Value of sine of complex numbers

Question

I stumbled upon a problem with evaluating the sine function for complex arguments. I know that in general I can use $$ \sin(ix)=\frac{1}{2i}(\exp(-x)-\exp(x))=i\sinh(x). $$ But I could also write the sine function as the imaginary part of the exponential function as $$\sin(ix)=\text{Im}(\exp(i(ix)))=\text{Im}(\exp(-x))=0$$ where Im is the imaginary part. Well, apparently I am not allowed to write it like that, but I don't see why. Could you give me a hint what went wrong here?

Answer 1

This, of course, uses three interconnected formulas: $e^{ix}= cos(x)+ i sin(x)$, $cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$, and $sin(x)= \frac{e^{ix}- e^{-ix}}{2}$

Your error is that you are assuming that the imaginary part of $e^{ix}$ is "i sin(x)". That is true only if itself is real. If x is not real the $i sin(x)$ is not imaginary because sin(x) is not real.

Answer 2

First, note that $$\mbox{for }y\in\mathbb{R}\ \mbox{we have}\ \sin{y}=\mathrm{Im}\{e^{iy}\}.$$ Stating $\mathrm{Im}\left\{\mathrm{exp}(-x)\right\}=0$ means you assume $x\in\mathbb{R}$. However, then $ix$ is complex, such that you cannot apply the above rule.

Answer 3

For $x\in\mathbb{R}$ we have $$e^{ix}= \cos x+ i \sin x,$$ $$\cos ix= \frac{e^{-x}+ e^{x}}{2}=\cosh x,$$ and $$\sin i x= \frac{e^{-x}- e^{-x}}{2i}=i\sinh x$$

Thus, $$ \cos ix+ i \sin ix=\cosh x-\sinh = \frac{e^{x}+ e^{-x}}{2} - \frac{e^{x}- e^{-x}}{2}=e^x$$ so, $$\sin ix = i({\cos ix- e^x})= i({\cosh x- e^x})=i\sinh x,$$ i.e. $\sin ix$ is an imaginary number and $Re(\sin ix) =0$.