# Got $e^{-x}=\cos(ix)$ from Euler's formula. Where is my mistake?

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?

• $i\sin (ix) = i(i \sinh x) = -\sinh x$ Feb 6 at 17:16
• Why should $\sin(ix)$ be real? Feb 6 at 17:16
• 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. Feb 6 at 17:33
• Since $\cos ix=\cosh x$ and $\sin ix=i\sinh x$, $\cos ix+i\sin ix=\cosh x-\sinh x$.
– J.G.
Feb 6 at 17:42
• 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. 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 $$\cos(\mathrm{i}x)+\mathrm{i}\sin(\mathrm{i}x)$$.