Exercício Resolvido de Funções Complexas
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c) \( \operatorname{senh}(-2+i) \)
O seno hiperbólico é dado por
\[ \bbox[#99CCFF,10px]
{\operatorname{senh}z=\frac{\operatorname{e}^{z}-\operatorname{e}^{-z}}{2}}
\]
\[
\begin{align}
\operatorname{senh}(-2+i) &=\frac{\operatorname{e}^{(-2+i)}-\operatorname{e}^{-(-2+i)}}{2}=\frac{\operatorname{e}^{-2}.\operatorname{e}^{i}-\operatorname{e}^{2}.\operatorname{e}^{-i}}{2}=\\
&=\frac{1}{2}\left(\operatorname{e}^{-2}.\operatorname{e}^{i}-\operatorname{e}^{2}.\operatorname{e}^{-i}\right)
\end{align}
\]
aplicando a fórmula de De Moivre
\[ \bbox[#99CCFF,10px]
{\operatorname{e}^{i\theta }=\cos \theta +i\operatorname{sen}\theta}
\]
\[
\begin{align}
\operatorname{senh}(-2+i) &=\frac{1}{2}\left[\operatorname{e}^{-2}(\cos1+i\operatorname{sen}1)-\operatorname{e}^{2}(\cos1-i\operatorname{sen}1)\right]\text{=}\\
&=\frac{1}{2}\left[\operatorname{e}^{-2}\cos1+i\operatorname{e}^{-2}\operatorname{sen}1-\operatorname{e}^{2}\cos1+i\operatorname{e}^{2}\operatorname{sen}1\right]=\\
&=\frac{1}{2}(\operatorname{e}^{-2}-\operatorname{e}^{2})\cos1+i\frac{1}{2}(\operatorname{e}^{-2}+i\operatorname{e}^{2})\operatorname{sen}1 =\\
&=\frac{\operatorname{e}^{-2}-\operatorname{e}^{2}}{2}\cos1+i\frac{\operatorname{e}^{-2}+\operatorname{e}^{2}}{2}\operatorname{sen}1=\\
&=-\frac{\operatorname{e}^{2}-\operatorname{e}^{-2}}{2}\cos1+i\frac{\operatorname{e}^{2}+\operatorname{e}^{-2}}{2}\operatorname{sen}1
\end{align}
\]
O cosseno hiperbólico e o seno hiperbólico são dados por
\[ \bbox[#99CCFF,10px]
{\cosh z=\frac{\operatorname{e}^{z}+\operatorname{e}^{-z}}{2}}
\]
\[ \bbox[#99CCFF,10px]
{\operatorname{senh}z=\frac{\operatorname{e}^{z}-\operatorname{e}^{-z}}{2}}
\]
\[ \bbox[#FFCCCC,10px]
{\operatorname{senh}(-2+i)=-\operatorname{senh}2\cos 1+i\cosh2\operatorname{sen}1}
\]
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Fisicaexe - Exercícios Resolvidos de Física de Elcio Brandani Mondadori está licenciado com uma Licença Creative Commons - Atribuição-NãoComercial-Compartilha Igual 4.0 Internacional .