Examples of conformal holomorphic maps.

1. $f(z)=e^z$, which means component-wise that $f(x,y)=(e^x\cos (y),e^x\sin y)$. The derivative is again nowhere vanishing, thus it is locally conformal.
   Here we also introduced coordinate systems on ${\mathbb{R}}^2$ to visualize what is going on. We look at the images of grid lines on the cartesian plane and see what happens to the images.

Now we consider vertical strips of the complex plane (that is strips of the form $(-r,r)\times (y_1,y_2)$), we consider what happens as $r\to \mathrm{\infty }$. The image under the exponential map, the limiting case gives us a sector (for $y_2-y_1<2\pi$ ) of the complex plane which depends on $y_1$ and $y_2$. if $y_2-y_1=2\pi$, then we get a slit plane, with the slit depending on the shift of $y_1$ and $y_2$.

We use this observation to conclude that the exponential map is not globally injective and thus not 1;;conformal.
2. Let $f(z)=\frac{z-i}{z+i}$ be a holomorphic function on $H=\{z:im(z)>0\}$. Then we note that the image of this map is actually $\mathbb{D}$ (at least the interior). We further note that this map is injective, and surjective on $int(\mathbb{D})$. Finally, $f^{\prime }(z)=\frac{2i}{(z+i{)}^2}$ which is non-zero for $z\in H$.

If there are two open sets $U,V\subseteq \mathbb{C}$ such that there is a conformal mapping $f:U\to V$, then we say that $U$ and $V$ are conformally equivalent.