Consider the map $w=f(z)$ defined by $$w=\dfrac{1}{2}\Big(3z+\dfrac{1}{z}\Big).$$
This map can map the unit circle to a ellipse centered at origin with major axis of length $4$ along the real line, and with minor axis of length $2$ along the imaginary line.
Then, I am curious where this map is conformal.
The way I did this question was that:
A holomorphic function is conformal if its first derivative never vanishes.
However, $f(z)$ has a pole at $z=0$. (this singularity is not removable as $\lim_{z\rightarrow 0}(z-0)f(z)\neq 0$ and by Riemann's theorem on removable singularity.)
Thus, we have cannot have $z=0$ in our region.
So now $f(z)$ is holomorphic in $\mathbb{C}\setminus\{0\}$. Then, we compute the first derivative which is $$f'(z)=\dfrac{1}{2}\Big(3-\dfrac{1}{z^{2}}\Big).$$
The root of $f'(z)=0$ is $z=\pm\dfrac{1}{\sqrt{3}}$.
Therefore, $f(z)$ is conformal in $\mathbb{C}\setminus\{0,\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}}\}$.
However, this argument has a problem that not all the conformal map is holomorphic, so we may not need to exclude $z=0$.
But then I cannot argument with derivative since it may not be holomorphic.
Alternatively, the only idea in my mind is to argument that $f(z)$ preserves angles in a certain region, but I don't know how to proceed.
Or I misunderstood the concept of conformal mapping? It seems that I mixed several concepts…
Any corrections or ideas are greatly appreciated!
Best Answer
The function doesn't even exist at $z=0$, which means that it certainly cannot be conformal there.