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Operations on Complex Numbers in Polar Form

This lesson covers how to multiply and divide complex numbers written in polar form by combining their moduli and angles, and how De Moivre's theorem extends this to raise a complex number to any whole-number power without expanding a binomial.

Introduction

Once a complex number is written in polar form, \( z = r(\cos\theta + i\sin\theta) \), multiplying, dividing, and raising it to a power become much simpler than working with the rectangular form \(a + bi\). Instead of expanding brackets and collecting real and imaginary parts, you combine the moduli and angles directly. If you need a refresher on converting a complex number into this form first, see angle and absolute value of complex numbers.

Multiplying Complex Numbers in Polar Form

Let \( z_1 = r_1(\cos\theta_1 + i\sin\theta_1) \) and \( z_2 = r_2(\cos\theta_2 + i\sin\theta_2) \). Multiplying these out and applying the angle sum identities for sine and cosine collapses everything into one clean rule:

\( z_1 z_2 = r_1 r_2\big[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\big] \)

In words: multiply the moduli (the \(r\) values) and add the angles. Geometrically, multiplying two complex numbers stretches or shrinks a vector by the product of the two lengths and rotates it by the sum of the two angles, as shown below.

The product vector's length is r one times r two, and its angle is theta one plus theta two.

Dividing Complex Numbers in Polar Form

Division works the same way, but in reverse: divide the moduli and subtract the angles.

\( \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\big[\cos(\theta_1-\theta_2) + i\sin(\theta_1-\theta_2)\big] \), for \( r_2 \ne 0 \)

This rule comes from multiplying the numerator and denominator by the conjugate factor needed to make the denominator real, but in polar form you never need to do that expansion — the subtraction of angles does the work automatically.

De Moivre's Theorem

Raising a complex number to a whole-number power is really just repeated multiplication of the same number by itself. Since each multiplication adds the angle again and multiplies the modulus again, the pattern for \(n\) repeats is:

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

This is De Moivre's theorem, and it saves you from expanding a binomial like \((a+bi)^5\) term by term. The \(r^n\) part behaves exactly like the exponents product rule, since raising \(r\) to the \(n\)th power is the same as multiplying \(r\) by itself \(n\) times, once for every copy of \(z\) in the product.

Worked Example: Multiplying and Dividing

Let \( z_1 = 4(\cos 50^\circ + i \sin 50^\circ) \) and \( z_2 = 2(\cos 20^\circ + i\sin 20^\circ) \).

Multiplication: \( z_1 z_2 = (4)(2)\big[\cos(50^\circ+20^\circ) + i\sin(50^\circ+20^\circ)\big] = 8(\cos 70^\circ + i\sin 70^\circ) \)

Division: \( \dfrac{z_1}{z_2} = \dfrac{4}{2}\big[\cos(50^\circ-20^\circ) + i\sin(50^\circ-20^\circ)\big] = 2(\cos 30^\circ + i\sin 30^\circ) \)

Worked Example: De Moivre's Theorem

Find \( \big[2(\cos 15^\circ + i\sin 15^\circ)\big]^4 \).

Apply the theorem directly: \( r^n = 2^4 = 16 \) and \( n\theta = 4(15^\circ) = 60^\circ \), so the result is \( 16(\cos 60^\circ + i\sin 60^\circ) \). Substituting the known values for \(\cos 60^\circ\) and \(\sin 60^\circ\) converts this back to standard form:

\( 16\left(\dfrac{1}{2} + i\dfrac{\sqrt3}{2}\right) = 8 + 8\sqrt3\,i \)

These polar-form rules apply no matter how the original complex number was given. If a problem hands you numbers already in rectangular form, review adding and subtracting complex numbers to see how those operations compare, then convert to polar form only when you need to multiply, divide, or raise a number to a power — that is where the polar rules truly save time over expanding everything by hand.

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