I am stuck on a problem. The problem sheet states that I have to simplify the following expression: $$\frac{2\pi\textrm{i}\textrm{e}^{\textrm{i}\pi a/3}\textrm{e}^{-2\pi\textrm{i}/3}}{1-\textrm{e}^{2\pi\textrm{i}a/3}\textrm{e}^{2\pi\textrm{i}/3}}$$
So that it is "manifestly" real if the parameter $a$ is also real. I don't understand what is being asked. Does it mean that I have to modify it until it becomes obvious that the expression results in a real number if a is replaced by a real number, too? What would be the best course of action?
Thanks.
If we multiply numerator and denominator with $e^{-\pi i(a+1)/3}$, we get $$ \frac{ 2\pi i e^{-2\pi i/3} e^{-\pi i /3} }{ e^{-\pi i(a+1)/3}-e^{\pi i(a+1)/3} } $$ We can now make use of $\sin(\varphi)=\frac{e^{i\varphi}-e^{-i\varphi}}{2i}$ and we get $$ \frac{ \pi e^{-\pi i} }{ -\sin(\pi (a+1)/3) } $$ Making use of $e^{-\pi i}=-1$, we get $$ \frac{ \pi }{ \sin(\pi (a+1)/3) } $$ from which we can directly see that the result is real if $a\in\mathbb{R}$ and $a\not\equiv 2\pmod{3}.$