NoteNextra-origin/content/Math416/Math416_L4.md

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}.