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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?

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    $\begingroup$ 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
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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: $$v(r,\theta)=\sin\theta(r-r^{-1})$$

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  • $\begingroup$ 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

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