Complex Numbers: Edge Cases and Subtle Traps (3)

Question

Find all values of $z$ satisfying $z^4 = -16$.

The trap: students take “fourth root of $-16$” and write $z = \pm 2i$, missing two of the four roots. Every degree-$n$ polynomial has $n$ roots in $\mathbb{C}$ — including the obvious ones and the not-so-obvious ones.

Solution — Step by Step

Why This Works

In $\mathbb{C}$, every polynomial of degree $n$ has exactly $n$ roots (counted with multiplicity) — the fundamental theorem of algebra. For $z^4 = w$ (where $w \neq 0$), there are always four distinct roots, equally spaced around a circle of radius $|w|^{1/4}$ in the complex plane.

The four roots of unity are $1, i, -1, -i$ (i.e., $e^{2\pi i k/4}$ for $k = 0, 1, 2, 3$). Any fourth roots of a complex number are obtained by taking one root and multiplying by each fourth root of unity.

Alternative Method

Factor $z^4 + 16 = 0$ as a product of quadratics over $\mathbb{R}$:

$z^4 + 16 = (z^2 + 4)^2 - (2\sqrt{2}z)^2 = (z^2 - 2\sqrt{2}z + 4)(z^2 + 2\sqrt{2}z + 4)$

Solve each quadratic:

$z^2 - 2\sqrt{2}z + 4 = 0 \implies z = \sqrt{2} \pm i\sqrt{2}$.

$z^2 + 2\sqrt{2}z + 4 = 0 \implies z = -\sqrt{2} \pm i\sqrt{2}$.

Same four roots. ✓

Common Mistake

The biggest trap: only finding two roots when looking for fourth roots. Students think $\sqrt[4]{-16} = \pm 2i$ (since $(2i)^4 = 16$ and $(-2i)^4 = 16$ — wait, that’s $-16$ check: $(2i)^4 = 16 i^4 = 16$, not $-16$). Actually neither $\pm 2i$ is a root of $z^4 = -16$ — students often get the sign wrong even on the “obvious” check. Use De Moivre’s theorem to get all four properly.

Another trap: forgetting the modulus — students get the angles right but write $z = e^{i\pi/4}$ instead of $z = 2e^{i\pi/4}$. The modulus is $|w|^{1/n} = 16^{1/4} = 2$.