Solution: $$\sin\left(\frac{\pi}{4}+2i\right) -\sin(2i)= \\ \frac{1}{2i}\left( e^{-2}\left( \cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right) -e^2\left( \cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right) \right) \right) \\ -\frac{e^{-2}}{2i}\left( \cos(0) +i\sin(0)\right)-\frac{e^2}{2i}\left(\cos(0) +i\sin(0)\right) = \\ \frac{e^{-2}}{2i} \left( \frac{\sqrt{2}-2}{2} +\frac{\sqrt{2}}{2}i \right) +\frac{e^{2}}{2i}\left( \frac{\sqrt{2}-2}{2} +\frac{\sqrt{2}}{2}i \right) = \\ (e^{-2}+e^2) \left(\frac{\sqrt{2}-2}{2}+\frac{\sqrt{2}}{2}i \right) $$
Firstly, is this the correct solution? Secondly, should I simplify the expression more so that it is in form $x+yi$ or leave it as it is? And also, I know that we can multiply the expression by $i$ to get rid of the $i$ on the denominator, can we do it with any power of $i$?
The result is not correct. Note,
$$\begin{align} & \sin\left(\frac{\pi}{4}+2i\right) -\sin(2i) \\ & =\frac{1}{2i}\left( e^{-2}\left( \cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right) -e^2\left( \cos\frac{\pi}{4}-i\sin\frac{\pi}{4} \right) \right) -\frac1{2i}\left(e^{-2}-e^{2}\right) \\ &=\frac{e^{-2}}{2i} \left( \frac{\sqrt{2}-2}{2} +\frac{\sqrt{2}}{2}i \right) +\frac{e^{2}}{2i}\left( \frac{2-\sqrt{2}}{2} +\frac{\sqrt{2}}{2}i \right) \\ &=\frac{1}{2i} \left( \frac{2-\sqrt{2}}{2} (e^2-e^{-2})+i\frac{\sqrt{2}}{2}(e^2+e^{-2}) \right)\\ & = \frac{\sqrt{2}}{2}\cosh(2)-i\frac{2-\sqrt{2}}{2}\sinh(2) \end{align}$$