Today I ended up solving
$$e^{i(\beta-\pi/2+\alpha)}+e^{i(\beta-\pi/2-\alpha)}+e^{i(-\beta+\pi/2+\alpha)}+e^{i(-\beta+\pi/2-\alpha)} = 0$$
with Euler's Identity. I rewrote the equation using $\cos$ and $\sin$ and solved the resulting two equations (one was luckily always true) with a case by case analysis.
However this was no fun, which is why I wanted to ask about a more direct way. I hate case by case solution searching, so I was hoping that there is some general tool to solve equations like the one above.
How would you approach such a problem.
Thanks in advance
ftiaronsem
Best Answer
It was a lot of fun for me! Here are the steps without much detail. It is a good exercise to check your understanding by supplying those details.
Let $\alpha + \beta = x$ and $\alpha - \beta = y$. The given equation is $$ e^{i(x-\pi/2)}+e^{i(-y-\pi/2)}+e^{i(y+\pi/2)}+e^{i(-x+\pi/2)} = 0 $$ which can be written as $$ \begin{eqnarray*} e^{i(x-\pi/2)}+e^{i(-x+\pi/2)} &=& -e^{i(-y-\pi/2)}-e^{i(y+\pi/2)} \\ -i e^{ix}+ie^{-ix} &=& ie^{-iy}-i e^{iy} \\ e^{ix}-e^{-ix} &=& e^{iy}- e^{-iy} \\ 2i \sin x &=& 2i \sin y \\ \sin x &=& \sin y \\ y &=& n \pi + (-1)^{n} x \quad (n \text{ is an integer}) \\ \alpha + \beta &=& n \pi + (-1)^{n} (\alpha - \beta). \\ &\vdots& \end{eqnarray*} $$ As Michael's comment points out, you have not mentioned what you are solving for, so I am unable to proceed further. But it is not hard to massage the above solution into the desired form.
Update (based on Michael's comments below). We can simplify this expression based on whether $n$ is odd or even. Suppose $n = 2m$ is even. Then, we have $$ \alpha + \beta = 2m \pi + \alpha - \beta \quad\implies\quad \beta = m \pi. $$ On the other hand, if $n = 2m+1$ is odd, then $$ \alpha + \beta = (2m+1) \pi - (\alpha - \beta) \quad\implies\quad \alpha = \left(m + \frac12 \right) \pi. $$
So the complete set of solutions are: $$ \left \{ ( \alpha, m \pi ) \mid \alpha \in \mathbb R, m \in \mathbb Z \right\} \bigcup \left \{ \left( \frac { (2m +1) \pi}{2}, \beta \right) \mid \beta \in \mathbb R, m \in \mathbb Z \right\} . $$