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The Imaginary Unit and 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 recorded value

Write $\frac{6-\sqrt{-28}}{2}$ in standard form $a+bi$ , with any real radicals simplified.
1. Hint 1
  A square root of a negative number contributes an imaginary term.
2. Hint 2
  Write $\sqrt{-28}$ as $i\sqrt{28}$ , then divide both numerator terms by $2$ .
Answer

$3-i\sqrt7$ .

Full solution

Convert the negative radical before simplifying it.
$\sqrt{-28}=i\sqrt{28}$
Then simplify the real radical:
$i\sqrt{28}=2i\sqrt7$

Divide each term by the nonzero real denominator.
$\frac{6-2i\sqrt7}{2}=3-i\sqrt7$

Answer

$3-i\sqrt7$ .

Key idea

A real denominator divides both parts of a complex number.
Problem 1: Mark complete
Problem 2 A position in a cycle

What is the smallest integer $n$ greater than $30$ for which $i^n=-i$ ?
1. Hint 1
  The power depends on the exponent remainder after division by $4$ .
2. Hint 2
  The value $-i$ occupies remainder $3$ in the cycle.
Answer

$n=31$ .

Full solution

The value $-i$ occurs at exponents with remainder $3$ .
$31=4(7)+3$
Since $31$ is already the first integer greater than $30$ , it is the smallest possible answer.
$i^{31}=i^3$
The last expression equals $-i$ .

Answer

$n=31$ .

Key idea

The four-step cycle can find an exponent as well as evaluate a power.
Problem 2: Mark complete
Problem 3 Two real entries

Write the ordered pair consisting of the real part first and the imaginary part second for $-\sqrt3+\frac{i}{5}$ .
1. Hint 1
  Each entry is a real number, even though the original number is complex.
2. Hint 2
  The coefficient multiplying $i$ is the imaginary part.
Answer

$(-\sqrt3,\frac15)$ .

Full solution

Standard form separates the two real coefficients.
$z=-\sqrt3+\left(\frac15\right)i$
Thus the real part is $-\sqrt3$ and the imaginary part is $\frac15$ ; reconstructing from that pair returns the given number.

Answer

$(-\sqrt3,\frac15)$ .

Key idea

The imaginary part is the real coefficient of $i$ , not the whole imaginary term.
Problem 3: Mark complete
Problem 4 Two labels for one number

A card is labeled $u+(u^2-6)i$ , where $u$ is real. A second card is labeled $2-2i$ . Determine whether some value of $u$ makes the labels equal, and give that value if it exists.
1. Hint 1
  Both real and imaginary parts must match for the same value of $u$ .
2. Hint 2
  The real parts fix $u$ ; then check that value in the imaginary part.
Answer

Yes; $u=2$ .

Full solution

The real parts require $u=2$ .
At this value the imaginary coefficient is
$u^2-6=4-6$
This is $-2$ , as required.
Both labels are $2-2i$ , so the value works and the real-part equation makes it unique.

Answer

Yes; $u=2$ .

Key idea

A proposed equality of complex numbers must pass both part checks with the same parameters.
Problem 4: Mark complete
Problem 5 A signed output

A calculation produces $T=i^6\sqrt{-3}\sqrt{-12}$ . Find $T$ in standard form.
1. Hint 1
  Separate the power of $i$ from the radical conversions before combining them.
2. Hint 2
  The radicals become $i\sqrt3$ and $2i\sqrt3$ ; their product contains a factor of $i^2$ .
Answer

$6+0i$ , or $6$ .

Full solution

The cycle gives $i^6=-1$ .
Convert both roots before multiplying: $\sqrt{-3}=i\sqrt3$ and
$\sqrt{-12}=2i\sqrt3.$
Their product is
$\left(i\sqrt3\right)\left(2i\sqrt3\right)=6i^2$
Thus the radical product is $-6$ , and
$T=(-1)(-6)$
The result is $6$ , a real complex number.

Answer

$6+0i$ , or $6$ .

Key idea

Convert negative radicals first so that each factor of $i^2$ supplies its correct sign.
Problem 5: Mark complete
Problem 6 Recovering an input

For a positive real number $k$ , the expression $\sqrt{-k}$ equals $3i\sqrt{11}$ . Find $k$ and check that the convention $\sqrt{-k}=i\sqrt k$ selects this value of the square root.
1. Hint 1
  Squaring the given number reveals the radicand.
2. Hint 2
  After finding $k$ , compare $i\sqrt{k}$ with the stated expression.
Answer

$k=99$ ; $i\sqrt{99}=3i\sqrt{11}$ .

Full solution

Square the right side using $i^2=-1$ .
$(3i\sqrt{11})^2=-99$
Therefore $-k=-99$ , so $k=99$ .
The prescribed root is
$i\sqrt{99}=3i\sqrt{11}$
Its real radical coefficient is positive, so it is the chosen root rather than its opposite.

Answer

$k=99$ ; $i\sqrt{99}=3i\sqrt{11}$ .

Key idea

Squaring finds the radicand, while the radical convention distinguishes the two square roots.
Problem 6: Mark complete
Problem 7 An equation record

A record lists the equations $3t=2$ and $s^2=-3$ . Solve both, taking $t$ rational and $s$ complex, and explain how each equation illustrates an enlargement beyond a smaller familiar number system.
1. Hint 1
  An equation may fail in one number system while succeeding in a larger one.
2. Hint 2
  Ask whether the first equation has an integer solution and whether the second has a real one.
Answer

$t=\frac23$ ; $s=\pm i\sqrt3$ . Integers enlarge to rationals; reals enlarge to complex numbers.

Full solution

Dividing by $3$ gives
$t=\frac23$
This rational value solves an equation with no integer solution.
For the second equation,
$s=\pm i\sqrt3$
Each candidate squares to $-3$ , while a real square is at least zero.
The first enlargement supplies fractions and the second supplies nonreal numbers, while the arithmetic laws continue to hold.

Answer

$t=\frac23$ ; $s=\pm i\sqrt3$ . Integers enlarge to rationals; reals enlarge to complex numbers.

Key idea

Enlarging a number system supplies solutions without discarding the smaller system or its arithmetic.
Problem 7: Mark complete
Problem 8 A repeating comparison

A student says $i^{n+8}=i^n$ for every positive integer $n$ . Is the claim correct? Justify your answer.
1. Hint 1
  Use an exponent law to separate the extra eight factors.
2. Hint 2
  Eight factors of $i$ contain two complete four-factor cycles.
Answer

Correct for every positive integer $n$ .

Full solution

Exponent addition separates the expression.
$i^{n+8}=i^n i^8$
Since $i^8=(i^4)^2=1$ , the extra factor changes nothing, giving
$i^{n+8}=i^n.$

Answer

Correct for every positive integer $n$ .

Key idea

Adding a multiple of four to an exponent preserves its power of $i$ .
Problem 8: Mark complete
Problem 9 A self-consistent equation

Find every real number $a$ for which $(ai)^2=a$ .
1. Hint 1
  Expand $(ai)^2$ using $i^2=-1$ before setting it equal to $a$ .
2. Hint 2
  The resulting equation in $a$ factors.
Answer

$a=0$ or $a=-1$ .

Full solution

Expanding gives
$\begin{aligned} (ai)^2&=a^2i^2 \\ &=-a^2 \end{aligned}$
Setting this equal to $a$ gives $-a^2=a$ , so $a^2+a=0$ .
Factoring, $a(a+1)=0$ , giving $a=0$ or $a=-1$ .
Both values check: at $a=0$ , both sides are $0$ ; at $a=-1$ , $(-i)^2=-1$ , matching $a=-1$ .

Answer

$a=0$ or $a=-1$ .

Key idea

Squaring $ai$ always produces $-a^2$ , a real number, since $i^2=-1$ regardless of the real coefficient $a$ .
Problem 9: Mark complete
Problem 10 A pair of settings

The labels $(v-1)+(v+1)i$ and $2+7i$ are proposed for the same real value of $v$ . Determine whether some real value of $v$ makes them equal, and justify your decision.
1. Hint 1
  Different expressions can have the needed values at the same input.
2. Hint 2
  Match each part separately and see whether the two conditions agree.
Answer

No real value of $v$ makes them equal.

Full solution

Matching real parts gives $v-1=2$ , so $v=3$ .
Matching imaginary parts gives $v+1=7$ , so $v=6$ .
Since $3\ne6$ , no real value of $v$ satisfies both conditions at once, so the labels can never represent the same complex number.

Answer

No real value of $v$ makes them equal.

Key idea

Two labels agree only when a single value satisfies both the real-part and imaginary-part conditions at once.
Problem 10: Mark complete

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