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Arithmetic of 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: Core (core-course level)

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Problem 1 of 10

Problem 1 A target reading

Find the complex number $z$ satisfying $\frac{z+2i}{3}=1-i$ . Give $z$ in standard form.
1. Hint 1
  Undo the real scaling before undoing the added imaginary term.
2. Hint 2
  Multiply both sides by $3$ , then isolate $z$ .
Answer

$z=3-5i$ .

Full solution

Multiply both sides by the nonzero constant $3$ .
$z+2i=3-3i$
Subtracting $2i$ gives $z=3-5i$ .
Check: adding $2i$ gives $3-3i$ , and dividing by $3$ returns $1-i$ .

Answer

$z=3-5i$ .

Key idea

An equation with complex addition and real scaling is solved by undoing those operations in reverse order.
Problem 1: Mark complete
Problem 2 A product with an offset

Simplify $(2+i)(1-3i)-4i$ to standard form.
1. Hint 1
  Expand the product before combining it with the remaining term.
2. Hint 2
  Two of the four product terms combine with the last $-4i$ term.
Answer

$5-9i$ .

Full solution

Expand the product first.
$(2+i)(1-3i)=2-6i+i-3i^2$
Replacing $i^2$ by $-1$ gives $2-5i+3=5-5i$ , and subtracting $4i$ gives $5-9i$ .

Answer

$5-9i$ .

Key idea

Expanding fully before combining with an extra term avoids losing a cross term.
Problem 2: Mark complete
Problem 3 An unknown adjustment

Find the complex number $w$ for which $\overline{w}=i(2-i)$ . Give $w$ in standard form.
1. Hint 1
  First put the given conjugate into standard form.
2. Hint 2
  Conjugating twice restores the starting number.
Answer

$w=1-2i$ .

Full solution

Expand the right side.
$i(2-i)=2i-i^2$
This is $1+2i$ , so taking the conjugate gives $w=1-2i$ .
Its conjugate is $1+2i$ , checking the equation.

Answer

$w=1-2i$ .

Key idea

Conjugation reverses itself, so a known conjugate determines the original number.
Problem 3: Mark complete
Problem 4 An output rule

A device uses the rule $F(z)=(1-2i)z+3$ . What input produces the output $4+3i$ ? Give the input in standard form and check it.
1. Hint 1
  Undo the addition before undoing the multiplication.
2. Hint 2
  The coefficient $1-2i$ is nonzero; its conjugate makes a real denominator.
Answer

$z=-1+i$ ; $F(-1+i)=4+3i$ .

Full solution

Subtract $3$ and divide by the nonzero coefficient.
$z=\frac{1+3i}{1-2i}$
Multiplying numerator and denominator by $1+2i$ gives numerator $-5+5i$ and denominator $5$ .
Hence $z=-1+i$ .

Check the product first:
$(1-2i)(-1+i)=1+3i$
Adding $3$ returns $4+3i$ .

Answer

$z=-1+i$ ; $F(-1+i)=4+3i$ .

Key idea

The conjugate allows a complex multiplication to be undone through division.
Problem 4: Mark complete
Problem 5 Two linked readings

Let $z=2-i$ and $w=-1+2i$ . Calculate $(z+w)\overline{(z-w)}$ in standard form.
1. Hint 1
  Find the sum and difference first, keeping their roles separate.
2. Hint 2
  Conjugate the entire difference, then expand the resulting product.
Answer

$6i$ .

Full solution

The sum is $z+w=1+i$ and the difference is $z-w=3-3i$ , whose conjugate is $3+3i$ .
Expanding and using $i^2=-1$ ,
$\begin{aligned} (1+i)(3+3i) &= 3+6i+3i^2 \\ &= 3+6i-3 \\ &= 6i \end{aligned}$

Answer

$6i$ .

Key idea

A conjugate applies to the completed number, including all terms of a difference.
Problem 5: Mark complete
Problem 6 A reference fraction

For a nonzero complex number $z$ , define $H(z)=\frac{\overline z}{z}$ . Find $H(7+4i)$ in standard form, then find $H(7-4i)$ .
1. Hint 1
  Each denominator is nonzero, so conjugate rationalization is valid.
2. Hint 2
  The two inputs are conjugates, so compare the two numerator squares.
Answer

$H(7+4i)=\frac{33}{65}-\frac{56}{65}i$ ; $H(7-4i)=\frac{33}{65}+\frac{56}{65}i$ .

Full solution

For the first input, rationalizing gives
$H(7+4i)=\frac{(7-4i)^2}{65}$
The square is $33-56i$ , so the value is $\frac{33}{65}-\frac{56}{65}i$ .

For the conjugate input,
$H(7-4i)=\frac{(7+4i)^2}{65}$
The square is $33+56i$ , giving $\frac{33}{65}+\frac{56}{65}i$ .
Multiplying each result by its original denominator returns its numerator.

Answer

$H(7+4i)=\frac{33}{65}-\frac{56}{65}i$ ; $H(7-4i)=\frac{33}{65}+\frac{56}{65}i$ .

Key idea

For this conjugate pair of inputs, the two outputs turned out to be conjugates of each other too.
Problem 6: Mark complete
Problem 7 A real product condition

Let $t$ be real and $z=t+i$ . Find every $t$ for which $z\overline z=z+\overline z$ , and give the corresponding $z$ .
1. Hint 1
  Both sides become real expressions in $t$ .
2. Hint 2
  The product is a sum of squares, while the sum keeps twice the real part.
Answer

$t=1$ ; $z=1+i$ .

Full solution

The product is $t^2+1$ and the sum is $2t$ , so
$t^2-2t+1=0$
This is $(t-1)^2=0$ , giving $t=1$ .
For $z=1+i$ , both $z\overline z$ and $z+\overline z$ equal $2$ , so the condition holds.

Answer

$t=1$ ; $z=1+i$ .

Key idea

Conjugate sums and products can turn a complex condition into a real equation.
Problem 7: Mark complete
Problem 8 A cancellation claim

A learner claims $z-\overline z$ is real for every complex number $z$ . Decide whether the claim is correct, and state exactly when the difference is real.
1. Hint 1
  Write $z=a+bi$ with $a$ and $b$ real.
2. Hint 2
  Subtraction cancels the real parts, so inspect the remaining imaginary coefficient.
Answer

False; $z-\overline z$ is real exactly when $z$ is real, and then it is $0$ .

Full solution

Using $z=a+bi$ gives
$z-\overline z=2bi$
This is real exactly when $2b=0$ , or $b=0$ .
For example, $z=i$ gives difference $2i$ , disproving the universal claim.
When $z$ is real, the difference is zero.

Answer

False; $z-\overline z$ is real exactly when $z$ is real, and then it is $0$ .

Key idea

The difference of conjugates isolates twice the imaginary term.
Problem 8: Mark complete
Problem 9 A rearranged quotient

For every nonzero complex number $z$ , a student writes $\frac{z}{z\overline z}=\frac1{\overline z}$ . Is this valid? Explain, including why both denominators are nonzero.
1. Hint 1
  A cancellation is valid when the canceled factor is nonzero.
2. Hint 2
  Write $z=a+bi$ and examine $z\overline z=a^2+b^2$ .
Answer

Valid for every $z\ne0$ .

Full solution

Write $z=a+bi$ with $a,b$ real, not both zero, since $z\ne0$ .
Then $\overline z=a-bi$ , which is nonzero for the same reason $z$ is.
Moreover
$z\overline z=a^2+b^2>0.$
Canceling the nonzero factor $z$ gives
$\frac{z}{z\overline z}=\frac1{\overline z}$
Multiplying either expression by $\overline z$ gives $1$ , so they agree.

Answer

Valid for every $z\ne0$ .

Key idea

Cancellation in complex fractions uses the same nonzero condition as in real fractions.
Problem 9: Mark complete
Problem 10 A product comparison

Can two non-real complex numbers that are not conjugates have a positive real product? Give an example or explain why none exists.
1. Hint 1
  What must the real and imaginary parts of a product satisfy for the product to be a positive real number?
2. Hint 2
  Try two pure imaginary numbers with opposite signs and unequal sizes.
Answer

Yes; $i$ and $-3i$ are not conjugates and have product $3$ .

Full solution

The conjugate of $i$ is $-i$ , so $-3i$ is not its conjugate.
Their product is
$i(-3i)=-3i^2$
Replacing $i^2$ by $-1$ gives $3$ , which is positive and real.

Answer

Yes; $i$ and $-3i$ are not conjugates and have product $3$ .

Key idea

A sufficient condition for a positive real product need not be a necessary condition.
Problem 10: Mark complete

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