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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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Are you trying to solve for $\beta$, or for $\alpha$, or for something else, or what? –  Michael Hardy Nov 5 '11 at 15:16
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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
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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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I agree with the other answers, in that it is beneficial to solve equations like this by hand every now and then, as it reinforces one's knowledge of identities. However, you did not ask for the answer to the equation, but rather, for:

some general tool to solve equations like the one above...

The only general tools I could think of were:

  • coming here, or a similar site/ individual, and getting the equation solved for you- which is a glib non-answer, OR
  • submission as a query to Wolfram Alpha. By using what is returned, you might expedite arrival at an identity-based, "long hand" solution. That surely won't be a general aid, but it might be useful in some instances. This was Wolfram Alpha's response when I submitted your equation. Keep in mind that there was no way to specify what term I wanted to solve for. Or rather, no way that I could determine, using the free web application. Given such additional input, perhaps Wolfram Alpha would be helpful.

EDIT: Numerical analysis is another option to consider. Also, there are books of commonly encountered equations, of the sort that appear frequently when using complex analysis for solving problems in engineering and the physical sciences. That resource is often more trouble than not, as it is difficult to find an entry that matches your particular equation closely enough to help (at least, that was my experience with such things).

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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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