Exercício Resolvido de Funções Complexas
a)
\( \operatorname{sen}i=i\operatorname{senh}1 \)
Seno é dado por
\[ \bbox[#99CCFF,10px]
{\operatorname{sen}z=\frac{\operatorname{e}^{iz}-\operatorname{e}^{-iz}}{2i}}
\]
\[
\begin{align}
\operatorname{sen}i&=\frac{\operatorname{e}^{i.i}-\operatorname{e}^{-i.i}}{2i}=\frac{\operatorname{e}^{i^{2}}-\operatorname{e}^{-i^{2}}}{2i}=\frac{\operatorname{e}^{-1}-\operatorname{e}^{-(-1)}}{2i}.\frac{i}{i}\text{=}\\
&=i\frac{\operatorname{e}^{-1}-\operatorname{e}^{1}}{2i^{2}}=-i\frac{\operatorname{e}^{1}-\operatorname{e}^{-1}}{-2}=i\frac{\operatorname{e}^{1}-\operatorname{e}^{-1}}{2}
\end{align}
\]
\[ \bbox[#FFCCCC,10px]
{\operatorname{sen}i=i\operatorname{senh}1}
\]