Notes - Complex Analysis MT23, Conformal maps

The angle between any two $C^1$ paths $\gamma _ 1, \gamma _ 2$ passing through $z _ 0$ is equal to the angle between the $C^1$ paths $T \circ \gamma _ 1$ and $T \circ \gamma _ 2$ at $T(z _ 0)$.

Suppose:

• $f : U \to \mathbb C$ is a holomorphic map
• $f’(z _ 0) \ne 0$
• $z _ 0 \in U$

Then $f$ is conformal at $z _ 0$.

There is a bijective conformal transformation between them.

Suppose:

• $U \subset \mathbb C$, open, connected, simply-connected, proper subset
• $z _ 0 \in U$

Then there exists a unique bijective conformal transformation $f : U \to \mathbb D$ such that $f(z _ 0) = 0$, $f’(z _ 0) > 0$.

• Möbius maps, since you can use them to map lines to circles.
• The exponential, since you can turn rectangular regions into annuli
• The logarithm,
• $z \mapsto z^2$ and $z \mapsto \sqrt z$

Recall that $f$ is conformal if

The angle between any two $C^1$ paths $\gamma _ 1, \gamma _ 2$ passing through $z _ 0$ is equal to the angle between the $C^1$ paths $f \circ \gamma _ 1$ and $f \circ \gamma _ 2$ at $T(z _ 0)$.

Let $\gamma _ 1, \gamma _ 2$ be $C^1$ paths such that $\gamma _ 1(0) = z _ 0$, $\gamma _ 2(0) = z _ 0$, and then we get paths $\eta _ 1$ and $\eta _ 2$ defined as $\eta _ 1 = f \circ \gamma _ 1$ and $\eta _ 2 = f \circ \gamma _ 2$.

Now we look at the derivative of each transformed path. For $i = 1, 2$:

\[\begin{aligned} \eta_i'(0) &= \lim_{h \to 0} \frac{f(\gamma_i(0 + h)) - f(\gamma_i(0))}{h} \\\\ &= \lim_{h \to 0} \frac{f(\gamma_i(h)) - f(z_0)}{\gamma_i(h) - z_0} \cdot \frac{\gamma_i(h) - z_0}{h} \\\\ &= \lim_{h \to 0} \frac{f(\gamma_i(h)) - f(z_0)}{\gamma_i(h) - z_0} \cdot \frac{\gamma_i(h) - \gamma_i(0)}{h} \\\\ &= f'(z_0) \gamma_i'(0) \end{aligned}\]

where we are not dividing by zero anywhere since $\gamma _ i(h) \ne 0$ near $z _ 0$. Then, to make the argument of the complex number clearer, write

\[\theta := \arg f'(z_0)\]

Then if

\[\begin{aligned} \phi_1 &:= \arg \gamma'_1 (0) \\\\ \phi_2 &:= \arg \gamma'_2(0) \end{aligned}\]

and

\[\begin{aligned} \psi_1 &:= \arg \eta'_1 (0) \\\\ \psi_2 &:= \arg \eta'_2(0) \end{aligned}\]

We see that

\[\begin{aligned} \psi_1 &= \theta + \phi_1 \\\\ \psi_2 &= \theta + \phi_2 \end{aligned}\]

Hence the angle between $\eta _ 1$ and $\eta _ 2$ at $f(z _ 0)$ is the same as the angle between $\gamma _ 1$ and $\gamma _ 2$ at $z _ 0$.

Suppose:

• $U, V$ are domains
• $G : U \to V$ conformal map (so in particular holomorphic)
• $v : V \to \mathbb R$ harmonic

Then:

• $u : U \to \mathbb R$ defined by $u = v \circ G$ is harmonic on $U$.

Since being harmonic is a local property, given $z _ 0 \in U$, it suffices to show that $v \circ G$ is harmonic on a disc $B(z _ 0, r) \subseteq U$ for some $r$.

Let $w _ 0 = G(z _ 0)$, i.e. the corresponding point in $V$. Then by continuity of $G$, $\exists \delta, \varepsilon$ such that

\[G(B(z_0, \delta)) \subseteq B(w_0, \varepsilon)\]

Since $v$ is harmonic and $B(w _ 0, \varepsilon)$ is simply connected, we have a theorem that says $\exists f : V \to \mathbb C$ such that $f$ is holomorphic and

\[v = \text{Re}(f)\]

on $B(w _ 0, \varepsilon)$ (the fact that the theorem that says this requires the domain to be simply connected is why we need to do all this stuff about taking discs, rather than just concluding $v = \text{Re}(f)$ on the whole domain). But then on $B(z _ 0, \delta)$, we have

\[u = v \circ G = \text{Re}(f \circ G)\]

Since $f \circ G$ is holomorphic, we know that $u$ is the real part of a holomorphic function, so is harmonic.