Simplifying expression of algebra 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,
$$\frac{1}{3+n\pi i} \left( \cos(n\pi) + i\sin (n\pi) \right )\frac{e^3-e^{-3}}{2}$$
Here is my attempt and I'm not quite sure where I am going wrong,
$$\frac{1}{2(3+n\pi i)} \left( -e^{-(3+n\pi i )}+ e^{3+n\pi i }\right)$$
$$\frac{1}{(3+n\pi i)} \left( -e^{-n\pi i }+ e^{n\pi i }\right) \frac{e^3 - e^{-3}}{2}$$
$$\frac{1}{(3+n\pi i)} \left( -e^{-n\pi i }+ e^{n\pi i }\right) \sinh3$$
$$\frac{1}{(3+n\pi i)} \left( { -[\cos(n\pi) - i\sin(n\pi)] }+ [cos(n\pi) + i\sin(n \pi)]\right) \sinh3$$
 A: The error is in the first step, where you are using an unknown rule of calculus. Use preferably obvious and detailed steps:
$$e^{-(3 + n\pi i)} = e^{-3} e^{-n\pi i} = e^{-3} ( \cos n\pi - i \sin n\pi)$$
$$e^{(3 + n\pi i)} = e^{3} e^{n\pi i} = e^{3} ( \cos n\pi + i \sin n\pi)$$
etc.
A: The mistake is in the first line of your attempt:

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

The two expressions are not the same (multiply the second one out and check for yourself). In particular, you cannot separate the powers of $e$ in this manner.
Instead, observe that
$$
\frac{e^{3+n\pi i} - e^{-(3 + n\pi i)}}{2} = \sinh(3 + n \pi i).
$$
Now, use the formula for $\sinh(z)$, for $z = x+iy$,
$$
\sinh(z) = \sinh(x)\cos(y) + i\cosh(x)\sin(y)
$$
to get
$$
\begin{align*}
\frac{-1}{2(3+n\pi i)}(e^{-(3+n\pi i)}-e^{3+n\pi i}) &= \frac{1}{3+n\pi i}(\sinh(3)\cos(n\pi) + i \cosh(3)\sin(n\pi))\\
&=\frac{1}{3+n\pi i}\left( \frac{e^{3}-e^{-3}}{2} \right) \cos(n \pi) \\
&=\frac{1}{3+n\pi i}\left( \frac{e^{3}-e^{-3}}{2} \right) (\cos(n \pi) + i \sin (n\pi)).
\end{align*}
$$
