<s>[INST] Complex number, logarithm power proof Proof that
i)$Log(1+i)^2=2*Log(1+i)$
ii)$Log(-1+i)^2\neq2*Log(-1+i)$
What I did
i)By definition $z^a=e^{a\log z}$, so if $z=(1+i)$ and $a=2$
$$Log(1+i)^2=Log(e^{2\log(i+1)})=2*log(i+1)$$
But I do not know how to prove the second item
ii)$(-1+i)^2=(1-2i-1)=-2i\rightarrow Log(-1+i)^2=Log(-2i)=2Log(-i)\neq2Log(-1+i)$
I can do it?
 [/INST] *

*Using $1 + i = \sqrt{2} \, e^{\pi i/4}$ then
\begin{align}
\ln(1 + i)^{2} &= \ln(2 \, e^{\pi i/2}) = 2 \left( \frac{1}{2} \ln(2 \, e^{\pi i/2}) \right) = 2 \, \ln(\sqrt{2} \, e^{\pi i/4}) = 2 \ln(1+i) \\
\end{align}

*Consider this possible way:
\begin{align}
(-1 + i)^{2} = 1 - 2i + i^{2} = - 2i   
\end{align}
now 
\begin{align}
\ln(-1+i)^{2} &= \ln(-2i) = \ln( 2 \, e^{-\pi i/2}) = \ln(2) - \frac{\pi i}{2} 
\end{align}
and $-1 + i = - \sqrt{2} \, e^{-\pi i/4} = \sqrt{2} \, e^{\pi i - \pi i/4} = \sqrt{2} \, e^{3 \pi i/4}$ for which
\begin{align}
2 \, \ln(-1+i) = 2 \, \ln(\sqrt{2}) + 2 \, \frac{3\pi i}{4} = \ln(2) + \frac{3\pi i}{2}.
\end{align}
Since $\ln(2) - \frac{\pi i}{2} \neq \ln(2) - \frac{3\pi i}{2}$ then it can be said that
\begin{align}
\ln(-1+i)^{2} \neq 2 \, \ln(-1+i)
\end{align} 

</s>