Solve for $z$ in $z^5=32$ This was the last question on my Year 11 Complex Numbers/Matrices Exam
Name all 5 possible values for $z$ in the equation $z^5=32$
I could only figure out $2$. How would I go about figuring this on paper?

Try $z_k = 2 e^{i \frac{2 k\pi}{5}}$.
To see where this comes from write $(r e^{i \theta} )^5 = r^5 e^{i 5\theta} = 32$. Taking absolute values gives $r = 2$, then you need to solve $e^{i 5\theta} = 1$. This requires $5 \theta = 2 k\pi$, and there are only 5 distinct solutions (modulo $2 \pi$) for $\theta$.
