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Math416Complex Variables (Lecture 3)

Math416 Lecture 3

Differentiation of functions in complex variables

Differentiability

Definition 2.1 of differentiability in complex variables

Suppose GG is an open subset of C\mathbb{C}. (very important, f(z0)f'(z_0) cannot be define unless z0z_0 belongs to an open set in which ff is defined.)

A function f:GCf:G\to \mathbb{C} is differentiable at z0Gz_0\in G if

f(z0)=limzz0f(z)f(z0)zz0f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}

exists.

Or equivalently,

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

Recall that u:GRu:G\to \mathbb{R} is differentiable at z0Gz_0\in G if and only if there exists a complex number (x,y)C(x,y)\in \mathbb{C} such that a function

R(x,y)=u(x,y)(u(x0,y0)+ux(x0,y0)(xx0)+uy(x0,y0)(yy0))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)(x0,y0)R(x,y)(x,y)(x0,y0)=lim(x,y)(x0,y0)R(x,y)(xx0)2+(yy0)2=0.\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.

R(x,y)R(x,y) is the immediate result of mean value theorem applied to uu at (x0,y0)(x_0,y_0).

Theorem from 4111?

If uu is differentiable at (x0,y0)(x_0,y_0), then ux(x0,y0)\frac{\partial u}{\partial x}(x_0,y_0) and uy(x0,y0)\frac{\partial u}{\partial y}(x_0,y_0) exist.

If ux(x0,y0)\frac{\partial u}{\partial x}(x_0,y_0) and uy(x0,y0)\frac{\partial u}{\partial y}(x_0,y_0) exist and one of them is continuous at (x0,y0)(x_0,y_0), then uu is differentiable at (x0,y0)(x_0,y_0).

lim(x,y)(x0,y0)R(x,y)(x,y)(x0,y0)=lim(x,y)(x0,y0)u(x,y)u(x0,y0)ux(x0,y0)(xx0)uy(x0,y0)(yy0)(xx0)2+(yy0)2=lim(x,y)(x0,y0)u(x,y)u(x0,y0)ux(x0,y0)(xx0)uy(x0,y0)(yy0)(xx0)2+(yy0)2\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)=ux(x,y)a(x,y)=\frac{\partial u}{\partial x}(x,y) and b(x,y)=uy(x,y)b(x,y)=\frac{\partial u}{\partial y}(x,y).

We can write R(x,y)R(x,y) as

R(x,y)=u(x,y)u(x0,y0)a(x,y)(xx0)b(x,y)(yy0).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)(x0,y0)R(x,y)(xx0)2+(yy0)2=0\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)(x0,y0)a(xx0)(xx0)2+(yy0)2=0\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)(x0,y0)b(yy0)(xx0)2+(yy0)2=0\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 proceed similarly. Define

S(x,y)=v(x,y)v(x0,y0)vx(x0,y0)(xx0)vy(x0,y0)(yy0).S(x,y)=v(x,y)-v(x_0,y_0)-\frac{\partial v}{\partial x}(x_0,y_0)(x-x_0)-\frac{\partial v}{\partial y}(x_0,y_0)(y-y_0).

Then the differentiability of vv at (x0,y0)(x_0,y_0) guarantees that

lim(x,y)(x0,y0)S(x,y)(xx0)2+(yy0)2=0.\lim_{(x,y)\to (x_0,y_0)}\frac{|S(x,y)|}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0.

Moreover, considering the definition of the complex derivative of f=u+ivf=u+iv, if we approach z0=x0+iy0z_0=x_0+iy_0 along different directions we obtain

f(z0)=ux(x0,y0)+ivx(x0,y0)=vy(x0,y0)iuy(x0,y0).f'(z_0)=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0) =\frac{\partial v}{\partial y}(x_0,y_0)-i\frac{\partial u}{\partial y}(x_0,y_0).

Equating the real and imaginary parts of these two expressions forces

ux(x0,y0)=vy(x0,y0),uy(x0,y0)=vx(x0,y0).\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).

Theorem 2.6 (The Cauchy-Riemann equations):

If f=u+ivf=u+iv is complex differentiable at z0Gz_0\in G, then uu and vv are real differentiable at (x0,y0)(x_0,y_0) and

ux(x0,y0)=vy(x0,y0),uy(x0,y0)=vx(x0,y0).\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).

Some missing details:

The Cauchy-Riemann equations are necessary and sufficient for the differentiability of ff at z0z_0.

This states that a function ff is complex differentiable at z0z_0 if and only if uu and vv are real differentiable at (x0,y0)(x_0,y_0) and the Cauchy-Riemann equations hold at (x0,y0)(x_0,y_0). That is f(z0)=ux(x0,y0)+ivx(x0,y0)=vy(x0,y0)iuy(x0,y0)f'(z_0)=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial y}(x_0,y_0)-i\frac{\partial u}{\partial y}(x_0,y_0).

And uu and vv have continuous partial derivatives at (x0,y0)(x_0,y_0).

And let c=ux(x0,y0)c=\frac{\partial u}{\partial x}(x_0,y_0) and d=vx(x0,y0)d=\frac{\partial v}{\partial x}(x_0,y_0).

Then f(z0)=c+idf'(z_0)=c+id, is holomorphic at z0z_0.

Holomorphic Functions

Definition 2.8 (Holomorphic functions)

A function f:GCf:G\to \mathbb{C} is holomorphic (or analytic) at z0Gz_0\in G if it is complex differentiable at z0z_0.

Note that the true definition of analytic function is that can be written as a convergent power series in a neighborhood of each point in its domain. We will prove that these two definitions are equivalent to each other in later sections.

Example:

Suppose f:GCf:G\to \mathbb{C} where f=u+ivf=u+iv and fx=ux+ivx\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}, fy=uy+ivy\frac{\partial f}{\partial y}=\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}.

Define z=12(xiy)\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right) and zˉ=12(x+iy)\frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right).

Suppose ff is holomorphic at zˉ0G\bar{z}_0\in G (Cauchy-Riemann equations hold at zˉ0\bar{z}_0).

Then fzˉ(zˉ0)=0\frac{\partial f}{\partial \bar{z}}(\bar{z}_0)=0.

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

i.e. aC\forall a\in \mathbb{C}, limzazmamza=(za)(zm1+zm2a++am1)za=mam1\lim_{z\to a}\frac{z^m-a^m}{z-a}=\frac{(z-a)(z^{m-1}+z^{m-2}a+\cdots+a^{m-1})}{z-a}=ma^{m-1}.

So polynomials are holomorphic on C\mathbb{C}.

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

Definition 2.9 (Complex partial differential operators)

Let f:GCf:G\to \mathbb{C}, f=u+ivf=u+iv, be a function defined on an open set GCG\subset \mathbb{C}.

Define:

xf=ux+ivx,yf=uy+ivy.\frac{\partial}{\partial x}f=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x},\quad \frac{\partial}{\partial y}f=\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}.

And

zf=12(xiy)f,zˉf=12(x+iy)f.\frac{\partial}{\partial z}f=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f,\quad \frac{\partial}{\partial \bar{z}}f=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f.

This definition of partial differential operators on complex functions is consistent with the definition of partial differential operators on real functions.

xf=zf+zˉf,yf=i(zfzˉf).\frac{\partial}{\partial x}f=\frac{\partial}{\partial z}f+\frac{\partial}{\partial \bar{z}}f,\quad \frac{\partial}{\partial y}f=i\left(\frac{\partial}{\partial z}f-\frac{\partial}{\partial \bar{z}}f\right).

Curves in C\mathbb{C}

Definition 2.11 (Curves in C\mathbb{C})

A curve γ\gamma in GCG\subset \mathbb{C} is a continuous map of an interval IRI\in \mathbb{R} into GG. We say γ\gamma is differentiable if t0I\forall t_0\in I, γ(t0)=limtt0γ(t)γ(t0)tt0\gamma'(t_0)=\lim_{t\to t_0}\frac{\gamma(t)-\gamma(t_0)}{t-t_0} exists.

If γ(t0)\gamma'(t_0) is a point in C\mathbb{C}, then γ(t0)\gamma'(t_0) is called the tangent vector to γ\gamma at t0t_0.

Definition of regular curves in C\mathbb{C}

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

Definition of angle between two curves

Let γ1,γ2\gamma_1,\gamma_2 be two curves in GCG\subset \mathbb{C} with γ1(t0)=γ2(t0)=z0\gamma_1(t_0)=\gamma_2(t_0)=z_0 for some t0I1I2t_0\in I_1\cap I_2.

The angle between γ1\gamma_1 and γ2\gamma_2 at z0z_0 is the angle between the vectors γ1(t0)\gamma_1'(t_0) and γ2(t0)\gamma_2'(t_0). Denote as arg(γ2(t0))arg(γ1(t0))=arg(γ2(t0)γ1(t0))\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0)).

Theorem 2.12 of conformality

Suppose f:GCf:G\to \mathbb{C} is holomorphic function on open set GCG\subset \mathbb{C} and γ1,γ2\gamma_1,\gamma_2 are regular curves in GG with γ1(t0)=γ2(t0)=z0\gamma_1(t_0)=\gamma_2(t_0)=z_0 for some t0I1I2t_0\in I_1\cap I_2.

If f(z0)0f'(z_0)\neq 0, then the angle between γ1\gamma_1 and γ2\gamma_2 at z0z_0 is the same as the angle between the vectors f(z0)γ1(t0)f'(z_0)\gamma_1'(t_0) and f(z0)γ2(t0)f'(z_0)\gamma_2'(t_0).

Lemma of function of a curve and angle

If f:GCf:G\to \mathbb{C} is holomorphic function on open set GCG\subset \mathbb{C} and γ\gamma is differentiable curve in GG with γ(t0)=z0\gamma(t_0)=z_0 for some t0It_0\in I.

Then,

(fγ)(t0)=f(γ(t0))γ(t0).(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 fγ1f\circ \gamma_1 and fγ2f\circ \gamma_2 at z0z_0 is

arg[(fγ2)(t2)(fγ1)(t1)]=\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)

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