I need to find the solution to $e^\frac{1}{z} = \displaystyle{\frac{e^{1+i}}{\sqrt 2}}$
I've tried replacing $\frac{1}{z}$ with $w$ so $e^w = \displaystyle{\frac{e^{1+i}}{\sqrt 2}}$, but I still don't have any idea how to deal with the RHS, especially the $\sqrt{2}$.
Apparently the solution is the RHS can be turned into $e^{1+\tfrac{\pi}{4}i}$ but I have no idea how. Perhaps by turning the $\sqrt{2}$ into a multiple of $e$ by taking the log?
Any ideas?
We have $$\Large e^{1\over z}=e^{1+i-\ln \sqrt 2}$$therefore $$\dfrac{1}{z}=1-\ln \sqrt 2+i(1+2k\pi)$$which means that $$z=\dfrac{1-\ln\sqrt 2-i(1+2k\pi)}{(1-\ln\sqrt 2)^2+(1+2k\pi)^2}$$