Find a solution to the equation $\tan(z)=7i$ which satisfies the condition $0<\Re(z)< \pi$} We use the $\sin(z)=\frac{e^{zi}-e^{-zi}}{2i}$ and $\cos(z)=\frac{e^{xi}+e^{-xi}}{2}$.
Here is what I have done. Are both the reasoning and solution of the problem correct?
\begin{align*} LHS&=\tan(z) \\ &=\frac{\sin(z)}{\cos(z)} \\ &=\frac{e^{iz}-e^{iz}}{i\left(e^{iz}+e^{-iz}\right)} \\ &=i\frac{1-e^{2iz}}{1+e^{2iz}} \end{align*} So, \begin{equation} \tan(z)=7i \iff i\frac{1-e^{2iz}}{1+e^{2iz}}=7i \end{equation} This can be simplified to \begin{equation} e^{2iz}=-\frac{3}{4} \end{equation} Now, we use the definition $\log(z):=\log(|z|)+\arg(z) i$ and the fact $0<\Re(z)<\pi$ to get: \begin{equation} 2iz=-\log \left(\frac{4}{3}\right)+\pi i \iff z=\frac{\pi}{2}+\frac{i}{2}\log\left( \frac{4}{3}\right) \end{equation} Thus, the answer is $z=\frac{\pi}{2}+\frac{i}{2}\log\left( \frac{4}{3}\right)$.