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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

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5  
Are you trying to solve for $\beta$, or for $\alpha$, or for something else, or what? –  Michael Hardy Nov 5 '11 at 15:16

4 Answers 4

up vote 7 down vote accepted

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: $$ \left \{ ( \alpha, m \pi ) \mid \alpha \in \mathbb R, m \in \mathbb Z \right\} \bigcup \left \{ \left( \frac { (2m +1) \pi}{2}, \beta \right) \mid \beta \in \mathbb R, m \in \mathbb Z \right\} . $$

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In case $n$ is even, you've got $\alpha+\beta=n\pi+\alpha -\beta$, and $\alpha$ cancels, giving $\beta=n\pi-\beta$, so $\beta=n\pi/2$. In case $n$ is odd, you've got $\beta$ cancelling and a similar thing with $\alpha$. –  Michael Hardy Nov 5 '11 at 16:02
    
@Michael I updated my answer. Thanks for your comment! –  Srivatsan Nov 5 '11 at 16:12
    
Thanks for this very good answer :-) –  ftiaronsem Nov 6 '11 at 18:58

It is good to remember other variants of Euler's formula $(\exp(i \theta) = \cos(\theta) + i \sin(\theta))$ are helpful as well. For instance, $$\cos(\theta) = \frac{\exp(i \theta) + \exp(-i \theta)}{2}$$ $$\sin(\theta) = \frac{\exp(i \theta) - \exp(-i \theta)}{2i}$$ Your problem can be simplified as follows. Group the first two terms to get $$\exp(i(\beta - \pi/2 + \alpha)) + \exp(i(\beta - \pi/2 - \alpha)) = \exp(i (\beta - \pi/2)) \times (\exp(i \alpha) + \exp(-i \alpha))$$ $$= 2 \exp(i(\beta - \pi/2)) \cos(\alpha)$$ Group the last two terms to get $$\exp(i(-\beta + \pi/2 + \alpha)) + \exp(i(-\beta + \pi/2 - \alpha)) = \exp(i(-\beta + \pi/2)) \times (\exp(i \alpha) + \exp(-i \alpha))$$ $$= 2 \exp(i(-\beta + \pi/2)) \cos(\alpha)$$ Now add the two to get, $$\exp(i(\beta - \pi/2 + \alpha)) + \exp(i(\beta - \pi/2 - \alpha)) + \exp(i(-\beta + \pi/2 + \alpha)) + \exp(i(-\beta + \pi/2 - \alpha))$$ $$= 2 \exp(i(\beta - \pi/2)) \cos(\alpha) + 2 \exp(-i(\beta - \pi/2)) \cos(\alpha)$$ $$= 2 \cos(\alpha) (\exp(i(\beta - \pi/2)) \cos(\alpha) + \exp(-i(\beta - \pi/2)) \cos(\alpha))$$ $$= 2 \cos(\alpha) \times 2 \cos(\pi/2 - \beta) = 4 \cos(\alpha) \sin(\beta)$$

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$$e^{i(\beta-\pi/2+\alpha)}+e^{i(\beta-\pi/2-\alpha)}+e^{i(-\beta+\pi/2+\alpha)}+e^{i(-\beta+\pi/2-\alpha)} = 0$$

Suppose you're trying to solve for $\beta$. Then you've got $$ e^{i\beta}e^{i(-\pi/2+\alpha)}+e^{i\beta}e^{i(-\pi/2-\alpha)}+e^{-i\beta}e^{i(\pi/2+\alpha)}+e^{-i\beta}e^{i(\pi/2-\alpha)} = 0 $$ $$ e^{i\beta}e^{i(-\pi/2+\alpha)} + e^{i\beta}e^{i(-\pi/2-\alpha)} + \frac{1}{e^{i\beta}}e^{i(\pi/2+\alpha)} + \frac{1}{e^{i\beta}}e^{i(\pi/2-\alpha)} = 0 $$ $$ e^{i\beta} \left(e^{i(-\pi/2+\alpha)} + e^{i(-\pi/2-\alpha)}\right) + \frac{1}{e^{i\beta}} \left(e^{i(\pi/2+\alpha)} + e^{i(\pi/2-\alpha)}\right) = 0 $$ Multiplying both sides by $e^{i\beta}$, we get: $$ e^{2i\beta} \left(e^{i(-\pi/2+\alpha)} + e^{i(-\pi/2-\alpha)}\right) + \left(e^{i(\pi/2+\alpha)} + e^{i(\pi/2-\alpha)}\right) = 0 $$ $$ e^{2i\beta} \left(e^{i(-\pi/2+\alpha)} + e^{i(-\pi/2-\alpha)}\right) = -\left(e^{i(\pi/2+\alpha)} + e^{i(\pi/2-\alpha)}\right) $$ $$ e^{2i\beta} = \frac{-\left(e^{i(\pi/2+\alpha)} + e^{i(\pi/2-\alpha)}\right)}{e^{i(-\pi/2+\alpha)} + e^{i(-\pi/2-\alpha)}} = \frac{-e^{i(\pi/2+\alpha)} - e^{i(\pi/2-\alpha)}}{e^{i(-\pi/2+\alpha)} + e^{i(-\pi/2-\alpha)}} $$ Multiplying both the top and bottom by $e^{i\pi/2}$, we get $$ e^{2i\beta} = \frac{-e^{i(\pi+\alpha)} - e^{i(\pi-\alpha)}}{e^{i\alpha} + e^{-i\alpha}} $$ Then recalling that $e^{i\pi}=-1$, we get $$ e^{2i\beta} = \frac{e^{i\alpha} + e^{-i\alpha}}{e^{i\alpha} + e^{-i\alpha}} =1. $$

So $2\beta = 0\pm 2\pi n$ and $\alpha$ could be anything that doesn't make the denominator $0$.

Unless $\alpha = \pm\pi/2$ modulo $2\pi$, since then the denominators are $0$, in which case $\beta$ could be anything.

If you're trying to solve for $\alpha$ rather than for $\beta$, the method is similar.

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$$e^{i(\beta-\pi/2+\alpha)} + e^{i(\beta-\pi/2-\alpha)}+e^{i(-\beta+\pi/2 + \alpha)} + e^{i(-\beta+\pi/2-\alpha)} = 0$$

Let's try yet another approach. Remember that $e^{\pm i\pi/2} = \pm i$ and $-i=1/i$. We have $$e^{-i\pi/2}e^{i(\beta+\alpha)} + e^{-i\pi/2}e^{i(\beta-\alpha)} + e^{i\pi/2}e^{i(-\beta + \alpha)} + e^{i\pi/2}e^{i(-\beta-\alpha)} = 0$$ $$-ie^{i(\beta+\alpha)} -ie^{i(\beta-\alpha)} + ie^{i(-\beta + \alpha)} + ie^{i(-\beta-\alpha)} = 0$$ $$-ie^{i(\beta+\alpha)} -ie^{i(\beta-\alpha)} + ie^{-i(\beta - \alpha)} + ie^{-i(\beta+\alpha)} = 0$$ $$\frac{e^{i(\beta+\alpha)}}{i} + \frac{e^{i(\beta-\alpha)}}{i} - \frac{e^{-i(\beta - \alpha)}}{i} - \frac{e^{-i(\beta+\alpha)}}{i} = 0$$ The first and fourth terms add up to $2\sin(\beta+\alpha)$. The second and third add up to $2\sin(\beta-\alpha)$.

So $$ \sin(\alpha+\beta) = \sin(\alpha-\beta). $$ $$ \alpha+\beta = \alpha-\beta + 2\pi n\quad \text{ or }\quad \alpha+\beta = \pi - (\alpha-\beta) +2\pi n. $$ In the first case the $\alpha$s cancel; in the second case the $\beta$s cancel. So either $$ \alpha = \pi n\text{ and }\beta\text{ could be anything} $$ or $$ \beta = \pi n\text{ and }\alpha\text{ could be anything} $$

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