If $$z=re^{i\theta}$$ write $$f(z)=z+\frac{1}{z}$$ as $$f(z)=u(r,\theta)+iv(r,\theta)$$What i did is $$z=re^{i\theta}=r(\cos\theta+i\sin\theta)\space and\space f(z)=z+z^{-1}\space so$$ $$f(z)=r(\cos\theta+i\sin\theta)+\frac{r(\cos\theta-i\sin\theta)}{r²(\cos²\theta+\sin²\theta)}=r³(\cos\theta+i\sin\theta)+r(\cos\theta+i\sin\theta)$$ $$=r³\cos\theta+r\sin\theta+ir³\sin\theta+ir\sin\theta=r\cos\theta(r²+1)+ir\sin\theta(r²+1)$$

It makes sense that?

  • 1
    $\begingroup$ what is sen? Do you mean sin? $\endgroup$ – Irrational Person Mar 12 '15 at 21:59
  • $\begingroup$ yes is sin, sorry. $\endgroup$ – Roland Mar 12 '15 at 22:01
  • $\begingroup$ Also, use \sin and \cos. $\endgroup$ – Demosthene Mar 12 '15 at 22:01

As you correctly noted: $$\dfrac{1}{z}=z^{-1}$$ And with $z=r(\cos\theta+i\sin\theta)$, we have: $$z^{-1}=r^{-1}(\cos\theta+i\sin\theta)^{-1}$$ Now, use De Moivre's formula to write: $$z^{-1}=r^{-1}\big(\cos{(-\theta)}+i\sin{(-\theta)}\big)$$ Now, we have: $$z+z^{-1}=r\cos\theta+ir\sin\theta+r^{-1}\cos\theta-ir^{-1}\sin\theta$$ $$f(z)=\cos\theta(r+r^{-1})+i\sin\theta(r-r^{-1})$$ $$f(z)=u(r,\theta)+iv(r,\theta)$$ with: $$u(r,\theta)=\cos\theta(r+r^{-1})$$ and: $$v(r,\theta)=\sin\theta(r-r^{-1})$$

  • $\begingroup$ what I did is wrong? $\endgroup$ – Roland Mar 12 '15 at 22:06
  • $\begingroup$ Yes, let me elaborate. $\endgroup$ – Demosthene Mar 12 '15 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.