I'm learning about conformal mappings into polygons in a class,(undergrad complex analysis) and am having trouble understanding one of the examples given in my book. (Stein & Shakarchi) Here it is:
For $z\in\mathbb{H}$(the open upper half plane), define \begin{align*} f(z) := \int_{0}^{z}\frac{1}{(1-\zeta^{2})^{1/2}}d\zeta \end{align*} The example goes on to talk about the boundary behavior of this function, which I'm o.k. with. It also says that $f$ is the inverse of $\sin(z)$, and thus is conformal from $\mathbb{H}$ to a "half-strip" described in the book. Is there another way we can see that $f$ is conformal?
The connection between conformal mapping and complex variables is about as fundamental an idea in complex variables as there is. This connection is the first example mentioned on the Wikipedia page for conformal map.
This basic observation is that a differentiable function behaves like it's linear approximation on a small scale. In symbols, for $z$ close to $z_0$, we have $$f(z) \approx f(z_0) + f'(z_0)(z-z_0)$$ with an error proportional to $|z-z_0|^2$. Thus, for $z$ close to $z_0$, a function that is differentiable in a neighborhood of $z_0$ with $f'(z_0)\neq0$ looks approximately affine, that is it looks like a function of the form $g(z)=az+b$. Now, multiplication by a complex number $a$ scales by $|a|$ and rotates by $\arg(a)$. Adding a complex number $b$ simply shifts the whole picture. These operations preserve angle measure so that an affine function does as well and a differentiable function with non-vanishing derivative does as well.