Still very new to complex variables and complex analysis so apologies for the probably very simple question.
What is the value(s) of this:
$$(1+i)^i$$ How to go about it?
Also I wanted to confirm that for $$\log(-i)$$ I got $$\ln1 + i\left(\frac{3\pi}{2}+2\pi k\right), k =....-1, 0, 1, 2.... $$.
Thanks again and any pointing in the right direction will also be greatly appreciated.
$$(1+i)^i=\exp{(i \cdot\ln{(1+i)})}=\exp{(i \cdot (\frac{\ln(2)}{2}+\frac{i\pi}{4}))}=\exp{ (\frac{\ln(2)}{2}i-\frac{\pi}{4})}$$ $$=e^{-\frac{\pi}{4}}(\cos{(\frac{\ln{(2)}}{2})}+i\cdot \sin{(\frac{\ln{(2)}}{2})})$$
By using the facts that $x=\exp{(\ln{(x)})}$; $\ln{(x^n)}=n \ln{(x)}$ and that $\exp{(i x)}=\cos{(x)}+i \cdot \sin{(x)}$.
For $\ln{(-i)}$; $$\ln{(-i)}=\ln{(e^{-\frac{i \pi}{2}+2\pi k})}=-\frac{i\pi}{2}+2\pi k$$ where $k \in \mathbb{Z}$.
$$ \log(1+i) = \frac12\log2 + \frac{8n+1}{4}i\pi $$ so $$ i\log(1+i) = \frac{i}2\log2 - \frac{8n+1}{4}\pi $$ and $$ (1+i)^i = \exp(i\log(1+i)) = e^{-\frac{(8n+1)\pi}{4}}\left(\cos \frac{\log2}2+ i\sin\frac{\log2}2\right) $$
