Let \begin{equation*} \begin{split} f \colon & \mathbb C \setminus \left\{0\right\} \to \mathbb C \\ & z \mapsto \frac{1}{2}\left( z+ \frac{1}{z}\right) \end{split} \end{equation*} I am asked to find the image of $\partial B_r := \{z \in \mathbb C : \vert z \vert = r\}$ (where $r >0$) under $f$.

Let me show you what I've done: hope it's correct.

Let $\vert z \vert = r$. Then \begin{equation*} \begin{split} \vert f(z) \vert & = \sqrt{f(z) \overline{f(z)}} = \sqrt{\frac{1}{2}\left( z+ \frac{1}{z}\right)\frac{1}{2}\left( \overline{z}+ \frac{1}{\overline{z}}\right)} = \\ & = \frac{1}{2}\sqrt{\left( z\overline{z} + \frac{z}{\overline{z}} + \frac{\overline{z}}{z} + \frac{1}{z\overline{z}}\right)} = \\ & = \frac{1}{2}\sqrt{\left( r^2 + 2\Re{\frac{z}{\overline{z}}} + \frac{1}{r^2} \right)} \end{split} \end{equation*}

Now we calculate $$ \Re{\frac{z}{\overline{z}}} = \Re{\frac{x+iy}{x-iy}} = \Re{\frac{x^2+y^2 + 2ixy}{x^2+y^2}} = 1 $$ hence we get \begin{equation*} \begin{split} \vert f(z) \vert & = \frac{1}{2}\sqrt{\left( r^2 + 2\Re{\frac{z}{\overline{z}}} + \frac{1}{r^2} \right)} = \\ & = \frac{1}{2}\sqrt{\left( r^2 + 2 + \frac{1}{r^2} \right)} = \frac{r^2+1}{2r} =: k. \end{split} \end{equation*}

So we can conclude that $$ f(\partial B_{r}) = \partial B_{k} $$

What do you think? Is it correct? Thanks for your help.

  • 1
    $\begingroup$ When you calculate $2\Re \frac{z}{\bar z}=\Re\frac{x+iy}{x-iy}$ What you get is $\Re \frac{x^2-y^2+2ixy}{x^2+y^2}$. So your calculation is wrong, then try to change everything accordingly. $\endgroup$ – uforoboa Aug 3 '12 at 14:14
  • $\begingroup$ You are right, sorry for my bad mistake. Now I try to fix it. $\endgroup$ – Romeo Aug 3 '12 at 14:24
  • $\begingroup$ Unfortunately, I don't know how to fix it. Have you got any ideas? Thanks. $\endgroup$ – Romeo Aug 3 '12 at 16:15

We have, writing $a+ib:=re^{it}$ that \begin{align} f(re^{it})&=\frac 12\left(a+ib+\frac 1{a+ib}\frac{a-ib}{a-ib}\right)\\ &=\frac 12\left(a+ib+\frac{a-ib}{r^2}\right)\\ &=\frac 12\left(a\left(1+\frac 1{r^2}\right)+ib\left(1-\frac 1{r^2}\right)\right)\\ &=\frac 12\left(r\cos \theta\frac{r^2+1}{r^2}+ir\sin\theta\frac{r^2-1}{r^2}\right)\\ &=\frac 1{2r}(\cos\theta (r^2+1)+i\sin\theta(r^2-1)). \end{align} Do you recognize curve in the plane of the form $\gamma(t)=(A\cos t,B\sin t)$, where $A$ and $B$ are fixed?

  • 1
    $\begingroup$ Thanks for your answer. Now it's clear: the image of $\partial B_r$ is an ellipse with semiaxes $A=\frac{r^2+1}{2r}$ and $B=\frac{r^2-1}{2r}$. Well it was an easy problem, I'm sorry since I wasn't able to conclude it by myself. Anyway, Davide, thank you very much for your help. $\endgroup$ – Romeo Aug 3 '12 at 21:53
  • 1
    $\begingroup$ Just in case you wanted some more information, this is the Joukowski transform -- en.wikipedia.org/wiki/Joukowsky_transform . $\endgroup$ – snar Aug 3 '12 at 22:38

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