Let $z = re^{i\theta}$ with $r^2e^{i\theta} + 3re^{-i\theta} = 0$. Find the number of solutions.
Step-by-Step Solution
Key Concept: Convert to exponential form and solve for both modulus and argument separately
Given $z = re^{i\theta}$, the equation becomes $r^2e^{i\theta} + 3re^{-i\theta} = 0$, which simplifies to $r^2e^{i\theta} = -3re^{-i\theta}$. Dividing by $re^{-i\theta}$ (assuming $r \neq 0$), we get $re^{2i\theta} = -3$, so $r = 3$ and $e^{2i\theta} = -1$. This gives $2\theta = \pi + 2\pi k$, so $\theta = \frac{\pi}{2} + \pi k$ for $k = 0, 1, 2, 3, 4$. Each value of $\theta$ gives a distinct complex number $z = 3e^{i\theta}$, yielding 5 solutions.
Correct Answer: 5