Operations with complex numbers in polar form

Product and quotient: moduli and arguments separately

The Argand diagram shows each number as a point, and multiplying by a complex number moves that point by a scaling and a rotation.

Conjugate, negative and rotations

The conjugate $\bar z$ is the reflection of $z$ in the real axis: same modulus, argument $-\theta$. The negative $-z$ is a half-turn about the origin: same modulus, argument $\theta + 180^\circ$.

To change from polar to rectangular form, use $a = r\cos\theta$ and $b = r\sin\theta$, with exact values such as $\cos 60^\circ = \tfrac12$, $\sin 60^\circ = \tfrac{\sqrt3}{2}$, $\cos 30^\circ = \tfrac{\sqrt3}{2}$, $\sin 30^\circ = \tfrac12$, $\cos 45^\circ = \sin 45^\circ = \tfrac{\sqrt2}{2}$.

Multiplying by $1_{90^\circ} = i$ keeps the length and turns a vector $90^\circ$ anticlockwise. Back to the ferry: if its first leg is $10_{30^\circ}$ (in km), the new leg is $10_{30^\circ}\cdot 1_{90^\circ} = 10_{120^\circ}$. In general, multiplying by $1_\theta$ is a pure rotation through $\theta$.

De Moivre's Theorem: powers in polar form

Raising to a power means multiplying a number by itself several times. If you apply the product rule $n$ times, the modulus is multiplied by itself $n$ times (it becomes $r^n$) and the argument is added to itself $n$ times (it becomes $n\alpha$).

Proof by induction (Higher Level)

We prove $(\cos\alpha+i\sin\alpha)^n = \cos n\alpha+i\sin n\alpha$ for $n \in \mathbb{N}$.

Step 1, $n = 1$: both sides equal $\cos\alpha + i\sin\alpha$, so it is true.

Step 2, assume true for $n = k$: $(\cos\alpha+i\sin\alpha)^k = \cos k\alpha+i\sin k\alpha$.

Step 3, show it is true for $n = k+1$: $(\cos\alpha+i\sin\alpha)^{k+1} = (\cos k\alpha+i\sin k\alpha)(\cos\alpha+i\sin\alpha) = (\cos k\alpha\cos\alpha - \sin k\alpha\sin\alpha) + i(\sin k\alpha\cos\alpha + \cos k\alpha\sin\alpha) = \cos(k+1)\alpha + i\sin(k+1)\alpha$.

Step 4: it is true for $n=1$, and true for $k$ implies true for $k+1$. Hence it is true for all $n \in \mathbb{N}$.

Use: trigonometric identities

Expand $(\cos\alpha + i\sin\alpha)^2$ in two ways and equate the real and imaginary parts. By De Moivre it equals $\cos 2\alpha + i\sin 2\alpha$. By multiplying out it equals $(\cos^2\alpha - \sin^2\alpha) + i(2\sin\alpha\cos\alpha)$.

The inverse: reciprocal modulus, negative argument

nth roots: $n$ solutions arranged in a polygon

An equation such as $x^2=4$ has two real solutions, $\pm2$. With complex numbers, $z^n = w$ has exactly $n$ solutions, and the points representing them on the Argand diagram are always arranged symmetrically, forming the vertices of a regular polygon with $n$ sides centred at the origin.

More on roots

Two complex numbers are equal when their moduli are equal and their arguments are equal up to a multiple of $360^\circ$ ($2\pi$). This is how you solve equations such as $z^3 = \bar z^{\,2}$: write $z = r_\theta$, so $\bar z = r_{-\theta}$, then equate moduli ($r^3 = r^2$, so $r = 1$ or $r = 0$) and arguments ($3\theta = -2\theta + 360^\circ k$). Do not forget $r = 0$: it gives the solution $z = 0$.

If the equation is $(z - c)^n = w$, put $u = z - c$, solve $u^n = w$ for the $n$ roots, then add $c$ back: $z = u + c$ (the polygon is just shifted).

The $n$th roots of unity solve $z^n = 1$: modulus $1$, arguments $\dfrac{360^\circ k}{n}$. Their sum is $0$: by the geometric series formula $1 + \omega + \dots + \omega^{n-1} = \dfrac{1-\omega^n}{1-\omega} = 0$ where $\omega = 1_{360^\circ/n}$, since $\omega^n = 1$ and $\omega \ne 1$. Geometrically, the vectors are symmetric about the origin and cancel.

The polygon has circumradius $R = \sqrt[n]{s}$, so each side has length $2R\sin\dfrac{180^\circ}{n}$ and the perimeter is $2nR\sin\dfrac{180^\circ}{n}$. For an equilateral triangle inscribed in a circle of radius $R$, area $= \tfrac{3\sqrt{3}}{4}R^2$.

Back to the start: the ferry's turn is multiplication by $\cos 90^\circ + i\sin 90^\circ = i$, a rotation with no change of length.

Summary