Complex Numbers: Arithmetic with i

A number that squares to −1

Every real number squared is zero or positive. So the equation

$$x^2 = -1$$

has no real solution. For a long time that ended the matter. Then mathematicians noticed that allowing a solution — and naming it — broke nothing and made a great deal work, so they defined one:

$$i^2 = -1$$

With $i$ in hand, the square root of any negative number can be written:

$$\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3i \qquad \sqrt{-7} = i\sqrt{7}$$

A complex number has a real part and an imaginary part:

$$a + bi \qquad a, b \text{ real}$$

Number	Real part	Imaginary part
$3 + 2i$	$3$	$2$
$-5i$	$0$	$-5$
$7$	$7$	$0$

The last row matters: every real number is a complex number with imaginary part $0$. The complex numbers extend the reals rather than replacing them.

Adding and subtracting

Treat $i$ like a variable and combine like parts — real with real, imaginary with imaginary:

$$(3 + 2i) + (1 - 5i) = 4 - 3i$$ $$(4 - i) - (2 + 3i) = 2 - 4i$$

The subtraction distributes the minus to both parts of $2 + 3i$, exactly as with polynomials.

Multiplying

Multiply as you would two binomials, then use $i^2 = -1$:

$$(2 + 3i)(4 - i) = 8 - 2i + 12i - 3i^2$$ $$= 8 + 10i - 3(-1) = 11 + 10i$$

The only new step is replacing $i^2$ with $-1$. Everything else is the distributive property that algebra already runs on — which is the whole reason defining $i$ was safe.

The powers of i cycle

$$i^1 = i \qquad i^2 = -1 \qquad i^3 = i^2 \cdot i = -i \qquad i^4 = (i^2)^2 = 1$$

After $i^4 = 1$, the pattern repeats: $i^5 = i$, $i^6 = -1$, and so on.

Remainder when the power is divided by $4$	$0$	$1$	$2$	$3$
value	$1$	$i$	$-1$	$-i$

Find $i^{23}$. $23 = 4 \times 5 + 3$, so $i^{23} = i^3 = -i$.

Why the conjugate makes division work

The conjugate of $a + bi$ is $a - bi$. Their product is always real:

$$(a + bi)(a - bi) = a^2 - abi + abi - b^2 i^2 = a^2 + b^2$$

The middle terms cancel, and $-b^2 i^2$ becomes $+b^2$.

That is exactly what division needs. A fraction with $i$ in the denominator cannot be written as $a + bi$ until the denominator is real, so multiply top and bottom by the conjugate of the bottom:

$$\frac{5 + i}{2 - i} = \frac{(5 + i)(2 + i)}{(2 - i)(2 + i)} = \frac{10 + 5i + 2i + i^2}{4 + 1} = \frac{9 + 7i}{5}$$ $$= \frac{9}{5} + \frac{7}{5}i$$

This is the same move as rationalizing a denominator that holds a square root, as in special right triangles — multiplying by something that turns the bottom into a plain number.

Worked examples

Common mistakes

Practice problems