3
$\begingroup$

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.

  • 0
    Unfortunately, I don't know how to fix it. Have you got any ideas? Thanks.2012-08-03

1 Answers 1

1

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
    Just in case you wanted some more information, this is the Joukowski transform -- http://en.wikipedia.org/wiki/Joukowsky_transform .2012-08-03