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 | $(a+bi) + (c+di) = (a+c) + (b+d)i$ | $(2+3i)+(1-i) = 3+2i$ |
| Subtraction | $(a+bi) - (c+di) = (a-c) + (b-d)i$ | $(2+3i)-(1-i) = 1+4i$ |
| Multiplication | $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$ | $(1+i)(2-i) = 3+i$ |
| Conjugate of $z = a+bi$ | $\bar{z} = a-bi$ | $\overline{2+3i} = 2-3i$ |
| Reciprocal | $\frac{1}{a+bi} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i$ | $\frac{1}{1+i} = \frac{1}{2} - \frac{1}{2}i$ |
| 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 + bi) + (c + di) = (a + c) + (b + d)i $Add real parts and imaginary parts separately.
Subtraction follows the same pattern:
$ (a + bi) - (c + di) = (a - c) + (b - d)i $Example 1. Compute $(2 + 3i) + (-3 - i)$ and $(2 + 3i) - (-3 - i)$.
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 + bi)(c + di) = ac + adi + bci + bdi^2 = ac + adi + bci - bd $Then group real and imaginary parts.
Example 2. Compute $(2 + 3i)(-3 - i)$.
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 $z = a + bi$ is:
$\bar{z} = a - bi$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 + bi)(a - bi) = a^2 - abi + abi - 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 $3 - 5i$ and compute its product with the original number.
Solution. The conjugate is $\overline{3 - 5i} = 3 + 5i$.
$ (3 - 5i)(3 + 5i) = 9 + 15i - 15i - 25i^2 = 9 + 25 = 34 $
Reciprocal and Division
For any complex number $z = a + bi$ 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 $a - bi$:
$ \frac{1}{a + bi} = \frac{a - bi}{(a + bi)(a - bi)} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}\,i $Derivation of the reciprocal formula (click to expand)
Let $z = a + bi \neq 0$. We want $z^{-1} = x + yi$ such that $z \cdot z^{-1} = 1$. Expanding:
$zz^{-1} = (ax - by) + (ay + bx)i = 1 + 0i$This gives two equations:
$ax - by = 1 \quad \text{and} \quad bx + ay = 0$Solving simultaneously yields:
$x = \frac{a}{a^2 + b^2}, \quad y = \frac{-b}{a^2 + b^2}$So $z^{-1} = \dfrac{a}{a^2 + b^2} - \dfrac{b}{a^2 + b^2}\,i$.
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 $\dfrac{2 + 3i}{1 + 4i}$.
Solution. Multiply numerator and denominator by the conjugate of the denominator, $1 - 4i$:
$ \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 $\dfrac{3 - i}{2 + 5i}$.
Solution. Multiply by the conjugate $2 - 5i$:
$ \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 $c + di$ is $c - di$, the denominator becomes $(c+di)(c-di) = c^2 + d^2$, which is a positive real number. Then simplify the numerator.
