What is conformal about this mapping?

Question

The mapping $w = f(z) = z - \frac{R^2}{z}$ transforms a circle in the z-plane to a vertical line in the $w$-plane. A conformal mapping is supposed to preserve angle. What angle is being preserved here (from circle to straight line)?

Answer 1

The angles that are being preserved are infinitesmal angles, i.e. angles between tangent vectors.

At a point $z_0$, you can measure the angles between pairs of vectors in the tangent space $T_{z_0}(\mathbb{C})$ based at $z_0$. At its image point $w_0=f(z_0)$, you can similarly measure the angles between pairs of vectors in the tangent space $T_{w_0}(\mathbb{C})$ based at $w_0$. Each of these two tangent spaces can be identified with $\mathbb{C}$ itself. The derivative map $$Df_{z_0} : \mathbb{C} \approx T_{z_0}(\mathbb{C}) \mapsto T_{w_0}(\mathbb{C}) \approx \mathbb{C} $$ is simply multiplication by the complex number $f'(z_0)$. Multiplication by $f'(z_0)$ is angle preserving. Hence, the angle between $u,v \in T_{z_0}(\mathbb{C})$ is equal to the angle between $Df_{z_0}(u),Df_{z_0}(v) \in T_{w_0}(\mathbb{C})$.

In your example, take any curve $C$ in the $z$ plane that intersects your circle, and let $z_0$ be any one of those intersection points. Then $w_0$ is an intersection point of $f(C)$ with your line. So what conformal means is that the infinitesmal angle at $z_0$ between $C$ and your circle is equal to the infinitesmal angle at $w_0$ between $f(C)$ and your line.

Or, with more information, if $u,v$ are tangent (respectively) to $C$ and your circle at $z_0$ then $Df_{z_0}(u)$ and $Df_{z_0}(v)$ are tangent (respectively) to $f(C)$ and your line at $w_0$, and the angle between $u,v$ equals the angle between $Df_{z_0}(u)$ and $Df_{z_0}(v)$.