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I am having trouble simplifying the following,

$$\frac{-1}{2(3+n\pi i)} \left( e^{-(3+n\pi i )}- e^{3+n\pi i }\right)$$

I am wondering how it reduces to the following, looking for someone to show me the steps,

$$\frac{1}{3+n\pi i} \left( \cos(n\pi) + i\sin (n\pi) \right )\frac{e^3-e^{-3}}{2}$$

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  • $\begingroup$ Use the fact that the exponential function maps sums into products as well as the formula $e^{ix}=\cos(x)+i\sin(x).$ $\endgroup$
    – José Carlos Santos
    Dec 5, 2017 at 20:11
  • $\begingroup$ but when i use that and take out $\frac{e^3-e^{-3}}{2}$ doesn't that leave $2e^{n\pi i?}$ $\endgroup$
    – jh123
    Dec 5, 2017 at 20:13
  • $\begingroup$ could you show me through steps? $\endgroup$
    – jh123
    Dec 5, 2017 at 20:14
  • $\begingroup$ I don't know where does that $2$ come from. Don't forget that $e^{n\pi i}=\cos(n\pi)+i\sin(n\pi)$. $\endgroup$
    – José Carlos Santos
    Dec 5, 2017 at 20:16
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    $\begingroup$ The “$e$ function” is normally called the natural exponential (function). $\endgroup$
    – gen-ℤ ready to perish
    Dec 6, 2017 at 1:35

2 Answers 2

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Your answer is overly complicated. $e^{n\pi i}=e^{-n\pi i}$ for all n. They are =1 for even n and =-1 for odd n. In your expression, the cos term has these values, while the sin term=0.

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Well, we know that $e^{ni} = \cos(n) + i\sin(n)$.

We also know that $\sin(-n) = -\sin(n)$ and $\cos(-n) = \cos(n)$, so we can figure that $e^{-ni} = \cos(n) - i\sin(n)$, and thus $e^{ni} - e^{-ni} = 2i\sin(n)$.

This is a classic extended definition of $sin$, where $sin(x) = \frac{ie^{-ix} - ie^{ix}}{2}$ (note that we multiplied by $\frac{i}{i}$).

So, I would simplify this to $\frac{i\sin(n\pi -3i)}{3+n\pi i}$, as a start.

Edit: Looking a little closer, I would actually opt for $\sinh $ instead of $\sin $. This is because $\sinh (ix) = i\sin (x) $ which simplifies nicely with what I gave above.

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