Rejects the second half by pointing at two unequal diagonal lengths. 2
Explains that a modulus records reach and not placement, so equal sides leave the diagonals free. 2

5. Two roots, located twice 17 points

Part A 5

Computes the discriminant from correctly identified coefficients and finds it negative. 2
Rewrites the root of the negative discriminant as $i$ times a real root and divides the WHOLE numerator by $2a$ . 2
Reports both roots in the requested form, with a real part and an imaginary part. 1

Part B 5

Uses the product of the roots as the squared modulus, dividing $c$ by $a$ rather than using $c$ alone. 2
Obtains the shared real part as half the sum of the roots and states the line it determines. 2
Checks both figures against the roots found in part A. 1

Part C 7

Derives equal moduli from the roots being a conjugate pair, whether by the formula or by reflection across the real axis. 3
Shows the line meets the circle exactly twice, using that the imaginary part is nonzero. 2
Follows both loci through $\Delta = 0$ , naming the single touching point there, and says what stops holding once the roots are real. 2

6. A root offered without any work 16 points

Part A 5

Squares the number correctly, with $25i^2$ resolved into the real part. 2
Distributes the $-2$ across both parts and collects real with real and imaginary with imaginary. 2
Reads the result back as a verdict on the claim, noting that both parts came out $0$ . 1

Part B 5

Gives the conjugate as the second root, flipping the sign of the imaginary part only. 1
States the real-coefficient hypothesis as what licenses the pairing, rather than treating it as automatic. 2
Checks the pair against the product of the roots, computed as a conjugate product. 2

Part C 6

Identifies the size comparison as a statement about moduli and evaluates the common value. 2
Rules out an ordering of the complex numbers by an argument, not by assertion. 3
Notes that equal moduli do not make the two numbers equal. 1

7. Two products that trade places 16 points

Part A 5

Forms each product with the conjugate on the correct factor, changing the sign of the imaginary part only. 2
Expands both products and resolves every $i^2$ into the real part. 2
Names the relationship between the two answers, that each is the conjugate of the other. 1

Part B 4

Computes both combinations part by part, with the minus sign reaching both parts of the second product. 2
Classifies each result, naming one as real and the other as pure imaginary. 2

Part C 7

Identifies the second product as the conjugate of the first, citing that conjugation passes through products and that conjugating twice returns the number. 3
Concludes that the sum is real from its equalling its own conjugate, rather than from the one worked example. 2
Shows the difference is the opposite of its own conjugate and deduces that only $0$ can be real. 2

8. Every number that cannot tell the two apart 17 points

Part A 5

Writes both distances with the modulus formula, subtracting $6$ from the real part only. 2
Squares both sides and cancels the squared terms correctly. 2
Reports the surviving condition and states that $b$ is unconstrained. 1

Part B 5

Names the set as a vertical line and locates it, rather than describing it only in symbols. 2
Gives two distinct members of the set. 1
Verifies one member by computing both moduli in the original equation. 2

Part C 7

Rejects the objection and identifies a positive constant on one side as what a circle equation requires. 2
Traces the shape back to the cancellation of the squared terms between the two sides. 3
Gives a modified equation whose solution set is a circle. 2

9. A number recovered from two totals 17 points

Part A 5

Turns each condition into a real equation, using $2a$ for the sum and $a^2 + b^2$ for the product. 2
Solves for both parts and keeps both signs of the imaginary part. 2
Reports both numbers and notes that they are a conjugate pair. 1

Part B 5

Expands the product in letters, reaching a middle coefficient that is the opposite of the sum and a constant that is the product. 2
Reports the quadratic with real coefficients and no $i$ remaining. 2
Identifies which coefficients were readable straight from the given totals. 1

Part C 7

Computes the sum and the product in general letters and observes that both are real for every $z$ . 2
Sets up the second question as the two demands that the sum and the product both be real. 2
Derives from those two demands that the partner must be the conjugate, using that the imaginary part of $z$ is nonzero. 3

10. Two unknown coefficients, and one equation that is really two 17 points

Part A 5

Squares the number correctly, with the $i^2$ resolved into the real part. 2
Distributes $b$ across both parts and gathers the expression into one real bracket and one imaginary bracket. 2
States that the brackets are the two parts because $b$ and $c$ are real. 1

Part B 5

Sets the real part and the imaginary part separately to zero, producing two real equations from one complex one. 2
Solves the pair correctly, using the imaginary equation first because it carries one unknown. 2
Checks the answer against the sum and the product of the roots. 1

Part C 7

States part-by-part equality with its hypothesis that the parts are real, rather than as an unconditional rule. 2
Points at the exact step where the reality of $b$ and $c$ is used. 2
Argues that non-real coefficients leave four real unknowns against two real equations, so the pair is undetermined, and supports it with an example or a count. 3

Multiple choice

Free response