Solve 2 cosh ⁡ z + sinh ⁡ z = i for z

$2\cosh (z)+\sinh (z)=i$
$e^z+e^{-z}+\frac{e^z-e^{-z}}{2}=i$
$2e^z+2e^{-z}+e^z-e^{-z}=2i$
$3e^z+e^{-z}=2i$
$3e^z+(e^z{)}^{-1}=2i$
Let $e^z=x$
$3x+x^{-1}=2i$
$3x^2+1=2ix$
$3x^2-2ix+1=0$
We solve equations of the form $a{x}^2+bx+c=0$ with
$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
By plugging in a=3, b=−2i and c=1:
$x_{1,2}=\frac{2i\pm \sqrt{-4-4\cdot 3\cdot 1}}{6}$
$=\frac{2i\pm \sqrt{-4-12}}{6}=\frac{2i\pm \sqrt{-16}}{6}=\frac{2i\pm 4i}{6}=\frac{i}{3}\pm \frac{2i}{3}$
The two roots are therefore
$x_1=i,x_2=-\frac{i}{3}$
Since we required $e^z=x$, $z=\text{Ln}(x)$ (Ln being the complex-valued logarithm). We are lazy and throw both values in Wolfram Alpha and get
$z_1=\ln (i)=\frac{i\pi }{2},z_2=\ln (-\frac{i}{3})=-\ln (3)-\frac{i\pi }{2}$
And now we have $z_1$ and $z_2$, which are the solutions to the initial equation. Notice that the solution to that complex logarithm has arbitrary multiples of $2\pi i$ in it