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I want to show that $f(z)=\frac{1}{2}\left(z+\frac{1}{z}\right)$ is a conformal map from the set of $z$ such that $0<|z|<1$ onto $\mathbb{C} \setminus [-1,1]$.

I find that $f'(z)=\frac{(z+1)(z-1)}{2z^2}$ so this means that $f$ is conformal except for $z=1$ or $z=-1$ which is ok since they're not in the domain of $f$. So $f$ is a conformal mapping.

Now for $z=re^{i\theta}$, we find $w=\frac{1}{2}\left(z+\frac{1}{z}\right)=re^{i\theta}+\frac{e^{-i\theta}}{r}$ which gives $u=(r+\frac{1}{r})\frac{cos(\theta)}{2}$ and $v=(r-\frac{1}{r})\frac{sin(\theta)}{2}$. When we compute we end up with $\frac{u^2}{(r+1/r)^2} + \frac{v^2}{(r-1/r)^2}=\frac{1}{4}$ which means that circles about the origin, i.e the ones such that $|z|<1$ are mapped to ellipses.

Finally if $[-1,1]$ were in the image then this means that the unit circle has been mapped by $f$ since would take $r=1$ which implies $-1<u<1$ ($v=0$ here).

Is the exercise complete or do I need to add more justification as to why $f$ maps the punctured interior of the unit disc to the whole plane?

EDIT: I corrected the typo in the equation of the ellipse, it should be $\frac{1}{4}$ like in Zarrax's answer

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  • $\begingroup$ It's unclear to me that you've shows that the function is onto. i.e. do you know that for every $z$ in $\mathbb{C}$ \ $[-1,1]$ there is a $z'$ with $0<|z'|<1$ which is mapped to $z$ by $f$?. It's possible that this is expressed in your second paragraph and I'm missing it. $\endgroup$ – enfield Jul 1 '11 at 23:25
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I think you still have to show that everything in ${\mathbb C} - [-1,1]$ is in the union of those ellipses. Or equivalently, for all $(u,v)$ in ${\mathbb C} - [-1,1]$ you need to show that there is some $r < 1$ such that ${\displaystyle \bigg({u \over (r + {1 \over r})}\bigg)^2 + \bigg({v \over (r - {1 \over r})}\bigg)^2 = {1 \over 4}}$. Since you can replace $r$ by ${1 \over r}$ you really just have to find some $r \neq 1$ that works.

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A quick way to show that the map is onto: Fix a complex number $w$ outside the real interval $[-1,1]$. If $z$ is a root of $f(z)=w$ (it is equivalent to a quadratic, so it has a root), then $1/z$ is the other root. It is impossible to have $|z|=1$, as then $f(z)$ is on the forbidden interval. Therefore either $z$ or $1/z$ is in the punctured open unit disk.

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  • $\begingroup$ Could you please explain why, if $z$ is a root of f(z)=w then 1/z is the other root. $\endgroup$ – Charith Jeewantha Apr 7 at 2:37
  • $\begingroup$ Oh.. I got it.. $z_0$ is a root of $f(z)=w$ implies $\frac{1}{2}(z_0+\frac{1}{z_0})=w$ and this is the same as $f(\frac{1}{z_0})$ $$ \\ Which is a luxury we get from this particular format of the function $\endgroup$ – Charith Jeewantha Apr 7 at 2:44

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