Welcome to our community

Be a part of something great, join today!

Billy 's question at Yahoo! Answers (Complex analysis)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Billy,

Using $z=r(\cos\theta+i\sin\theta)=re^{i\theta}$ and if $z\neq 0$ (i.e. $r\neq 0)$:
$$\begin{aligned}f(z)&=z+\frac{1}{z}\\&=re^{ i\theta}+\frac{1}{re^{i\theta}}\\&=re^{i\theta}+ \frac{1}{r}e^{-i\theta}\\&=r(\cos \theta+i\sin\theta)+ \frac{1}{r}(\cos \theta -i\sin\theta)\\&= \left (r+\frac{1}{r}\right)\cos\theta +i\left (r-\frac{1}{r}\right)\sin\theta\end{aligned}$$
 
Last edited: