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Imaginary 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 Matching powers

Find the smallest whole number $n$ greater than $100$ for which $i^n=i^7$ .
1. Hint 1
  Powers with matching remainders on division by $4$ have matching values.
2. Hint 2
  Find the remainder of $7$ , then inspect the whole numbers just above $100$ .
Answer

$n=103$ .

Full solution

The exponent $7$ has remainder $3$ on division by $4$ .
$i^7=-i$

Test the candidates in turn: $101$ leaves remainder $1$ and $102$ leaves remainder $2$ , so neither matches.
The next candidate does.
$103=4(25)+3$

Its remainder also gives $-i$ , so it satisfies the condition and is the smallest permitted exponent.

Answer

$n=103$ .

Key idea

Matching values of powers of $i$ can be found by matching exponent remainders.
Problem 1: Mark complete
Problem 2 A coefficient on the square

Solve $5x^2+180=0$ for all values of $x$ , real or imaginary.
1. Hint 1
  Isolate $x^2$ first; a negative value there is not a dead end once $i$ is available.
2. Hint 2
  Divide by the coefficient of $x^2$ before taking the square root of both sides, and keep both signs.
Answer

$x=\pm 6i$ , that is, $x=6i$ and $x=-6i$ .

Full solution

Subtract $180$ from both sides.
$5x^2=-180$

Divide both sides by $5$ to isolate the square.
$x^2=-36$

Take the square root of both sides, keeping both signs.
$x=\pm\sqrt{-36}$

The real square root of $36$ is $6$ , and the negative radicand supplies the factor $i$ .
$x=\pm 6i$

Check either value: squaring gives $-36$ , and the equation closes.
$5(-36)+180=0$

Answer

$x=\pm 6i$ , that is, $x=6i$ and $x=-6i$ .

Key idea

Dividing by $5$ reduces $5x^2+180=0$ to $x^2=-36$ , and an equation of that shape, $x^2=-k$ with $k>0$ , has the two solutions $\pm\sqrt{k}\,i$ .
Problem 2: Mark complete
Problem 3 Two terms, one value

Evaluate $(-5i)^2+\sqrt{-24}\,\sqrt{-54}$ .
1. Hint 1
  Work out the two terms separately, and notice that the second term multiplies two square roots of negative numbers.
2. Hint 2
  Rewrite each square root in the form $i\sqrt{\;}$ and simplify the radical left behind before multiplying, then replace every $i^2$ by $-1$ .
Answer

$-61$ .

Full solution

Squaring $-5i$ squares the real factor $-5$ and the unit $i$ separately, and $i^2=-1$ .
$(-5i)^2=-25$

In the second term both radicands are negative, so convert each root and simplify it before multiplying.
$\begin{aligned}\sqrt{-24}&=2i\sqrt{6}\\ \sqrt{-54}&=3i\sqrt{6}\end{aligned}$

The two factors of $i$ give $i^2=-1$ , and $\sqrt{6}\,\sqrt{6}=6$ .
$(2i\sqrt{6})(3i\sqrt{6})=-36$

Add the two real values.
$-25+(-36)=-61$

Multiplying the radicands first would give $\sqrt{1296}=36$ for the second term and the wrong total of $11$ .

Answer

$-61$ .

Key idea

Converting each square root of a negative to imaginary form before multiplying keeps the sign of the product right.
Problem 3: Mark complete
Problem 4 Repeated multiplication

A calculation starts at $-3i$ . Each step multiplies the current value by $i$ . What is the value after $9$ steps?
1. Hint 1
  Combine the repeated steps into a power of $i$ .
2. Hint 2
  Group eight of the nine factors into complete cycles of four.
Answer

$3$ .

Full solution

Nine steps multiply the starting value by $i^9$ .
$\text{Final value}=(-3i)i^9$

Two complete cycles leave one factor of $i$ .
$i^9=i$

Use $i^2=-1$ .
$(-3i)i=3$

The first step gives $3$ , and each additional four steps returns to that value, confirming the ninth step.

Answer

$3$ .

Key idea

Repeated multiplication by $i$ returns to its starting value after each four steps.
Problem 4: Mark complete
Problem 5 Three root factors

Write $\sqrt{-2}\sqrt{-3}\sqrt{-6}$ as a real multiple of $i$ .
1. Hint 1
  Convert every negative square root before combining factors.
2. Hint 2
  The three factors of $i$ give $i^3$ ; the positive radicands may be multiplied.
Answer

$-6i$ .

Full solution

Each negative square root contributes one factor of $i$ .
$\sqrt{-2}\sqrt{-3}\sqrt{-6}=i^3\sqrt{36}$

The real root is $6$ and $i^3=-i$ .
$i^3\sqrt{36}=-6i$

As a check, the first two factors give $-\sqrt{6}$ ; multiplying by $i\sqrt{6}$ also gives $-6i$ .

Answer

$-6i$ .

Key idea

Convert each negative root to imaginary form before combining several radical factors.
Problem 5: Mark complete
Problem 6 Equal expressions

Find all values of $x$ , real or imaginary, for which $2x^2-8=x^2-23$ .
1. Hint 1
  Collect the squared terms on one side and constants on the other.
2. Hint 2
  After isolating $x^2$ , keep both square roots.
Answer

$x=\pm i\sqrt{15}$ .

Full solution

Subtract $x^2$ and add $8$ on both sides.
$x^2=-15$

Take the square root of both sides, keeping both signs, which gives
$x=\pm\sqrt{-15}.$
The negative radicand supplies the factor $i$ .
$x=\pm i\sqrt{15}$

For either value, $x^2=-15$ .
The original left and right sides both equal $-38$ .

Answer

$x=\pm i\sqrt{15}$ .

Key idea

An equation involving squared terms may have two imaginary solutions after the terms are collected.
Problem 6: Mark complete
Problem 7 Specified root product

Find the positive real number $k$ such that $\sqrt{-5}\sqrt{-k}=-15$ .
1. Hint 1
  The product of two square roots of negative numbers carries a factor of $i^2$ , so it is negative.
2. Hint 2
  Convert the roots, isolate a positive real square root, and square.
Answer

$k=45$ .

Full solution

Because $k$ is positive, conversion gives
$\sqrt{-5}\sqrt{-k}=-\sqrt{5k}$

The required product therefore gives
$\sqrt{5k}=15$

Squaring and dividing by $5$ yields
$k=45$

Check: $i\sqrt{5}$ times $3i\sqrt{5}$ equals $-15$ .

Answer

$k=45$ .

Key idea

The sign from two factors of $i$ matters when reconstructing a radicand from a product.
Problem 7: Mark complete
Problem 8 Two opposite bases

For which whole numbers $n$ is $(-i)^n=i^n$ ? Justify your answer.
1. Hint 1
  Write $-i$ as $(-1)\cdot i$ , so the power splits into a power of $-1$ and a power of $i$ .
2. Hint 2
  No power of $i$ is zero, so compare the two sides separately in the even and the odd cases.
Answer

Exactly the even whole numbers $n$ , zero included, and no odd ones.

Full solution

Separate the base into its two factors with the power of a product, $(ab)^n=a^nb^n$ .
$(-i)^n=(-1)^n i^n$

When $n$ is even, $(-1)^n=1$ , so the right side reduces to $i^n$ and the two sides agree.
At $n=0$ each side is a nonzero base raised to the power zero, so both equal $1$ .

When $n$ is odd, $(-1)^n=-1$ , so the same identity gives
$(-i)^n=-i^n$

Agreement would then force $i^n=-i^n$ , and hence $i^n=0$ .
Every power of $i$ is one of $1$ , $i$ , $-1$ and $-i$ , none of them zero, so no odd exponent works.

The identity therefore holds for the even whole numbers and for no others.

Answer

Exactly the even whole numbers $n$ , zero included, and no odd ones.

Key idea

An even power removes the sign difference between $i$ and $-i$ , while an odd power preserves it.
Problem 8: Mark complete
Problem 9 A window of exponents

Find every whole number $n$ with $0\le n\le 20$ for which $i^n\sqrt{-4}$ is a positive real number.
1. Hint 1
  Rewrite the square root in imaginary form, so the expression becomes one power of $i$ times a real factor.
2. Hint 2
  A power of $i$ equals $1$ exactly when its exponent leaves remainder $0$ on division by $4$ ; apply that to the exponent you now have.
Answer

$n=3$ , $7$ , $11$ , $15$ and $19$ .

Full solution

Convert the square root first.
$\sqrt{-4}=2i$

The square root contributes one more factor of $i$ , so the exponent rises from $n$ to $n+1$ .
$i^n\sqrt{-4}=2i^{n+1}$

The factor $2$ is positive, so the value is a positive real exactly when $i^{n+1}=1$ , which happens when $n+1$ is a multiple of $4$ .

In the range $0\le n\le 20$ , the multiples of $4$ available to $n+1$ are $4$ , $8$ , $12$ , $16$ and $20$ .

Subtracting one from each gives the exponents $n=3$ , $7$ , $11$ , $15$ and $19$ .

Check $n=15$ : the expression becomes $2i^{16}$ , and $16$ is a multiple of $4$ , so the value is $2$ .

Answer

$n=3$ , $7$ , $11$ , $15$ and $19$ .

Key idea

Folding a square root of a negative into the power of $i$ turns a question about the value into a question about the exponent's remainder.
Problem 9: Mark complete
Problem 10 Two steps apart

For every whole number $n$ , is $i^{n+2}$ the opposite of $i^n$ ? Explain without listing separate exponent values.
1. Hint 1
  Split the exponent sum into a product.
2. Hint 2
  Use the value of the extra factor $i^2$ .
Answer

Yes, $i^{n+2}=-i^n$ .

Full solution

The exponent rule separates the two extra factors.
$i^{n+2}=i^n i^2$

Since $i^2=-1$ , this becomes
$i^{n+2}=-i^n$

The exponent rule also applies when $n=0$ , so the statement covers every whole number.

Answer

Yes, $i^{n+2}=-i^n$ .

Key idea

Increasing an exponent of $i$ by two reverses the sign of its value.
Problem 10: Mark complete

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