simplification of complex numbers

Question

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?

Answer 1

Note that $$e^{-\frac{5\pi i}{6}}=-\frac{\sqrt{3}}{2}-\frac{i}{2}$$ $$e^{\frac{3\pi i}{6}}=-i$$ and $$e^{\frac{\pi i}{6}}=\frac{\sqrt{3}}{2}-\frac{i}{2}$$

Answer 2

$$e^{-\pi i/6}\sum_{r=0}^2(e^{-\pi i/3})^r=\dfrac{-1-1}{e^{-\pi i/6}-e^{\pi i/6}}=\dfrac2{2i\sin\dfrac\pi6}=?$$

Answer 3

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 $\pi/3$.

Answer 4

$$ \begin{align} 2πi\left(1/e^{5πi/6}+1/e^{3πi/6}+1/e^{πi/6}\right) &=2πi\,\color{#C00}{e^{-3πi/6}}\left(\color{#090}{e^{-2πi/6}}+1+\color{#090}{e^{2πi/6}}\right)\\ &=2πi\,\color{#C00}{\frac1i}\left(1+\color{#090}{2\cos\left(\frac\pi3\right)}\right)\\ &=4\pi \end{align} $$