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Arithmetic with Complex Numbers: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 Signed counters

A collection contains three counters worth $1$ , seven worth $-1$ , four worth $i$ , and two worth $-i$ . Two counters worth $1$ and three worth $i$ are then removed. Write the value of the remaining collection in the form $a+bi$ .
1. Hint 1
  Counters worth real values and counters worth imaginary values never mix, so keep a running total of each kind on its own.
2. Hint 2
  Removing counters subtracts their combined value, and opposite counters cancel within each kind.
Answer

$-6-i$ .

Full solution

Before the removal the real counters total
$3(1)+7(-1)=-4$

The imaginary counters total
$4i+2(-i)=2i$

So the collection starts at $-4+2i$ , and the counters taken away are worth $2+3i$ together.

Subtract part by part.
$(-4+2i)-(2+3i)=-6-i$

Answer

$-6-i$ .

Key idea

Complex addition and subtraction combine the real and imaginary parts separately.
Problem 1: Mark complete
Problem 2 A product in standard form

Write $(2-7i)(3+4i)$ in standard form.
1. Hint 1
  The two factors multiply like a pair of binomials, and only one of the four terms does anything a real product would not.
2. Hint 2
  The two imaginary terms multiply to a term in $i^2$ ; replace that $i^2$ with $-1$ before collecting like terms.
Answer

$34-13i$ .

Full solution

Expand the product term by term.
$6+8i-21i-28i^2$

Replace $i^2$ with $-1$ , which turns $-28i^2$ into $+28$ .

Collect the real terms $6$ and $28$ , then the imaginary terms $8i$ and $-21i$ .
$(2-7i)(3+4i)=34-13i$

Answer

$34-13i$ .

Key idea

Expanding like a pair of binomials and replacing $i^2$ with $-1$ returns a product to standard form.
Problem 2: Mark complete
Problem 3 A quotient in standard form

Write $\dfrac{12-14i}{5-3i}$ in standard form, then check the result by multiplying it by $5-3i$ .
1. Hint 1
  A denominator with an imaginary part can be made real by multiplying above and below by one well chosen number.
2. Hint 2
  Use the conjugate $5+3i$ , then divide each part of the new numerator by the real denominator.
Answer

$3-i$ .

Full solution

The conjugate of the denominator is $5+3i$ , and it makes the denominator real.
$(5-3i)(5+3i)=34$

Multiply the numerator by $5+3i$ as well.
$(12-14i)(5+3i)=102-34i$

Divide each part by $34$ .
$\frac{102-34i}{34}=3-i$

The check multiplies back.
$(3-i)(5-3i)=12-14i$

Answer

$3-i$ .

Key idea

Multiplying above and below by the denominator's conjugate turns a quotient into standard form.
Problem 3: Mark complete
Problem 4 Two-stage calculation

A machine first adds $1-2i$ to its input, then multiplies the result by $2+i$ . Its output is $5+5i$ . Find the input in standard form.
1. Hint 1
  Undo the operations in reverse order.
2. Hint 2
  Divide by the nonzero number $2+i$ using its conjugate, then undo the addition.
Answer

$2+3i$ .

Full solution

Let the input be $z$ .
Before the multiplication, its value was
$z+1-2i=\frac{5+5i}{2+i}$

The denominator is nonzero.
Multiply numerator and denominator by $2-i$ .
$\frac{5+5i}{2+i}=\frac{15+5i}{5}$

This is $3+i$ .
Undo the first stage.
$z=(3+i)-(1-2i)$

Therefore $z=2+3i$ .
Checking forward gives $3+i$ after the addition and $5+5i$ after multiplication.

Answer

$2+3i$ .

Key idea

Reverse a sequence of complex operations in the opposite order to recover its input.
Problem 4: Mark complete
Problem 5 Two connected arrows

The figure shows two arrows placed tip to tail. Each arrow represents the complex number given by its horizontal change plus $i$ times its vertical change. Write the product of the two represented numbers in standard form.

Two arrows placed tip to tail on the complex plane.

Text description of this figure

A square grid on the complex plane. The horizontal real axis and the vertical imaginary axis each run from negative four to four, with tick marks and gridlines at every whole number and equal unit lengths on both axes; the whole numbers from negative four to four are printed along the lower edge of the grid for the real axis and along the left edge for the imaginary axis. The origin is marked and labeled O. A marked point A sits two units to the left of the imaginary axis and one unit above the real axis. A marked point B sits one unit to the right of the imaginary axis and three units below the real axis. One solid arrow runs from O to A, and a second solid arrow in a different color runs from A to B, starting where the first arrow ends. Only the letters O, A and B are shown: no coordinate pairs, no complex number labels and no other arrows.
1. Hint 1
  Read each arrow as a change, not as its endpoint alone.
2. Hint 2
  Subtract the starting coordinates from the ending coordinates for the second arrow, then multiply the two complex numbers.
Answer

$-2+11i$ .

Full solution

The first arrow runs from $(0,0)$ to $(-2,1)$ , so it represents $-2+i$ .
The second has horizontal change $1-(-2)=3$ and vertical change $-3-1=-4$ , so it represents $3-4i$ .

Expand the product of $-2+i$ and $3-4i$ .
$-6+11i-4i^2$

Replace $i^2$ with $-1$ .
$(-2+i)(3-4i)=-2+11i$

Answer

$-2+11i$ .

Key idea

A complex arrow represents its change in position, even when it does not start at the origin.
Problem 5: Mark complete
Problem 6 Adjacent inputs

Let $z=4-5i$ . Find $(z+1)^2-z^2$ in standard form, and check it using an expansion with $z$ left as a symbol.
1. Hint 1
  One route evaluates both squares; another expands before substituting.
2. Hint 2
  In the symbolic expansion the $z^2$ terms cancel.
Answer

$9-10i$ .

Full solution

Here $z+1=5-5i$ .
Square it.
$(5-5i)^2=-50i$

The other square is
$(4-5i)^2=-9-40i$

Subtracting gives $9-10i$ .
For a separate check, expand with $z$ unchanged.
$(z+1)^2-z^2=2z+1$

Substituting $z=4-5i$ into this shorter expression also gives $9-10i$ .

Answer

$9-10i$ .

Key idea

Expanding before substituting can reveal cancellation in expressions with complex numbers.
Problem 6: Mark complete
Problem 7 An imaginary denominator

Write $\dfrac{(1+4i)^2-5}{2i}$ in standard form.
1. Hint 1
  Simplify the squared expression and the subtraction before dividing.
2. Hint 2
  The denominator is nonzero; multiplication by its conjugate turns it into a real denominator.
Answer

$4+10i$ .

Full solution

Expand the square and subtract $5$ .
$(1+4i)^2-5=-20+8i$

The denominator $2i$ is nonzero.
Its conjugate is $-2i$ .
$\frac{-20+8i}{2i}=\frac{16+40i}{4}$

Splitting the fraction gives $4+10i$ .
Check by multiplying this result by $2i$ to recover $-20+8i$ .

Answer

$4+10i$ .

Key idea

Simplify the numerator fully before clearing an imaginary denominator.
Problem 7: Mark complete
Problem 8 A student's division

To write $\dfrac{8+11i}{6-i}$ in standard form, a student multiplies the numerator and the denominator by $6-i$ . Explain why that choice leaves the quotient no nearer standard form, and give the correct standard form.
1. Hint 1
  A quotient reaches standard form only once its denominator is a real number, so ask what this student's denominator becomes.
2. Hint 2
  Compare the two products $(6-i)(6-i)$ and $(6-i)(6+i)$ : only one of them clears the $i$ .
Answer

The student's factor leaves the denominator $35-12i$ , which is not real; the correct standard form is $1+2i$ .

Full solution

Multiplying above and below by the same number is legitimate, since it multiplies the fraction by $1$ , but it helps only when it makes the denominator real.
$(6-i)(6-i)=35-12i$

That denominator still has an imaginary part, so no progress was made.
The factor that does work is the conjugate $6+i$ , because a number times its conjugate is $a^2+b^2$ .
$(6-i)(6+i)=37$

Multiply the numerator by $6+i$ as well.
$(8+11i)(6+i)=37+74i$

Dividing each part by $37$ gives $1+2i$ .

Answer

The student's factor leaves the denominator $35-12i$ , which is not real; the correct standard form is $1+2i$ .

Key idea

A quotient is in standard form only once its denominator is real, and for $6-i$ the conjugate $6+i$ gets there while a second $6-i$ does not.
Problem 8: Mark complete
Problem 9 A reciprocal claim

A student claims that every complex number $z$ satisfying $z\overline z=1$ also satisfies $1/z=z$ . Decide whether the claim is correct. If it is not, say what $1/z$ does equal for every complex number $z$ with $z\overline z=1$ .
1. Hint 1
  The product condition ensures $z$ is nonzero.
2. Hint 2
  Divide the given equality by $z$ , and test the proposed claim with a nonreal value.
Answer

False; $1/z=\overline z$ . A counterexample is $z=i$ .

Full solution

Since $z\overline z=1$ , neither factor is zero.
Dividing by $z$ gives
$\frac{1}{z}=\overline z$

For $z=i$ , the conjugate is $-i$ and their product is $1$ .
However,
$\frac{1}{i}=-i$

This differs from $i$ , so the proposed equality fails.

Answer

False; $1/z=\overline z$ . A counterexample is $z=i$ .

Key idea

A complex number with conjugate product one has its conjugate as its reciprocal.
Problem 9: Mark complete
Problem 10 A negative real square

Let $z=a+bi$ with real $a$ and $b$ . A student says that if $z^2$ is a negative real number, then $a=0$ and $b\ne0$ . Is this necessarily true? Explain.
1. Hint 1
  A real square result has imaginary part zero.
2. Hint 2
  Use both the vanishing imaginary part and the negative real part to rule out one of the possibilities.
Answer

Yes; $a=0$ and $b\ne0$ .

Full solution

Expand the square.
$z^2=(a^2-b^2)+2abi$

For the result to be real, $2ab=0$ , so at least one of $a,b$ is zero.
If $b=0$ , its real part would be $a^2$ , which is not negative.
Therefore $b\ne0$ and $a=0$ .

With these conditions, the square is $-b^2$ , which is indeed negative.

Answer

Yes; $a=0$ and $b\ne0$ .

Key idea

A negative real square of a complex number forces the original number to be nonzero and pure imaginary.
Problem 10: Mark complete

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