Check whether a complex function is conformal or not.

complex numberscomplex-analysiscomplex-manifolds

Let
$$f_1(z) = \exp(\sin(z)),\quad f_2(z) = \exp(\log(z)),\quad f_3(z) = \exp(\tan(z))$$ be complex functions defined on an open unit disc centered at origin.

Now $f_1$ and $f_3$ is conformal as it is holomorphic and it's derivative is non zero in the given domain.

But is $f_2$ a conformal map? Derivative is non zero but $\log(z)$ is not defined at $z=0$.

Also I read, since a branch of $\log(z)$ is holomorphic and derivative is non zero, $\log(z)$ is conformal.

So got confused to conclude whether $f_2$ is conformal or not.

Please provide a clarity about these two.

  1. Is the above argument for telling $f_1$ and $f_3$ conformal correct ?

  2. What and how can we conclude whether $f_2$ is a conformal map or not.

Thanks in advance.

Best Answer

A holomorphic function $f$ with nonvanishing derivative is locally conformal. This means that each point $z_0$ in the domain of $f$ has a small neighborhood which is bijectively and conformally mapped onto a neighborhood of $f(z_0)$. This is the case for $f_1$ and $f_3$ in your example, after you have verified that $\sin$ and $\tan$ have no critical point in the unit disc $D$. (Of course $\exp$ has no critical points.)

The "function" $f_2$ is not defined as a holomorphic function in $D$, since "$\>\log\>$" has no ${\mathbb C}$-valued holomorphic representant there.

Whether $f_1$ and $f_3$ are actually "conformal maps" of $D$ to certain other regions requires further study. It is commonly accepted that a conformal map is not only locally conformal, but a bijective map between two regions. This means that you have to check whether $\sin(D)$, resp. $\tan(D)$, contains two points $w_i=\sin (z_i)$, $z_1\ne z_2$, differing by an integer multiple of $2\pi i$. This check is not so easy, and is maybe not required in your source problem.

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