When I was messing around with Euler's formula, I came across this: $$e^{ix}=\cos(x)+\sin(x)i$$ Then, let $x$ be an imaginary value, $ix$, so then: $$e^{i(ix)}=\cos(ix)+\sin(ix)i$$ which we can simplify to $$e^{-x}=\cos(ix)+\sin(ix)i,$$ but since $e^{-x}$ is a real value for all real inputs, then $\sin(ix)i$ must be equal to $0$. so that means that $$e^{-x}=\cos(ix)$$ This doesn't seem right, so could someone please point out where I made a mistake?

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    $\begingroup$ $i\sin (ix) = i(i \sinh x) = -\sinh x$ $\endgroup$
    – player3236
    Feb 6 at 17:16
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    $\begingroup$ Why should $\sin(ix)$ be real? $\endgroup$ Feb 6 at 17:16
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    $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$
    – Shaun
    Feb 6 at 17:33
  • $\begingroup$ Since $\cos ix=\cosh x$ and $\sin ix=i\sinh x$, $\cos ix+i\sin ix=\cosh x-\sinh x$. $\endgroup$
    – J.G.
    Feb 6 at 17:42
  • $\begingroup$ You assumed that both $\sin(ix)$ and $\cos(ix)$ are real. Based on this assumption, you have concluded that the only real part of $\cos(ix) + i \sin(ix)$ is $\cos(ix)$. The mistake in this reasoning is your initial assumption. $\endgroup$
    – Saeed
    Feb 6 at 23:14

If $a+bi=\lambda\in\Bbb R$, with $a,b\in\Bbb R$ then, indeed, you must have $a=\lambda$ and $b=0$. But, in the equality$$e^{-x}=\cos(ix)+\sin(ix)i,$$you have no reason to assume that $\cos(ix),\sin(ix)\in\Bbb R$.


You are right to say that $e^{-x}$ is a real value for all real inputs. But you seem to assume that $\sin(ix)$ is also a real value for all real inputs. Maybe this assumption is hasty...


$$e^{-x}=\cos(ix)+\sin(ix)i,$$ but since $e^{-x}$ is a real value for all real inputs, then $\sin(ix)i$ must be equal to $0$.

That does not follow unless you know that $\sin(ix)$ is real, and there's no reason to think that.

$$ \sin(ix) = \frac{e^{i(ix)} - e^{-i(ix)}}{2i} = \frac{e^{-x} - e^x}{2i}. $$


You assumed $\sin(ix)=0$, but according to the formula \begin{equation} \sin(x)=\mathrm{i}\frac{e^{-\mathrm{i}x}-e^{\mathrm{i}x}}{2}, \end{equation} ($\sin(x)$ is sometimes defined this way for all $x\in\mathbf{C}$), \begin{equation} \sin(\mathrm{i}x)=\mathrm{i}\frac{e^x-e^{-x}}{2}, \end{equation} and thus this is true if and only if $x=0$ (if $x=0$, $\cos(ix)=1$ and there is no contradiction).

I guess you assumed this because you know that a complex number is real if and only if its imaginary part is zero, but $\sin(\mathrm{i}x)$ is not the imaginary part of $\cos(\mathrm{i}x)+\mathrm{i}\sin(\mathrm{i}x)$.


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