# Simplifying $\frac{\sqrt[3]{e^{3i\pi}-e^{i\pi/3}}+\sqrt[3]{e^{3i\pi}+e^{i\pi/3}}}{\sqrt[3]{e^{3i\pi}-e^{i\pi/3}}-\sqrt[3]{e^{3i\pi}+e^{i\pi/3}}}$

How to simplify the expression? $$\frac{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}+\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}-\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}$$

We also have $$\left(-\frac 32 - i \frac{\sqrt 3}{2}\right)^{1/3}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}} \quad\text{and}\quad (-1)^{2\over 9}=\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}$$

WolframAlpha gives this result: $$-1 - \frac{2 i}{(3^{1/6} -i)}$$ How can we obtain it?

The problem was initially posed by K.Srinivasa Raghava.

• You can begin by getting rid of the $e^{3i\pi}=(e^{i\pi})^3=(-1)^3=-1$
– MasB
Commented Nov 3, 2020 at 15:36
• Of Course ! Thanks that simplifies it a lot! Commented Nov 3, 2020 at 18:58

$$\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}=\sqrt[3\,]{e^{\frac {2i\pi}{3}}}$$
$$-\frac {3}{2} - i \frac{\sqrt 3}{2}=-\sqrt 3 e^{i\pi/6}$$
$$\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}$$ $$=e^{\frac {4i\pi}{18}}(3^{1/6}e^{\frac {-9i\pi}{18}}+1)$$ $$=e^{\frac {4i\pi}{18}}(-i3^{1/6}+1)$$
$$\frac{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}+\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}-\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}$$ $$=\frac{(-i3^{1/6}+1)}{(i3^{1/6}+1)}= -1 - \frac{(2 i)}{(3^{1/6} + -i)} =\frac{-3^{1/6}-i}{3^{1/6}-i}$$