I have tried this is a couple of different ways and have different answers.

  1. $\sin(\pi/2+i\ln2)=\cos(i\ln2)=\cosh(\ln2)=\frac{e^{\ln2}+e^{-\ln2}}{2}=5/4$

  2. $\sin(\pi/2+i\ln2)={\rm Im}(e^{i(\pi/2+i\ln2)})={\rm Im}(ie^{-\ln2})={\rm Im}(1/2i)=1/2$

I have a feeling the first is correct, but I am not sure where the other could have gone wrong.

  • 2
    $\begingroup$ Something's fishy about all of these up-votes in just 20 minutes. $\endgroup$ – user384138 May 18 '17 at 8:09
  • $\begingroup$ @OpenBall It's not my doing if there is something going on! I just posted the question, went to a lecture, and am checking again now! I presume it is because the distinction of when the expansions/methods are valid is not taught well in schools(certainly I have never come across this) so it seems like an strange conundrum until you know one method is simply not valid. $\endgroup$ – Meep May 18 '17 at 9:28

My guess is that $\sin(x)=\mathrm{Im}(e^{ix})$ only works for $x\in \mathbb{R}$. Take for example the series representation of $$\sin(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$ Now if we insert $\sin(ix)$ with real $x$, then we see that $\sin(ix)\in\mathbb{C}$. On the other hand we see that $\sin(ix)=\mathrm{Im}(e^{i(ix)}) = \mathrm{Im}(e^{-x})=0$ which clearly contradicts our earlier finding. Only valid solution for complex $x$ would be $$\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$$


The first is correct. In the second, you are using that $\sin(x)=\text{im}(e^{ix}),$ but that only applies if $x$ is real, which is not the case here. We know that $e^{ix}=\cos(x)+i\sin(x)$, which makes it look like $\sin(x)$ is the imaginary part of $e^{ix}$, but what if $\cos(x)$ and $\sin(x)$ themselves are not real?


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