So I was bored, and decided to do some complex analysis for fun. After a while, I came up with this:$$e^{iz}=\sec(z)+i\csc(z)$$which I thought that I might be able to solve. Here is my attempt at doing so:
So we have$$e^{iz}=\sec(z)+i\csc(z)\implies\cos(z)+i\sin(z)=\sec(z)+i\csc(z)\\\implies1=\dfrac{(\sec(z)+i\csc(z))(\cos(z)-i\sin(z))}{\cos^2(z)+\sin^2(z)}\\\implies1=1-i(\tan(z)+\cot(z))+1\\\implies i\tan(z)+i\cot(z)=1\\\implies\tan(z)+\cot(z)=-i\\\implies\dfrac{\tan^2(z)+1}{\tan(z)}=-i\\\implies\sec^2(z)=-i\tan(z)\\\implies\dfrac1{\cos^2(z)}=-\dfrac{i\sin(z)}{\cos(z)}\\\implies i=\sin(z)\cos(z)\\\implies i=\dfrac12\sin(2z)\\\implies\dfrac12\arcsin(2i)=z$$or$$z\approx0+0.72181773758941i$$
My question
Is my solution to this seemingly difficult math equation correct, or what would I do to get the correct answer?
Best Answer
Use the identity: $e^{iz}=\cos z+i\sin z$, we have $$ \begin{aligned} e^{i z} & =\frac{1}{\cos z}+\frac{i}{\sin z} \\ & =\frac{2}{e^{i z}+e^{-i z}}-\frac{2}{e^{i z}-e^{-i z}} \\ & =\frac{-4 e^{-i z}}{e^{2 i z}-e^{-2 i z}} \end{aligned} $$ Rearranging and simplifying gives \begin{aligned} e^{3 i z} & =-3 e^{-i z} \\ \Leftrightarrow \quad e^{4 i z} & =-3 \\ \Leftrightarrow \quad 4 i z & =\ln (-3) \\ \Leftrightarrow \quad \quad z & =\ln 3+\ln (-1) \\ & =-\frac{i}{4}(\ln 3+i \pi+2 n \pi)\\ & =\frac{1}{4}[(2 n+1) \pi-i \ln 3] \end{aligned} where $n\in \mathbb{Z}$.
*checked by WA.