Complex Numbers: Common Mistakes and Fixes (9)

Question

Find all complex numbers $z$ satisfying $z^2 = 8 + 6i$. A student writes $z = \sqrt{8 + 6i}$ and stops. Why is that incomplete?

Solution — Step by Step

Why This Works

Every nonzero complex number $w$ has exactly $n$ distinct $n$-th roots, equally spaced on a circle of radius $|w|^{1/n}$ in the complex plane. For $n = 2$, the two roots are diametrically opposite — so they differ only by a sign.

The trick of using $|z|^2 = |z^2|$ adds a third equation that combines linearly with the first to crack $a^2$ and $b^2$ cleanly without quadratic-in-$b$ algebra.

Alternative Method

Polar form: $8 + 6i = 10\, e^{i\theta}$ where $\tan\theta = 3/4$, $\cos\theta = 4/5$, $\sin\theta = 3/5$. Square roots: $\sqrt{10}\, e^{i\theta/2}$ and $\sqrt{10}\, e^{i(\theta/2 + \pi)}$. Use half-angle identities: $\cos(\theta/2) = \sqrt{(1+4/5)/2} = 3/\sqrt{10}$, $\sin(\theta/2) = 1/\sqrt{10}$. So one root is $\sqrt{10}(3/\sqrt{10} + i/\sqrt{10}) = 3 + i$. The other is $-(3+i) = -3 - i$. Same answer.

Common Mistake