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Polar form

Any complex number $z=a+bi$ can be written using its modulus $r$ and argument $\theta$ :

z = r(\cos\theta+i\sin\theta), \qquad r=\sqrt{a^2+b^2}, \qquad \theta = \arctan\!\left(\dfrac ba\right)\ \text{(adjusted for quadrant)}

The quadrant adjustment matters — the raw arctangent only returns values in $(-\tfrac\pi2, \tfrac\pi2)$ , so you must add $\pi$ when $a<0$ .

De Moivre’s theorem

Polar form makes exponentiation almost trivial:

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

Compare this to expanding $(a+bi)^n$ directly with the binomial theorem — polar form avoids that entirely.

Roots of unity and nth roots

The $n$ complex $n$ th roots of unity are evenly spaced around the unit circle:

\omega_k = \cos\!\left(\dfrac{2\pi k}{n}\right) + i\sin\!\left(\dfrac{2\pi k}{n}\right), \qquad k=0,1,\ldots,n-1

They always sum to zero (for $n \geq 2$ ) — a fact worth remembering, since it frequently shortens problems that ask for a sum over all roots. To find all $n$ th roots of a general complex number, convert to polar form and divide the argument (plus multiples of $2\pi$ ) by $n$ .

Worked example

Problem: Find all solutions to $z^3 = -8$ .

Solution: Write $-8$ in polar form: $r=8$ , $\theta = \pi$ (since $-8$ is on the negative real axis). The cube roots have modulus $8^{1/3}=2$ and arguments $\dfrac{\pi + 2\pi k}{3}$ for $k=0,1,2$ :

\theta_0 = \dfrac\pi3, \qquad \theta_1 = \pi, \qquad \theta_2 = \dfrac{5\pi}3

So the three roots are $2\left(\cos\tfrac\pi3+i\sin\tfrac\pi3\right)=1+i\sqrt3$ , $2(\cos\pi+i\sin\pi)=-2$ , and $2\left(\cos\tfrac{5\pi}3+i\sin\tfrac{5\pi}3\right)=1-i\sqrt3$ .

Practice problems

1.Write $z = -1+i$ in polar form.

2.Use De Moivre’s theorem to compute $\left(\cos30°+i\sin30°\right)^6$ .

3.Find the sum of all 5th roots of unity.

4.Find $(1+i)^8$ using polar form.

Complex Numbers in Polar Form

Polar form

De Moivre’s theorem

Roots of unity and nth roots

Practice problems