$\tan(z)=i$, no solution

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?

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.

I would like to expand a bit on this answer, because the lack of a solution for $\tan z=i$ dovetails with the characteristics of the singularity of the function at infinity.

In the complex domain, all periodic functions except constants are singular at infinity. The periodicity rules out a pole, leaving an essential singularity or a nonisolated singularity as the only possibilities. The function $\tan z$, of course, shows the latter with its infinite and periodic sequence of poles.

Disappearing Act

The proofs given in various answers, including the one in this answer, also can be used to show that $\tan z\not=-i$. Therefore in any neighborhood of infinity we identify two function values in $\mathbb{C}$ that are not allowed. An essential singularity can have no more than one such forbidden value in the neighborhood, so this approach too indicates that the singularity of the tangent function at infinity is instead nonisolated. The two forbidden tangent values also show up as a pair of branch points, at $\pm i$, when we take the inverse of the tangent function.

But what if we could eliminate this roadblock by forcibly removing one of the forbidden values? To this end consider the function

$f(z)=1/(\tan z+i).$

Following the forbidden values of the the tangent function, we identify the forbidden values of $f$ as $1/(i+i)$ and $1/(-i+i)$, but the latter is of course a division by zero and does not exist. Therefore only a single function value, which simplifies to $-(1/2)i$, remains forbidden, and that opens the door to the singularity of $f$ at infinity possibly being isolated.

Render $f$ in terms of sizes and cosines:

$f(z)=\dfrac{\cos z}{\sin z+i\cos z}=\dfrac{\cos z}{\cos(\pi/2-z)+i\sin(\pi/2-z)}=\color{blue}{(\cos z)\exp(iz-i\pi/2)}$

The product form shown in blue, obtained with the aid of Euler's Formula, indicates that the singularity has indeed been isolated. In terms of the inverse function having a branch cut, this manipulation does not actually remove the branch cut from $f^{-1}(z)=\tan^{-1}[(1/z)-i]$, but it does force one of the branch points to infinity and thus out of $\mathbb{C}$ proper.

Revealing the Trick

When we relegate $\tan z$ to the denominator, we remove the poles of that function (or more precisely, turn the poles into removable singularities). But we must also adjust the denominator to avoid zero values in order to prevent another series of poles from keeping the singularity nonisolated. Setting the denominator to $\tan z+i$ in the function $f$ exploits the fact that $\tan z$ cannot assume the value $-i$ to achieve this end.

The reader may also wish to verify that $1/(\tan z-i)$ and $\tan z/(\tan z\pm i)$ show similar singularity-isolation properties.

Thus the nonexistence of both $i$ and $-i$ as tangent values in the complex plane, which blocks isolation of the singularity at infinity, also provides a path around this barrier.

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.