Exercício Resolvido de Funções Complexas
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f) \( \displaystyle (-3+4i)^{(1+i)} \)
Usando a expressão
\[ \bbox[#99CCFF,10px]
{u^{v}=\operatorname{e}^{v\;\text{Ln}u}}
\]
\[
\begin{gather}
(-3+4i)^{(1+i)}=\operatorname{e}^{(1+i)\;\operatorname{Ln}(-3+4i)} \tag{I}
\end{gather}
\]
A função multivalente do logaritmo é dada por
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\operatorname{Ln}z=\ln |z|+i(\operatorname{arg}(z)+2k\pi)} \tag{II}
\end{gather}
\]
O argumento do logaritmo é o número complexo da forma
\[
u=-3+4i
\]
O módulo é dado por
\[ \bbox[#99CCFF,10px]
{|z|=\sqrt{x^{2}+y^{2}}}
\]
\[
\begin{gather}
|u|=\sqrt{(-3)^{2}+4^{2}\;}\\
|u|=\sqrt{9+16\;}\\
|u|=\sqrt{25\;}\\
|u|=5 \tag{III}
\end{gather}
\]
O argumento é dado por
\[ \bbox[#99CCFF,10px]
{\theta=\operatorname{arg}(z)=\operatorname{arctg}\left(\frac{y}{x}\right)}
\]
\[
\begin{align}
\theta &=\operatorname{arg}(u)=\operatorname{arctg}\left(\frac{y}{x}\right)=\\
&=\operatorname{arctg}\left(\frac{4}{-3}\right)=\pi-\operatorname{arctg}\left(\frac{4}{3}\right) \tag{IV}
\end{align}
\]
Gráfico 1
substituindo os valores (III) e (IV) na expressão (II)
\[
\begin{gather}
\operatorname{Ln}(-3+4i)=\ln \left(5\right)+i\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi \right] \tag{V}
\end{gather}
\]
substituindo a expressão (V) na expressão (I)
\[
\begin{gather}
(-3+4i)^{(1+i)}=\operatorname{e}^{(1+i)\;\left\{\ln\left(5\right)+i\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi\right]\right\}}\\[5pt]
(-3+4i)^{(1+i)}=\operatorname{e}^{\left\{\ln\left(5\right)+i\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi \right]+i\;\ln\left(5\right)+i.i\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi\right]\right\}}\\[5pt]
(-3+4i)^{(1+i)}=\operatorname{e}^{\left\{\ln\left(5\right)+i\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi \right]+i\;\ln\left(5\right)+i^{2}\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi\right]\right\}}\\[5pt]
(-3+4i)^{(1+i)}=\operatorname{e}^{\left\{\ln\left(5\right)+i\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi \right]+i\;\ln\left(5\right)+(-1)\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi\right]\right\}}\\[5pt]
(-3+4i)^{(1+i)}=\operatorname{e}^{\left\{\ln\left(5\right)+i\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi \right]+i\;\ln\left(5\right)-\;\left[\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi\right]\right\}}\\[5pt]
(-3+4i)^{(1+i)}=\operatorname{e}^{\left\{\left[\ln\left(5\right)-\pi +\operatorname{arctg}\left(\frac{4}{3}\right)-2k\pi\right]+i\;\left[\ln \left(5\right)+\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi\right]\right\}}\\[5pt]
(-3+4i)^{(1+i)}=\operatorname{e}^{\left[\ln\left(5\right)-\pi +\operatorname{arctg}\left(\frac{4}{3}\right)-2k\pi\right]}+\operatorname{e}^{i\;\left[\ln\left(5\right)+\pi-\operatorname{arctg}\left(\frac{4}{3}\right)+2k\pi\right]}
\end{gather}
\]
Aplicando a Fórmula de Euler
\[ \bbox[#99CCFF,10px]
{\operatorname{e}^{i\theta}=\cos \theta +i\operatorname{sen}\theta}
\]
\[
\begin{split}
(3-4i)^{(1+i)}=\operatorname{e}^{\ln(5)}.\operatorname{e}^{\operatorname{arctg}\left(\frac{4}{3}\right)+(2k+1)\pi} & \left\{\cos \left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)+\pi\right]\text{+} \right. \\
& \left. \text{+} i\operatorname{sen}\left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)+\pi\right]\right\}
\end{split}
\]
Observação: Desenvolvendo o cosseno e o seno da soma.
\[
\cos (a+b)=\cos a\cos b-\operatorname{sen}a\operatorname{sen}b
\]
\[
\begin{align}
\cos\left[\underbrace{\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)}_{a}+\underbrace{\pi}_{b}\right] &=\cos \left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]\;\underbrace{\cos\left[\pi\right]}_{-1}-\\
&-\operatorname{sen}\left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]\underbrace{\operatorname{sen}\left[\pi\right]}_{0}
\end{align}
\]
\[
\cos \left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)+\pi\right]=-\cos \left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]
\]
\[
\operatorname{sen}(a+b)=\operatorname{sen}a\cos b+\operatorname{sen}b\cos a
\]
\[
\begin{align}
\operatorname{sen}\left[\underbrace{\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)}_{a}+\underbrace{\pi}_{b}\right] &=\operatorname{sen}\left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]\;\underbrace{\cos\left[\pi \right]}_{-1}\text{+}\\
& -\underbrace{\operatorname{sen}\left[\pi\right]}_{0}\cos \left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]
\end{align}
\]
\[
\operatorname{sen}\left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)+\pi\right]=-\operatorname{sen}\left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]
\]
\[
\begin{split}
(3-4i)^{(1+i)}=5\operatorname{e}^{\operatorname{arctg}\left(\frac{4}{3}\right)+(2k+1)\pi} &\left\{-\cos \left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]- \right. \\
& \left. -i\operatorname{sen}\left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]\right\}
\end{split}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{
\begin{split}
(3-4i)^{(1+i)}=-5\operatorname{e}^{\operatorname{arctg}\left(\frac{4}{3}\right)+(2k+1)\pi} & \left\{\cos \left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]\text{+} \right. \\
& \left. \text{+} i\operatorname{sen}\left[\ln\left(5\right)-\operatorname{arctg}\left(\frac{4}{3}\right)\right]\right\}
\end{split}
}
\end{gather}
\]
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