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Let $f(z)=z+\frac{1}{z}$ for $z\in \mathbb C$ with $z\ne 0$. Which of the following is always true?

(a) $f$ is an analytic on $\mathbb C$\ $\{0\}$

(b) $f$ is a conformal mapping on $\mathbb C$\ $\{0\}$

(c) None of the above

(a)$f(z)=z+\frac{1}{z}=z+\frac{\overline z}{|z|^2}$. On $|z|=1$ $f(z)=z+\overline z$. differentiationg w.r.t $\overline z$ which is not equall to $0$. So, $f$ can not be analytic. $|z|\ne1$ It satisfy C-R equation.

(b) $z+\frac{1}{z}=re^{i\theta}+\frac{e^{-i \theta}}{r}$. I know the theorem , Let $f$ be analytic on $D$ and $z_0 \in D$ such that $f'(z_0)\ne 0$. Then $f$ is conformal at $z_0$. I can see that $f'(z_0)=1-\frac{1}{z_{0}^2}$ which is never zero in $z\ne \pm 1$. f(z) is not analytic. I am not able to apply the theorem. Please help me

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  • $\begingroup$ $1 - \dfrac{1}{{z_0}^2}$ is zero for $z_0=\pm 1$ $\endgroup$
    – Dylan
    Dec 14, 2017 at 3:13
  • $\begingroup$ I apologize Dylan. $\endgroup$
    – user464147
    Dec 14, 2017 at 3:25

1 Answer 1

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(a) You can’t just rewrite the function in a new form and claim that it doesn’t work in this case. Why didn’t you just take the derivative? You get $1-\frac{1}{z^2}$, which exists everywhere on the domain.

(b) Take a look at the statement you made: $f’(z_0)=1-\frac{1}{z_0^2}$ which is never $0$. Is this true? And then given that $f$ is in fact analytic, try using the theorem.

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  • $\begingroup$ (b) I apologize for the error. I meant to say $ |z|\ne 0$ . For (a) I don't understand. What is the problem of considering in the unit circle? $\endgroup$
    – user464147
    Dec 14, 2017 at 3:30
  • $\begingroup$ Isn't it a rationalization(I don't know the exact name for this) in complex algebra? $\endgroup$
    – user464147
    Dec 14, 2017 at 3:54
  • $\begingroup$ You cannot take the derivative with respect to $\overline{z}$ if you're restricting to $|z|=1$, because there's only one direction you may travel in, which is around the circle. The tangent vector points in the direction of $iz$. You also know it's wrong because if you leave it as is, the derivative with respect to $\overline{z}$ IS $0$, even on $|z|=1$. So something is wrong, namely the idea that "taking the derivative with respect to $\overline{z}$" makes any sense looking just at the unit circle. $\endgroup$
    – Bob Jones
    Dec 14, 2017 at 18:36
  • $\begingroup$ You are right. Thank you very much. Now I have doubt on (b) at $z=\pm 1 f'(z)=0$ right. Then how can I use the theorem? $\endgroup$
    – user464147
    Dec 15, 2017 at 1:59
  • $\begingroup$ Everywhere else it's nonzero, so everywhere else the theorem tells us it's a conformal mapping, but near $1$ and $-1$, it's not injective so it's not a conformal map: $f(z)=f\left(\frac{1}{z}\right)$. $\endgroup$
    – Bob Jones
    Dec 15, 2017 at 4:40

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