f)
\( \text{Ln}\left(\dfrac{1+i}{\sqrt{2}}\right) \)
O logaritmo é dado por
\[ \bbox[#99CCFF,10px]
{\text{Ln}(z)=\ln |z|+i(\operatorname{arg}(z)+2k\pi )}
\]
O módulo é dado por
\[ \bbox[#99CCFF,10px]
{|z|=\sqrt{x^{2}+y^{2}}}
\]
\[
\begin{gathered}
z=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\\[5pt]
|z|=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\;}\\
|z|=\sqrt{\frac{1}{2}+\frac{1}{2}\;}\\
|z|=\sqrt{1\;}\\
|z|=1
\end{gathered}
\]
O argumento é dado por
\[ \bbox[#99CCFF,10px]
{\operatorname{arg}(z)=\operatorname{arctg}\left(\frac{y}{x}\right)}
\]
\[
\operatorname{arg}(z)=\operatorname{arctg}\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)=\operatorname{arctg}\left(1\right)=\frac{\pi}{4}
\]
\[
\text{Ln}(-1)=\underbrace{\ln (1)}_{0}+i\left(\frac{\pi }{4}+2k\pi\right)
\]
\[ \bbox[#FFCCCC,10px]
{\text{Ln}\left(\frac{1+i}{\sqrt{2}}\right)=\left(\frac{1}{4}+2k\right)\pi i}
\]