# Difference between revisions of "1990 AHSME Problems/Problem 22"

## Problem

If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, then the product of those solutions with $a>0$ is

$\text{(A) } -2\quad \text{(B) } 0\quad \text{(C) } 2i\quad \text{(D) } 4\quad \text{(E) } 16$

## Solution

This equation is $r^6e^{6\theta i}=2^6e^{(\pi\pm 2k\pi) i}$. Solving in the usual way, $r=2$ and $\theta\in\{\pm 30^\circ,\pm 90^\circ,\pm 150^\circ\}$.

Thus there are only two solutions with positive real part, and they are conjugates, so their product is $r^2=4$ which is $\fbox{D}$