<s>[INST] Solve $e^{2z} - 2e^z + 2= 0$ So I've started by looking at
\begin{align} e^{z} =  x:\\
x^2 - 2x + 2 = 0
\end{align}
Whose solutions should be: \begin{align}\ 1+i, 1-i \end{align}
Then I did the following:
\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{2}+\frac{i \cdot 1}{2}\\
e^x= {\sqrt{2}} \quad \text{and}\quad y&=\frac{\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{\pi}{4}\\
\text{Also,} \quad e^x(\cos y+\sin y)&=\cos(\frac{7\pi}{4})+i\sin\frac{7\pi}{4})=\frac{1}{2}-\frac{i \cdot 1}{2}\\
e^x= \sqrt{2} \quad \text{and}\quad y&=\frac{7\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{7\pi}{4}\\
\end{align}
Are the solutions the following? \begin{align} \sqrt{2} + \frac{\pi}{4} \quad \text{and} \quad \sqrt{2} + \frac{7\pi}{4} \end{align}
Forgive the shoddy formatting as this is my first ever post here. Thanks.
 [/INST] Yes.....
Once you get $e^z = 1 +i, 1-i$ you can take the following as a formula (assuming $a,b$ are real):
$a + bi= \sqrt{a^2 + b^2}e^{\arctan \frac ab i}= e^{\ln(a^2 + b^2)}e^{\arctan \frac ab i}= e^{\ln(a^2+b^2) + \arctan \frac ab i}$
[I suppose another way putting this is the principal natural log $\operatorname{Ln}(a+bi) = \ln(\sqrt {a^2 + b^2}) + \arctan \frac abi$. But as natural log is a "multivalued function" on complex numbers $\ln (a+bi) = \ln(\sqrt {a^2 + b^2}) + \arctan \frac abi + 2k\pi i$.]
so $z = \ln 2 + \arctan \pm 1 i = \ln 2 + \frac \pi 4i, \ln -\frac \pi 4 i$
.....
But as $\arg$ is congruent $\mod 2\pi i$ we can actually have an infinite number of solutions $z = \ln 2 +(\pm \frac \pi 4 + 2k\pi )i$ for any integer (positive or negative) $k$.
.....
Also you shouldn't use $x$ as a variable for two different meanings (although being in isolated steps of the proof there was no harm done).
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