NoteNextra-origin/pages/Math416/Math416_L3.md

Lecture 3

Differentiation of functions in complex variables

Differentiability

Definition of differentiability in complex variables

Suppose G is an open subset of \mathbb{C}.

A function f:G\to \mathbb{C} is differentiable at \zeta_0\in G if

\lim_{\zeta\to \zeta_0}\frac{f(\zeta)-f(\zeta_0)}{\zeta-\zeta_0}

exists.

Or equivalently,

We can also express the f as f=u+iv, where u,v:G\to \mathbb{R} are real-valued functions.

Recall that u:G\to \mathbb{R} is differentiable at \zeta_0\in G if and only if there exists a complex number (x,y)\in \mathbb{C} such that a function

R(x,y)=u(x,y)-\left(u(x_0,y_0)+\frac{\partial u}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial u}{\partial y}(x_0,y_0)(y-y_0)\right)

satisfies

\lim_{(x,y)\to (x_0,y_0)}\frac{|R(x,y)|}{|(x,y)-(x_0,y_0)|}=\lim_{(x,y)\to (x_0,y_0)}\frac{|R(x,y)|}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0.

Theorem from 4111?

If u is differentiable at (x_0,y_0), then \frac{\partial u}{\partial x}(x_0,y_0) and \frac{\partial u}{\partial y}(x_0,y_0) exist.

If \frac{\partial u}{\partial x}(x_0,y_0) and \frac{\partial u}{\partial y}(x_0,y_0) exist and one of them is continuous at (x_0,y_0), then u is differentiable at (x_0,y_0).

\begin{aligned}
\lim_{(x,y)\to (x_0,y_0)}\frac{|R(x,y)|}{|(x,y)-(x_0,y_0)|}&=\lim_{(x,y)\to (x_0,y_0)}\frac{|u(x,y)-u(x_0,y_0)-\frac{\partial u}{\partial x}(x_0,y_0)(x-x_0)-\frac{\partial u}{\partial y}(x_0,y_0)(y-y_0)|}{\sqrt{(x-x_0)^2+(y-y_0)^2}}\\
&=\lim_{(x,y)\to (x_0,y_0)}\frac{|u(x,y)-u(x_0,y_0)-\frac{\partial u}{\partial x}(x_0,y_0)(x-x_0)-\frac{\partial u}{\partial y}(x_0,y_0)(y-y_0)|}{\sqrt{(x-x_0)^2+(y-y_0)^2}}\\
\end{aligned}

Let a(x,y)=\frac{\partial u}{\partial x}(x,y) and b(x,y)=\frac{\partial u}{\partial y}(x,y).

We can write R(x,y) as

R(x,y)=u(x,y)-u(x_0,y_0)-a(x,y)(x-x_0)-b(x,y)(y-y_0).

So \lim_{(x,y)\to (x_0,y_0)}\frac{|R(x,y)|}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0 if and only if \lim_{(x,y)\to (x_0,y_0)}\frac{a(x-x_0)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0 and \lim_{(x,y)\to (x_0,y_0)}\frac{b(y-y_0)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0.

On the imaginary part, we have

...

Conclusion (The Cauchy-Riemann equations):

If f=u+iv is complex differentiable at \zeta_0\in G, then u and v are real differentiable at (x_0,y_0) and

\frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial y}(x_0,y_0),\quad \frac{\partial u}{\partial y}(x_0,y_0)=-\frac{\partial v}{\partial x}(x_0,y_0).

And u and v have continuous partial derivatives at (x_0,y_0).

And let c=\frac{\partial u}{\partial x}(x_0,y_0) and d=\frac{\partial v}{\partial x}(x_0,y_0).

Then f'(\zeta_0)=c+id, is holomorphic at \zeta_0.

Holomorphic Functions

Definition of holomorphic functions

A function f:G\to \mathbb{C} is holomorphic (or analytic) at \zeta_0\in G if it is complex differentiable at \zeta_0.

Example:

Suppose f:G\to \mathbb{C} where f=u+iv and \frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}, \frac{\partial f}{\partial y}=\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}.

Define \frac{\partial}{\partial \zeta}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right) and \frac{\partial}{\partial \bar{\zeta}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right).

Suppose f is holomorphic at \bar{\zeta}_0\in G (Cauchy-Riemann equations hold at \bar{\zeta}_0).

Then \frac{\partial f}{\partial \bar{\zeta}}(\bar{\zeta}_0)=0.

Note that \forall m\in \mathbb{Z}, \zeta^m is holomorphic on \mathbb{C}.

i.e. \forall a\in \mathbb{C}, \lim_{\zeta\to a}\frac{\zeta^m-a^m}{\zeta-a}=\frac{(\zeta-a)(\zeta^{m-1}+\zeta^{m-2}a+\cdots+a^{m-1})}{\zeta-a}=ma^{m-1}.

So polynomials are holomorphic on \mathbb{C}.

So rational functions p/q are holomorphic on \mathbb{C}\setminus\{z\in \mathbb{C}:q(z)=0\}.

Curves in \mathbb{C}

Definition of curves in \mathbb{C}

A curve \gamma in G\subset \mathbb{C} is a continuous map of an interval I into G. We say \gamma is differentiable if \forall t_0\in I, \gamma'(t_0)=\lim_{t\to t_0}\frac{\gamma(t)-\gamma(t_0)}{t-t_0} exists.

If \gamma'(t_0) is a point in \mathbb{C}, then \gamma'(t_0) is called the tangent vector to \gamma at t_0.

Definition of regular curves in \mathbb{C}

A curve \gamma is regular if \gamma'(t)\neq 0 for all t\in I.

Definition of 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)=\zeta_0 for some t_0\in I_1\cap I_2.

The angle between \gamma_1 and \gamma_2 at \zeta_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)).

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)=\zeta_0 for some t_0\in I_1\cap I_2.

If f'(\zeta_0)\neq 0, then the angle between \gamma_1 and \gamma_2 at \zeta_0 is the same as the angle between the vectors f'(\zeta_0)\gamma_1'(t_0) and f'(\zeta_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)=\zeta_0 for some t_0\in I.

Then,

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

If Lemma of function of a curve and angle holds, then the angle between f\circ \gamma_1 and f\circ \gamma_2 at \zeta_0 is

\begin{aligned}
\arg\left[(f\circ \gamma_2)'(t_2)(f\circ \gamma_1)'(t_1)\right]&=\cdots
\end{aligned}

Continue on Thursday. (Applying the chain rules)