Conformal Maps

Arc\Curve

Let $\gamma (t)$ be a curve (arc) with $\alpha \le t\le \beta$ contained in a region $\mathrm{\Omega }$.

Let $f(z)$ be continuous on $\mathrm{\Omega }$.
Then

defines an arc in the $w$–plane (the image of $\gamma$).

Assume $f(z)$ is analytic in $\mathrm{\Omega }$ and $\gamma (t)$ is piecewise smooth.

Then by the chain rule:

If $f^{\prime }(\gamma (t_0))\ne 0\text{for some }{t}_0\in [\alpha ,\beta ]$, then

taking arguments gives

Therefore, the tangent vector of the image curve is obtained by scaling and rotating the tangent vector of the original curve.

The amount of rotation is

and this same rotation is applied to any curve passing through the point.

Angle Between Curves

Is the angle between their tangent lines at the point they are intersecting

Example

Show directly (by computing the tangents and angles) that $f(z)=z^3$
preserves the angle between the curves

at their point of intersection.

1. Find the intersection point of the curves.
2. Compute the tangent vectors at that point.
3. Find the angle between those tangents.
4. Apply the map $f(z)$ and its derivative at the point.
5. Compare the new angle — it’s the same, so the angle is preserved.

We have

At the intersection point,

Tangents:

Angle between curves is the difference in their arguments:

so $|\theta |=\arctan \left(\frac{3}{4}\right)$

For $f(z)=z^3$ we have $f^{\prime }(z)=3z^2$, and

Then

Since $\arg (6i)=\frac{\pi }{2}$,

so

Conclusion: $f(z)=z^3$ preserves the angle between ${\gamma }_1$ and ${\gamma }_2$ at $z=1+i$.

Conformal

A function is conformal is it preserves angles between curves. More precisely, a smooth complex-valued function $f$ (cts with cts partial derivatives) is conformal at $z_0$ whenever the two curves ${\gamma }_1$ and ${\gamma }_2$ intersect at $z_0$ with non-zero tangents, $f({\gamma }_1),f({\gamma }_2)$ have non-zero tangents at $f(z_0)$ that intersect at the same angle.

Analytic then Conformal

If $f:U\to \mathbb{C}$ is analytic and if $z_0\in U$ st $f^{\prime }(z_0)\ne 0$ then $f$ is conformal at $z_0$

• $f$ must be defined on an open subset of the complex plane $\mathbb{C}$ (i.e., $f:\mathbb{C}\to \mathbb{C}$).
• $f$ must be analytic (complex differentiable) at $z_0$.
• The derivative must be non-zero ($f^{\prime }(z_0)\ne 0$)

Conformal vs Conformal Mapping

For

we have

Therefore, $f$ is conformal at every point of $\mathbb{C}$.

However, $f$ is not one-to-one on $\mathbb{C}$ since

Hence,

is not a conformal mapping (it fails injectivity).

If instead we take

then $f$ is one-to-one on $D$, and

is a conformal map.

Conformal then Analytic

If $f$ is angle-preserving at $z_0$ and has continuous partial derivatives at $z_0$, then $f$ is analytic at $z_0$.

Proof:

Let $z=z(t)$ be a smooth arc with $z(t_0)=z_0$ and define $w(t)=f(z(t))$.

Chain rule:

Divide by $z^{\prime }(t_0)\ne 0$:

Let $z^{\prime }(t_0)=r{e}^{i\theta }\Rightarrow \frac{\overline{z^{\prime }(t_0)}}{z^{\prime }(t_0)}=e^{-2i\theta }$

So:

Angle-preserving ⇒ $\arg \left(\frac{w^{\prime }(t_0)}{z^{\prime }(t_0)}\right)$ independent of $\theta$

Only possible if:

Now:

Let $f=u+iv$:

Substitute:

Match real and imaginary parts:

• $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$
• $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$

These are the Cauchy-Riemann equations ⇒ $f$ is analytic at $z_0$.

Conformal maps preserve Laplace’s equation

Let $f=u+iv$ be analytic and our conformal coordinate map and $f^{\prime }(z)\ne 0$. Define $H(x,y)=h(u(x,y),v(x,y))$.

We want to show:

Use product rule:

Group terms:

Similarly:

By Cauchy-Riemann:

So:

This implies $u$ and $v$ are harmonic.

So:

Now use:

So:

Conclusion:

If $h$ satisfies Laplace's equation:

So conformal (analytic) maps preserve harmonic functions.