Math416 Lecture 4

Review

Derivative of a complex function

\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)

\frac{\partial f}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\right)

Angle between two curves

Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$ .

The angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$ . Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$ .

Cauchy-Riemann equations

\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)

Continue on last lecture

Theorem of conformality

Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$ .

If $f'(z_0)\neq 0$ , then the angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the same as the angle between the vectors $f'(z_0)\gamma_1'(t_0)$ and $f'(z_0)\gamma_2'(t_0)$ .

Lemma of function of a curve and angle

If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=z_0$ for some $t_0\in I$ .

Then,

(f\circ \gamma)'(t_0)=f'(\gamma(t_0))\gamma'(t_0).

Looks like the chain rule.

Proof

We want to show that

\lim_{t\to t_0}\frac{(f\circ \gamma)(t)-(f\circ \gamma)(t_0)}{t-t_0}=f'(\gamma(t_0))\gamma'(t_0).

Notation:

A function $g(h)$ is $O(h)$ if $\exists C>0$ such that $|g(h)|\leq C|h|$ for all $h$ in a neighborhood of $0$ .

A function $g(h)$ is $o(h)$ if $\lim_{h\to 0}\frac{g(h)}{h}=0$ .

$f$ is differentiable if and only if $f(z+h)=f(z)+f'(z)h+\frac{1}{2}h^2f''(z)+o(h^3)$ as $h\to 0$ . (By Taylor expansion)

Since $f$ is holomorphic at $\gamma(t_0)=z_0$ , we have

f(z_0)=f(z_0)+(z-z_0)f'(z_0)+o(z-z_0)

This result comes from Taylor Expansion of the derivative of the function around the point $z_0$

and

f(\gamma(t_0))=f(\gamma(t_0))+f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))

So,

\begin{aligned} \lim_{t\to t_0}\frac{(f\circ \gamma)(t)-(f\circ \gamma)(t_0)}{t-t_0} &=\lim_{t\to t_0}\frac{\left[f(\gamma(t_0))+f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))\right]-f(\gamma(t_0))}{t-t_0} \\ &=\lim_{t\to t_0}\frac{f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))}{t-t_0} \\ &=\lim_{t\to t_0}\frac{f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))}{t-t_0} +\lim_{t\to t_0}\frac{o(\gamma(t)-\gamma(t_0))}{t-t_0} \\ &=f'(\gamma(t_0))\lim_{t\to t_0}\frac{\gamma(t)-\gamma(t_0)}{t-t_0} +0\\ &=f'(\gamma(t_0))\gamma'(t_0) \end{aligned}

Definition 2.12 (Conformal function)

A function $f:G\to \mathbb{C}$ is called conformal if it preserves the angle between two curves.

Theorem 2.13 (Conformal function)

If $f:G\to \mathbb{C}$ is conformal at $z_0\in G$ , then $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$ .

Example:

f(z)=z^2

is not conformal at $z=0$ because $f'(0)=0$ .

Lemma of conformal function

Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial z}(z_0)$ , $b=\frac{\partial f}{\partial \overline{z}}(z_0)$ .

Let $\gamma(t_0)=z_0$ . Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$ .

Proof

$f=u+iv$ , $u,v$ are real differentiable.

a=\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)

b=\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)

\gamma'(t_0)=\frac{d\alpha}{dt}+i\frac{d\beta}{dt}

\overline{\gamma'(t_0)}=\frac{d\beta}{dt}-i\frac{d\alpha}{dt}

\begin{aligned} (f\circ \gamma)'(t_0)&=\frac{\partial f}{\partial z}(\gamma(t_0))\gamma'(t_0)+\frac{\partial f}{\partial \overline{z}}(\gamma(t_0))\overline{\gamma'(t_0)} \\ &=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\alpha}{dt}+i\frac{d\beta}{dt}\right)\\ &+\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\beta}{dt}-i\frac{d\alpha}{dt}\right) \\ &=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\alpha}{dt}-\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\beta}{dt}\right]\\ &+i\left[\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\alpha}{dt}+\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\beta}{dt}\right] \\ &=\left[a+b\right]\frac{d\alpha}{dt}+i\left[a-b\right]\frac{d\beta}{dt} \\ &=\left[u_x+iv_x\right]\frac{d\alpha}{dt}+i\left[v_y-iu_y\right]\frac{d\beta}{dt} \\ &=a\gamma'(t_0)+b\overline{\gamma'(t_0)} \end{aligned}

Theorem of differentiability

Let $f:G\to \mathbb{C}$ be a function defined on an open set $G\subset \mathbb{C}$ that is both holomorphic and (real) differentiable, where $f=u+iv$ with $u,v$ real differentiable functions.

Then, $f$ is conformal at every point $z_0\in G$ if and only if $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$ .

Proof

We prove the equivalence in two parts.

( $\implies$ ) Suppose that $f$ is conformal at $z_0$ . By definition, conformality means that $f$ preserves angles (including their orientation) between any two intersecting curves through $z_0$ . In the language of real analysis, this requires that the (real) derivative (Jacobian) of $f$ at $z_0$ , $Df(z_0)$ , acts as a similarity transformation. Any similarity in $\mathbb{R}^2$ can be written as a rotation combined with a scaling; in particular, its matrix representation has the form

\begin{pmatrix} A & -B \\ B & A \end{pmatrix},

for some real numbers $A$ and $B$ . This is exactly the matrix corresponding to multiplication by the complex number $a=A+iB$ . Therefore, the Cauchy-Riemann equations must hold at $z_0$ , implying that $f$ is holomorphic at $z_0$ . Moreover, because the transformation is nondegenerate (preserving angles implies nonzero scaling), we must have $f'(z_0)=a\neq 0$ .

( $\impliedby$ ) Now suppose that $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$ . Then by the definition of the complex derivative, the first-order (linear) approximation of $f$ near $z_0$ is

f(z_0+h)=f(z_0)+f'(z_0)h+o(|h|),

for small $h\in\mathbb{C}$ . Multiplication by the nonzero complex number $f'(z_0)$ is exactly a rotation and scaling (i.e., a similarity transformation). Therefore, for any smooth curve $\gamma(t)$ with $\gamma(t_0)=z_0$ , we have

(f\circ\gamma)'(t_0)=f'(z_0)\gamma'(t_0),

and the angle between any two tangent vectors at $z_0$ is preserved (up to the fixed rotation). Hence, $f$ is conformal at $z_0$ .

For further illustration, consider the special case when $f$ is an affine map.

Case 1: Suppose

f(z)=az+b\overline{z}.

The Wirtinger derivatives of $f$ are

\frac{\partial f}{\partial z}=a \quad \text{and} \quad \frac{\partial f}{\partial \overline{z}}=b.

For $f$ to be holomorphic, we require $\frac{\partial f}{\partial \overline{z}}=b=0$ . Moreover, to have a nondegenerate (angle-preserving) map, we must have $a\neq 0$ . If $b\neq 0$ , then the map mixes $z$ and $\overline{z}$ , and one can check that the linearization maps the real axis $\mathbb{R}$ into the set $\{(a+b)t\}$ , which does not uniformly scale and rotate all directions. Thus, $f$ fails to be conformal when $b\neq 0$ .

Case 2: For a general holomorphic function, the lemma of conformal functions shows that if

(f\circ \gamma)'(t_0)=f'(z_0)\gamma'(t_0)

for any differentiable curve $\gamma$ through $z_0$ , then the effect of $f$ near $z_0$ is exactly given by multiplication by $f'(z_0)$ . Since multiplication by a nonzero complex number is a similarity transformation, $f$ is conformal at $z_0$ .

Harmonic function

Let $\Omega$ be a domain in $\mathbb{C}$ . A function $u:\Omega\to \mathbb{R}$

A domain is a connected open set.

Say $g:\Omega\to \mathbb{R} \text{ or } \mathbb{C}$ is harmonic if it satisfies the Laplace equation

\Delta g=\frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}=0.

Theorem of harmonic conjugate

Let $f=u+iv$ be holomorphic function on domain $\Omega\subset \mathbb{C}$ . Then $u$ and $v$ are harmonic functions on $\Omega$ .

Proof

\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.

Using the Cauchy-Riemann equations, we have

\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x\partial y}, \quad \frac{\partial^2 u}{\partial y^2}=-\frac{\partial^2 v}{\partial y\partial x}.

So,

\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0.

If $v$ is such that $f=u+iv$ is holomorphic on $\Omega$ , then $v$ is called harmonic conjugate of $u$ on $\Omega$ .

Example:

u(x,y)=x^2-y^2

is harmonic on $\mathbb{C}$ .

To find a harmonic conjugate of $u$ on $\mathbb{C}$ , we need to find a function $v$ such that

\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=2y, \quad \frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=2x.

Integrating, we get

v(x,y)=2xy+G(y)

\frac{\partial v}{\partial y}=2x+G'(y)=2x

So,

G'(y)=0 \implies G(y)=C

v(x,y)=2xy+C

is a harmonic conjugate of $u$ on $\mathbb{C}$ .

Combine $u$ and $v$ to get $f(x,y)=x^2-y^2+2xyi+C=(x+iy)^2+C=z^2+C$ , which is holomorphic on $\mathbb{C}$ .