Solve $e^{2z} - 2e^z + 2= 0$

Yes, the solutions of $x^2-2x+2=0$ are $1+i$ and $1-i$. So, the solutions of $e^{2z}-2e^z+2=0$ are all those numbers $z$ such that $e^z=1+i$ or that $e^z=1-i$. But$$1+i=\sqrt2\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)=\sqrt2e^{\pi i/4}. $$Therefore$$e^z=1+i\iff z=\log\sqrt2+\frac{\pi i}4+2\pi in,$$for some $n\in\Bbb Z$. The case of the equality $e^z=1-i$ is similar.