Unit 8 · Lesson 6 of 6

Complex numbers

About 25 minutes · 5 steps · one idea: \(i^2 = -1\)

You'll be able to: write square roots of negative numbers using \(i\), and add, subtract, multiply, and divide complex numbers, giving answers as \(a + bi\).

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.

\((2+\sqrt{3})(2-\sqrt{3})\)

A conjugate pair: \(2^2 - (\sqrt{3})^2 = 4 - 3 = 1\). The middle terms cancel. You'll use the same trick to divide complex numbers.

\(-3^2 + (-2)^2\)

\(-3^2\) means \(-(3^2) = -9\): the exponent goes on the 3 only. \((-2)^2 = 4\): the parentheses put the minus inside.

\(-9 + 4 = -5\).

Step 2 of 5The idea

No real number squares to a negative: \(3^2 = 9\) and \((-3)^2 = 9\). So mathematicians named a new number, the imaginary unit \(i\), with \(i^2 = -1\). That means \(\sqrt{-1} = i\).

A complex number is a number written \(a + bi\): a real part \(a\) and an imaginary part \(b\). For example, \(4 - 3i\) has real part 4 and imaginary part \(-3\).

Work with \(i\) like a variable, with one extra rule: whenever you see \(i^2\), replace it with \(-1\).

Roots of negatives: \(\sqrt{-a} = i\sqrt{a}\) for \(a \gt 0\). So \(\sqrt{-25} = 5i\). Do this first, before anything else.

Add or subtract: combine the real parts, then combine the imaginary parts.

Multiply: FOIL, then replace \(i^2\) with \(-1\).

Divide: multiply top and bottom by the conjugate of the bottom (\(c + di\) becomes \(c - di\)). The bottom turns into \(c^2 + d^2\), a real number.

Powers of \(i\) repeat every 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\). For \(i^n\), divide n by 4 and use the remainder.

Why this shows up on the placement test: they are the answers to quadratics like \(x^2 + 9 = 0\) (Unit 9), and College Algebra expects you to work with them.

Step 3 of 5Worked examples

Example 1: \((5-2i) - (3-6i)\), then \((4+i)(2-3i)\)

  1. Distribute the minus to both parts: \(5 - 2i - 3 + 6i\).
  2. Real parts: \(5 - 3 = 2\). Imaginary parts: \(-2i + 6i = 4i\). Answer: \(2 + 4i\).
  3. FOIL the product: \(8 - 12i + 2i - 3i^2\).
  4. Replace \(i^2\) with \(-1\): \(-3i^2 = -3(-1) = 3\). This is the step people forget.
  5. \(8 + 3 - 10i = 11 - 10i\).

Example 2: \(\dfrac{5+5i}{2+i}\), then \(i^{23}\)

  1. The conjugate of \(2+i\) is \(2-i\). Multiply top and bottom by it.
  2. Bottom: \((2+i)(2-i) = 4 - i^2 = 4 + 1 = 5\).
  3. Top: \((5+5i)(2-i) = 10 - 5i + 10i - 5i^2 = 10 + 5i + 5 = 15 + 5i\).
  4. Divide each part by 5: \(\dfrac{15 + 5i}{5} = 3 + i\).
  5. For \(i^{23}\): \(23 \div 4 = 5\) remainder 3, so \(i^{23} = i^3 = -i\). Every block of four is \(i^4 = 1\).

Step 4 of 5Practice

Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Type \(i\) as the letter i, like 3 - 2i.

\(\sqrt{-36}\)

\(\sqrt{-36} = i\sqrt{36} = 6i\).

Simplify \((3+4i) + (2-7i)\).

Real parts: \(3 + 2 = 5\). Imaginary parts: \(4i - 7i = -3i\). Answer: \(5 - 3i\).

Simplify \((6-2i) - (-1+5i)\).

Distribute the minus: \(6 - 2i + 1 - 5i\).

Real parts: \(6 + 1 = 7\). Imaginary parts: \(-2i - 5i = -7i\). Answer: \(7 - 7i\).

Simplify \((2+3i)(4-i)\).

FOIL: \(8 - 2i + 12i - 3i^2\).

Replace \(i^2\) with \(-1\): \(-3i^2 = 3\). So \(8 + 3 + 10i = 11 + 10i\).

\(\sqrt{-2} \cdot \sqrt{-8}\)

Write each with \(i\) first: \(i\sqrt{2} \cdot i\sqrt{8} = i^2\sqrt{16}\).

\(i^2 = -1\) and \(\sqrt{16} = 4\), so the answer is \(-4\) (not 4).

\(i^{30}\)

\(30 \div 4 = 7\) remainder 2, so \(i^{30} = i^2 = -1\).

Divide and write as \(a + bi\): \(\dfrac{4+2i}{1+i}\).

Multiply top and bottom by the conjugate \(1 - i\). Bottom: \(1 - i^2 = 1 + 1 = 2\).

Top: \((4+2i)(1-i) = 4 - 4i + 2i - 2i^2 = 4 - 2i + 2 = 6 - 2i\).

\(\dfrac{6 - 2i}{2} = 3 - i\).

Which is equal to \((3-2i)^2\)?

  1. \(13 - 12i\)
  2. \(5 + 12i\)
  3. \(9 - 12i\)
  4. \(5 - 12i\)

Write it as \((3-2i)(3-2i)\) and FOIL: \(9 - 6i - 6i + 4i^2\).

\(4i^2 = 4(-1) = -4\), so \(9 - 4 - 12i = 5 - 12i\). Answer: (d).

(a) treats \(i^2\) as \(+1\). (b) has the sign of the middle terms wrong. (c) leaves out the \(4i^2\) term.

Step 5 of 5Wrap-up

You can now write \(\sqrt{-a}\) as \(i\sqrt{a}\), do arithmetic with complex numbers by treating \(i\) like a variable and replacing \(i^2\) with \(-1\), divide using the conjugate, and simplify powers of \(i\).

Read or watch next: OpenStax Intermediate Algebra 2e, 8.8 Use the Complex Number System (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Algebra 2, The imaginary unit i (short videos). Quick rules: Radicals and complex numbers cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why do I multiply by the conjugate when I divide complex numbers?" It knows this course and can explain things another way.