Roots of Complex Numbers
Why This Matters
You already know that $\sqrt{9} = 3$ (and $-3$), giving two square roots of 9. But what about the cube roots of $-8$, or the fourth roots of $i$? In the complex-number world, every nonzero number has exactly $n$ distinct $n$th roots. De Moivre's Theorem is the key that unlocks them all.
1. Converting to Polar (Trigonometric) Form
Before we can use De Moivre's Theorem we need the complex number in polar form.
Any complex number $z = a + bi$ can be written as:
$$z = r(\cos\theta + i\sin\theta)$$
where
- Modulus: $r = |z| = \sqrt{a^2 + b^2}$ — the distance from the origin on the Argand diagram.
- Argument: $\theta = \arg(z)$ — the angle measured counter-clockwise from the positive real axis. Use $\theta = \arctan\!\left(\dfrac{b}{a}\right)$, adjusted for the correct quadrant.
⚠️ Quadrant Check: The formula $\arctan(b/a)$ only gives the correct angle when $a > 0$. If the number lies in Quadrant II or III (i.e., $a < 0$), you must add $\pi$ to the result. If it lies in Quadrant IV with $a > 0, b < 0$, add $2\pi$ if you want a positive angle.
| Rectangular | $r$ | $\theta$ | Polar Form |
|---|---|---|---|
| $1 + i$ | $\sqrt{2}$ | $\pi/4$ | $\sqrt{2}\,(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$ |
| $-1 + i\sqrt{3}$ | $2$ | $2\pi/3$ | $2(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3})$ |
| $-2$ | $2$ | $\pi$ | $2(\cos\pi+i\sin\pi)$ |