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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?

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  1. 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}
  2. 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}
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