Wolfram alpha says there are no solutions. Is this because of the following?

$$\tan(z)=i \Rightarrow \frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=i \Rightarrow \frac{e^{iz}-e^{-iz}}{(e^{iz}+e^{-iz})}=-1$$ Let $x=e^{iz}$. Then,

$$\frac{x-1/x}{x+1/x}=-1 \Rightarrow x-1/x=-x-1/x \Rightarrow 2x=0$$

Then, $$x=e^{iz} \Rightarrow \ln(0)=iz$$ but $\ln(0)$ does not exist so we have no solutions?

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    $\begingroup$ $e^{iz}\neq 0$ for all $z$ (no need to use the complex logarithm)... $\endgroup$ – Surb Jul 9 '18 at 20:00

Suppose $\tan (z)=i$. Then $\sin (z)=i\cos (z)$ from which $\sin^2 (z)+\cos^2 (z)=0$. Uh, actually $\sin^2 (z)+\cos^2 (z)=1$, so we have a little problem.


You can also use Euler's formula :

$$\tan(z)=i \implies \frac {\sin(z)}{\cos(z)}=i \implies \cos(z)+i\sin(z)=0 \implies e^{iz}=0$$

That's not possible.


Just to give a different explanation,

$$\tan z=i\implies\tan^2z+1=0\implies\sec^2z=0\implies\sec z=0\implies\cos z=\infty$$

but $\cos z$ is well defined for all $z\in\mathbb{C}$.


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