simplification of complex numbers How does $$2πi(1/e^{5πi/6} +1/e^{3πi/6}+1/e^{πi/6})$$ reduce to π/3.
i tried using exponential form but for some reason this does not give me the answer. what approaches could you take to simplifying this?
 A: When adding complex numbers, it is often easier to use rectangular rather than polar form. This is straightforward in this case since the angles we're dealing with are multiples of $\pi/6$ and thus have exact rectangular forms. Going off of @Henry's comment, converting to negative exponents will greatly simplify the process. Recall that 
$$
e^{i\theta} = \cos(\theta) + i\sin(\theta)
$$
So we have $e^{-5i\pi/6} = -\frac{\sqrt{3}}{2}-i\frac{1}{2}$, $e^{-3i\pi/6} = e^{-i\pi/2} = -i$, and $e^{-i\pi/6} = \frac{\sqrt{3}}{2}-i\frac{1}{2}$.
Then the expression is
$$
2\pi i(-2i) = 4\pi
$$
So in fact the expression does not simplify to $\pi/3$.
A: $$
\begin{align}
2πi\left(1/e^{5πi/6}+1/e^{3πi/6}+1/e^{πi/6}\right)
&=2πi\,\color{#C00}{e^{-3πi/6}}\left(\color{#090}{e^{-2πi/6}}+1+\color{#090}{e^{2πi/6}}\right)\\
&=2πi\,\color{#C00}{\frac1i}\left(1+\color{#090}{2\cos\left(\frac\pi3\right)}\right)\\
&=4\pi
\end{align}
$$
A: Note that $$e^{-\frac{5\pi i}{6}}=-\frac{\sqrt{3}}{2}-\frac{i}{2}$$
$$e^{\frac{3\pi i}{6}}=-i$$ and $$e^{\frac{\pi i}{6}}=\frac{\sqrt{3}}{2}-\frac{i}{2}$$
A: $$e^{-\pi i/6}\sum_{r=0}^2(e^{-\pi i/3})^r=\dfrac{-1-1}{e^{-\pi i/6}-e^{\pi i/6}}=\dfrac2{2i\sin\dfrac\pi6}=?$$
