Math Core

Complex numbers follow the same arithmetic rules as real numbers: addition and multiplication are commutative and associative, and multiplication distributes over addition. The only new fact is $i^2 = -1$ . Treat $i$ like a variable, and replace $i^2$ with $-1$ whenever it appears. That is enough to add, subtract, multiply and even divide.

Adding and subtracting

Combine the real parts with each other and the imaginary parts with each other, just as you combine like terms in $(3 + 5x) + (7 - 2x)$ .

(a + bi) + (c + di) = (a + c) + (b + d)i, \qquad (a + bi) - (c + di) = (a - c) + (b - d)i.

Worked example: Subtracting

Simplify $(3 + 5i) - (7 - 2i)$ .

Distribute the minus sign to both parts of the second number, then combine:

3 + 5i - 7 + 2i = (3 - 7) + (5 + 2)i = -4 + 7i.

In the complex plane, adding complex numbers works exactly like adding vectors: move $a$ units horizontally and $b$ units vertically, then $c$ more horizontally and $d$ more vertically.

Multiplying

Use the distributive property (for two binomials, that's FOIL), then replace $i^2$ with $-1$ and combine like terms.

Worked example: Multiplying two complex numbers

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

\begin{aligned} (2 - 3i)(4 + i) &= 8 + 2i - 12i - 3i^2 \\ &= 8 - 10i - 3(-1) \\ &= 11 - 10i \end{aligned}

Common mistake

The step people forget is the sign change from $i^2$ . The last term above is $-3i^2 = -3(-1) = +3$ , which is added to the real part. Leaving $i^2$ in the answer, or treating it as $+1$ , gives the wrong real part. A final answer in standard form should contain no powers of $i$ at all.

Squaring works the same way. Use $(a + b)^2 = a^2 + 2ab + b^2$ :

(1 + 2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i.

Complex conjugates

Some products come out purely real. Look at what happens when you multiply $3 + 4i$ by $3 - 4i$ :

(3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 + 16 = 25.

The imaginary terms cancel, and the $-16i^2$ turns into $+16$ .

Definition

Complex conjugate

The complex conjugate of $a + bi$ is $a - bi$ : same real part, opposite imaginary part. Their product is always a nonnegative real number:

(a + bi)(a - bi) = a^2 + b^2.

This is the complex version of the difference of squares, $(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2$ . Notice that $a^2 + b^2$ is the square of the absolute value $|a + bi|$ from the last lesson. In the complex plane, the conjugate is the reflection of a point across the real axis.

Dividing

A quotient such as $\dfrac{5 + i}{2 - 3i}$ isn't in standard form, because $i$ appears in the denominator. You fixed a similar problem with radicals by multiplying by a conjugate, and the same idea works here.

Dividing complex numbers

To write $\dfrac{a + bi}{c + di}$ in standard form, multiply the numerator and the denominator by the conjugate of the denominator, $c - di$ . The denominator becomes the real number $c^2 + d^2$ , and then you divide each part of the numerator by it.

Worked example: Dividing

Write $\dfrac{5 + i}{2 - 3i}$ in standard form.

The conjugate of $2 - 3i$ is $2 + 3i$ .

\begin{aligned} \frac{5 + i}{2 - 3i} \cdot \frac{2 + 3i}{2 + 3i} &= \frac{10 + 15i + 2i + 3i^2}{2^2 + 3^2} \\ &= \frac{10 + 17i - 3}{13} \\ &= \frac{7 + 17i}{13} = \frac{7}{13} + \frac{17}{13}i \end{aligned}

You can check a quotient by multiplying back: $(2 - 3i)\left(\dfrac{7}{13} + \dfrac{17}{13}i\right) = \dfrac{(2 - 3i)(7 + 17i)}{13} = \dfrac{14 + 34i - 21i - 51i^2}{13} = \dfrac{65 + 13i}{13} = 5 + i$ . It matches the numerator.

When the denominator is pure imaginary, such as $\dfrac{4 - 6i}{2i}$ , the conjugate of $2i$ is $-2i$ . Multiplying top and bottom by just $i$ also works:

\frac{4 - 6i}{2i} \cdot \frac{i}{i} = \frac{4i - 6i^2}{2i^2} = \frac{6 + 4i}{-2} = -3 - 2i.

Putting it together

Worked example: A mixed expression

Simplify $i(3 - i) + (2 + i)^2$ .

Work each piece, then combine:

\begin{aligned} i(3 - i) &= 3i - i^2 = 1 + 3i \\ (2 + i)^2 &= 4 + 4i + i^2 = 3 + 4i \\ \text{sum} &= (1 + 3) + (3 + 4)i = 4 + 7i \end{aligned}

Tip

Complex conjugates have tidy shortcuts. The sum of a number and its conjugate is twice the real part, $(a + bi) + (a - bi) = 2a$ , and their product is $a^2 + b^2$ . Both are real. You'll use exactly these two facts in the next lesson, when complex solutions of quadratic equations show up in conjugate pairs.

Practice

Practice 1

Simplify $(4 - 6i) - (-1 + 9i)$ and write it as $a + bi$ . Enter $(a, b)$ .

Enter a point like (2, -3)

Practice 2

Simplify $3i(2 - 5i)$ and write it as $a + bi$ . What is the real part $a$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Simplify $(3 + 4i)(3 - 4i)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Simplify $(2 + 5i)(1 - 3i)$ and write it as $a + bi$ . Enter $(a, b)$ .

Enter a point like (2, -3)

Practice 5

Simplify $(4 - i)^2$ and write it as $a + bi$ . What is the imaginary part $b$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

What is the complex conjugate of $-2 + 7i$ ?

Practice 7

Write $\dfrac{10 - 5i}{1 + 2i}$ in the form $a + bi$ . Enter $(a, b)$ .

Enter a point like (2, -3)

Practice 8

Write $\dfrac{3 + 2i}{4 - i}$ in the form $a + bi$ . What is the imaginary part $b$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.