I'm trying to solve $\lim _{z \rightarrow e^{\pi i/3}} (z-e^{\pi i/3}) \left ( \frac{z}{z^3+1} \right) $. I am given the answer as $1/6-i \sqrt3/6$. I have an answer, but it doesn't match this, and I'm not sure if it is a matter of algebra to rearrange the answer, or if I'm making a mistake somewhere. Here is what I've done:
Plugging in $e^{\pi i/3}$ for z, I get $\frac{0}{0}$ because $e^{\pi i/3}-e^{\pi i/3}=0$ and $\left(e^{\pi i/3} \right)^3+1=0$. So applying L'Hopital's rule, the new limit to work with is, $\lim _{z \rightarrow e^{\pi i/3}} \frac{2z-e^{\pi i/3}}{3z^2}$.
Now, $e^{\pi i/3}$ can be plugged in so that the limit is equal to $$\frac{e^{\pi i/3}}{3(e^{\pi i/3})^2}=\frac{1}{3e^{\pi i/3}}$$
Note that\begin{align}\frac1{3e^{\pi i/3}}&=\frac{e^{-\pi i/3}}3\\&=\frac 13\left(\frac12-\frac{\sqrt3}2i\right)\\&=\frac16-\frac{\sqrt3}6i.\end{align}