Complex Numbers

Square roots of negative numbers have no answer in $\mathbb{R}$. Complex numbers supply one, and the payoff is enormous: every polynomial equation of degree $n$ acquires exactly $n$ roots.

The imaginary unit

$$i = \sqrt{-1} \qquad \text{so} \qquad i^2 = -1$$

A complex number is $z = x + yi$, with $x$ the real part $\operatorname{Re}(z)$ and $y$ the imaginary part $\operatorname{Im}(z)$ — note that $\operatorname{Im}(z)$ is the real number $y$, not $yi$.

Powers of $i$ cycle with period 4:

$$i^1 = i \qquad i^2 = -1 \qquad i^3 = -i \qquad i^4 = 1$$

To evaluate a high power, divide the exponent by 4 and use the remainder: $i^{27} = i^{24}i^3 = i^3 = -i$.

Arithmetic with complex numbers

Addition and subtraction work componentwise:

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

Multiplication expands as normal, then $i^2 = -1$:

$$(3+2i)(1-5i) = 3 - 15i + 2i - 10i^2$$ $$= 3 - 13i + 10 = 13 - 13i$$

Two complex numbers are equal exactly when their real parts match and their imaginary parts match. Equating real and imaginary parts turns one complex equation into two real ones — the standard method for finding unknowns.

Conjugates and division

The conjugate of $z = x + yi$ is $\bar{z} = x - yi$: the sign of the imaginary part flips.

$$z\bar{z} = (x+yi)(x-yi) = x^2 + y^2$$

which is always real and non-negative. For $z = 3+2i$:

$$z\bar{z} = 9 + 4 = 13$$

That is what makes division possible: multiply top and bottom by the conjugate of the denominator.

Simplify $\dfrac{3+2i}{1-i}$.

$$\frac{3+2i}{1-i} \times \frac{1+i}{1+i} = \frac{3 + 3i + 2i + 2i^2}{1 - i^2} = \frac{3 + 5i - 2}{2} = \frac{1+5i}{2}$$

So the answer is $\frac12 + \frac52 i$.

The Argand diagram

Plot $z = x + yi$ as the point $(x, y)$: the real part horizontally, the imaginary part vertically. Addition of complex numbers is then vector addition.

The modulus is the distance from the origin:

$$|z| = \sqrt{x^2+y^2}$$

$$|3+4i| = \sqrt{9+16} = 5 \qquad |5-12i| = \sqrt{25+144} = 13$$

The argument $\arg z$ is the angle from the positive real axis, measured anticlockwise, conventionally taken in $(-\pi, \pi]$.

$$\arg(1+i) = \frac{\pi}{4} \qquad \arg(-1+i) = \frac{3\pi}{4}$$

Modulus–argument form

$$z = r(\cos\theta + i\sin\theta) \qquad \text{where } r = |z|, \ \theta = \arg z$$

$$1 + i = \sqrt2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$$

This form makes multiplication and division remarkably simple:

$$|z_1z_2| = |z_1||z_2| \qquad \arg(z_1z_2) = \arg z_1 + \arg z_2$$ $$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \qquad \arg\!\left(\frac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$$

Multiplying multiplies the lengths and adds the angles — a rotation combined with a stretch.

De Moivre's theorem

$$\big[r(\cos\theta + i\sin\theta)\big]^n = r^n\big(\cos n\theta + i\sin n\theta\big)$$

Find $(1+i)^8$.

$$1+i = \sqrt2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$$

$$(1+i)^8 = (\sqrt2)^8\left(\cos 2\pi + i\sin 2\pi\right) = 16(1 + 0) = 16$$

Check directly: $(1+i)^2 = 2i$, so $(1+i)^4 = (2i)^2 = -4$, and $(1+i)^8 = 16$ ✓

De Moivre also generates identities. Expanding $(\cos\theta + i\sin\theta)^3$ and comparing real parts gives

$$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$

Roots of polynomials

With complex numbers available, every quadratic has two roots.

Solve $x^2 - 4x + 13 = 0$.

$$x = \frac{4 \pm \sqrt{16-52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$$

Notice the two roots are conjugates. That is guaranteed:

The condition matters. $x^2 - (1+i)x = 0$ has roots $0$ and $1+i$, with no conjugate in sight — because its coefficients are not all real.

For a cubic with real coefficients, one complex root forces its conjugate, and the third root must then be real.