Arithmetic Operations on Complex Numbers

Adding and multiplying complex numbers works just like with real numbers, with one extra rule: whenever $i^{2}$ appears, replace it with $- 1$ . Division uses the complex conjugate to clear the imaginary part from the denominator.

Quick Reference

Operation	Formula	Example
Addition	$(a + b i) + (c + d i) = (a + c) + (b + d) i$	$(2 + 3 i) + (1 - i) = 3 + 2 i$
Subtraction	$(a + b i) - (c + d i) = (a - c) + (b - d) i$	$(2 + 3 i) - (1 - i) = 1 + 4 i$
Multiplication	$(a + b i) (c + d i) = (a c - b d) + (a d + b c) i$	$(1 + i) (2 - i) = 3 + i$
Conjugate of $z = a + b i$	$\bar{z} = a - b i$	$\overset{―}{2 + 3 i} = 2 - 3 i$
Reciprocal	$\frac{1}{a + b i} = \frac{a}{a^{2} + b^{2}} - \frac{b}{a^{2} + b^{2}} i$	$\frac{1}{1 + i} = \frac{1}{2} - \frac{1}{2} i$
Division	$\frac{z}{w} = z \cdot w^{- 1}$	$\frac{2 + i}{1 + i} = \frac{3}{2} - \frac{1}{2} i$

Addition and Subtraction

Definition. The sum of two complex numbers is defined by:

(a + b i) + (c + d i) = (a + c) + (b + d) i

Add real parts and imaginary parts separately.

Subtraction follows the same pattern:

(a + b i) - (c + d i) = (a - c) + (b - d) i

Example 1. Compute $(2 + 3 i) + (- 3 - i)$ and $(2 + 3 i) - (- 3 - i)$ .

Solution.

\begin{aligned} (2 + 3i) + (-3 - i) &= (2 + (-3)) + (3 + (-1))i = -1 + 2i \\ (2 + 3i) - (-3 - i) &= (2 - (-3)) + (3 - (-1))i = 5 + 4i \end{aligned}

Multiplication

Definition. The product of two complex numbers is defined by:

$ (a + bi)(c + di) = (ac - bd) + (ad + bc)i $

This is obtained by expanding as for real numbers and replacing $i^{2}$ with $- 1$ .

You do not need to memorize this formula. Simply expand using the distributive law:

(a + b i) (c + d i) = a c + a d i + b c i + b d i^{2} = a c + a d i + b c i - b d

Then group real and imaginary parts.

Example 2. Compute $(2 + 3 i) (- 3 - i)$ .

Solution.

\begin{aligned} (2 + 3i)(-3 - i) &= 2(-3) + 2(-i) + 3i(-3) + 3i(-i) \\ &= -6 - 2i - 9i - 3i^2 \\ &= -6 - 11i - 3(-1) \\ &= -6 + 3 - 11i \\ &= -3 - 11i \end{aligned}

Complex numbers satisfy the same fundamental algebraic properties as real numbers.

Algebraic properties: commutativity, associativity, distributivity (click to expand)

For any complex numbers $z_{1}$ , $z_{2}$ , $z_{3}$ :

Commutativity: $z_{1} + z_{2} = z_{2} + z_{1}$ and $z_{1} z_{2} = z_{2} z_{1}$ .
Associativity: $(z_{1} + z_{2}) + z_{3} = z_{1} + (z_{2} + z_{3})$ and $(z_{1} z_{2}) z_{3} = z_{1} (z_{2} z_{3})$ .
Distributivity: $z_{1} (z_{2} + z_{3}) = z_{1} z_{2} + z_{1} z_{3}$ .

These follow directly from the definitions and the corresponding properties of real numbers.

The Complex Conjugate

Definition. The complex conjugate of $z = a + b i$ is:

\bar{z} = a - b i

The conjugate is obtained by changing the sign of the imaginary part.

A key property: multiplying a complex number by its conjugate always gives a real number:

(a + b i) (a - b i) = a^{2} - a b i + a b i - b^{2} i^{2} = a^{2} + b^{2}

This works because the imaginary terms cancel out, leaving only $a^{2} + b^{2} \geq 0$ .

Example 3. Find the conjugate of $3 - 5 i$ and compute its product with the original number.

Solution. The conjugate is $\overset{―}{3 - 5 i} = 3 + 5 i$ .

(3 - 5 i) (3 + 5 i) = 9 + 15 i - 15 i - 25 i^{2} = 9 + 25 = 34

Reciprocal and Division

For any complex number $z = a + b i$ with $z \neq 0$ , the reciprocal (or multiplicative inverse) is the unique complex number $z^{- 1}$ satisfying $z \cdot z^{- 1} = 1$ .

To find it, multiply the numerator and denominator by the conjugate $a - b i$ :

\frac{1}{a + b i} = \frac{a - b i}{(a + b i) (a - b i)} = \frac{a}{a^{2} + b^{2}} - \frac{b}{a^{2} + b^{2}} i

Derivation of the reciprocal formula (click to expand)

Let $z = a + b i \neq 0$ . We want $z^{- 1} = x + y i$ such that $z \cdot z^{- 1} = 1$ . Expanding:

z z^{- 1} = (a x - b y) + (a y + b x) i = 1 + 0 i

This gives two equations:

a x - b y = 1 and b x + a y = 0

Solving simultaneously yields:

x = \frac{a}{a^{2} + b^{2}}, y = \frac{- b}{a^{2} + b^{2}}

So $z^{- 1} = \frac{a}{a^{2} + b^{2}} - \frac{b}{a^{2} + b^{2}} i$ .

If $z$ and $w$ are complex numbers and $w \neq 0$ , the quotient is defined by:

\frac{z}{w} = z \cdot w^{- 1}

In practice, divide by multiplying numerator and denominator by the conjugate of the denominator:

\frac{z}{w} = \frac{z \cdot \bar{w}}{w \cdot \bar{w}}

The denominator $w \cdot \bar{w} = | w |^{2}$ is always a positive real number, so this converts the division into multiplication.

Example 4. Compute $\frac{2 + 3 i}{1 + 4 i}$ .

Solution. Multiply numerator and denominator by the conjugate of the denominator, $1 - 4 i$ :

\begin{aligned} \frac{2 + 3i}{1 + 4i} &= \frac{(2 + 3i)(1 - 4i)}{(1 + 4i)(1 - 4i)} \\ &= \frac{2 - 8i + 3i - 12i^2}{1 + 16} \\ &= \frac{2 + 12 - 5i}{17} \\ &= \frac{14}{17} - \frac{5}{17}\,i \end{aligned}

Example 5. Compute $\frac{3 - i}{2 + 5 i}$ .

Solution. Multiply by the conjugate $2 - 5 i$ :

\begin{aligned} \frac{3 - i}{2 + 5i} &= \frac{(3 - i)(2 - 5i)}{(2 + 5i)(2 - 5i)} \\ &= \frac{6 - 15i - 2i + 5i^2}{4 + 25} \\ &= \frac{6 - 5 - 17i}{29} \\ &= \frac{1}{29} - \frac{17}{29}\,i \end{aligned}

Consistency with Real Number Arithmetic

When the four fundamental operations are applied to complex numbers whose imaginary parts are zero, the results are identical to the usual operations on real numbers. Complex numbers truly extend, rather than replace, the real number system.

Frequently Asked Questions

How do I divide complex numbers?

Multiply both the numerator and denominator by the conjugate of the denominator. Since the conjugate of $c + d i$ is $c - d i$ , the denominator becomes $(c + d i) (c - d i) = c^{2} + d^{2}$ , which is a positive real number. Then simplify the numerator.

Why does multiplying by the conjugate eliminate the imaginary part in the denominator?

The product

(a + b i) (a - b i) = a^{2} + b^{2}

has no imaginary part because the cross terms

+ a b i

and

- a b i

cancel. This is the same mechanism as the difference of squares factorization:

(a + b) (a - b) = a^{2} - b^{2}

, here with

b

replaced by

b i

and using

i^{2} = - 1

What is the additive inverse of a complex number?

The additive inverse of

z = a + b i

- z = - a - b i

. This follows from multiplying by

- 1

(- 1) (a + b i) = - a - b i

. Verify:

z + (- z) = (a + b i) + (- a - b i) = 0

Do complex numbers form a field?

Yes. The complex numbers form a field: addition and multiplication satisfy commutativity, associativity, and distributivity; there is an additive identity ($0$) and multiplicative identity ($1$); every complex number has an additive inverse; and every non-zero complex number has a multiplicative inverse. No further algebraic extension is needed for polynomial equations.

Arithmetic Operations on Complex Numbers

Arithmetic Operations on Complex Numbers

Quick Reference

Addition and Subtraction

Multiplication

The Complex Conjugate

Reciprocal and Division

Consistency with Real Number Arithmetic

Frequently Asked Questions

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