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Cauchy's Integral Theorem

Integration along curves

Parameterized Curve

A complex function \(z(t)\) is said to be parametrized curve, if it maps a closed interval \([a,b]\subset \mathbb{R}\) to \(D\subset \mathbb{C}\).

A parametrized curve is said to be smooth, if \(z'(t)\) exists, \(z'(t)\neq 0, \forall t\in [a,b]\) and is continuous on \([a,b]\). At point \(t=a,b\), we intepret continuity and differentiability as one-side limits. (here readers could just view \(z\) as a function vector with respect to variable \(t\)). Similar situation for piecewise-smooth.

Equivalence of parametrized curve

Two parametrized curves \(z:[a,b]\mapsto \mathbb{C}\) and \(\tilde{z}:[c,d]\mapsto \mathbb{C}\) are said to ba equivalent, if there exsits a continuously differentiable bijection \(t\) \((t'(s)>0)\) which maps \([c,d]\) to \([a,b]\), such that

\[ \tilde{z}(s)=z(t(s)). \]

The above equivalence defines an equivalent family of \(z(t)\), the difference of which are the definition domains, varying among any closed interval in \(\mathbb{R}\). They are all the same in the range, so we denote the whole of them as \(\gamma\subset \mathbb{C}\), called smooth curve. Thus, a smooth curve could determine incountable parametrized curves.

The condition \((t'(s)>0)\) guarantee the direction of the curve. Later we would see that it determines the sign of integral.

Reversing the direction

For a given smooth curve \(\gamma\), and its one parametrized curve \(z:[a,b]\mapsto \mathbb{C}\), we could obtain one of its reverse curve \(\gamma^-\), by

\[ z^-=z(b+a-t). \]

so \(z^-:[a,b]\mapsto \mathbb{R}\). In fact, we could formulate infinite number of these equivalent curve, while the above one guarantees the definition domain is kept same as \(z\).

Simple curve, closed curve

A smooth curve is said to be simple, if it is not self-intersecting. Or in mathematical language, \(z(t)\neq z(s)\) unless \(s=t\).

A smooth curve is said to be closed, if \(z(a)=z(b)\) for any of its parametrizations.

For the following parts, we call the above piecewise-smooth curve a curve for brevity.

Example. Some examples of curves.

(i) Circle. Set like \(C_r(z_0)=\{z\in\mathbb{C}:|z-z_0|=r\}.\) are called circle, with its usual parametrization (positive and negative orientation respectively)

\[ z(t)=z_0+re^{i\theta},\quad t\in [0,2\pi], \]
\[ z(t)=z_0+re^{-i\theta},\quad t\in [0,2\pi]. \]

In the following chapters, we denote \(C\) as a general positive oriented circle.

Riemann integral of complex function f

Assume complex function \(f\) is defined on \(\gamma\), where \(\gamma\) is not necessarily smooth, but must be length-calculatable. For one parametrization \(z:[a,b]\mapsto \mathbb{C}\), we have partition

\[ \Delta_n: z(a)=z_0<z_1<\cdots<z_n=z(b) \]

Choose an arbitrary point \(\xi_k\) on arc from \(z_{k-1}\) to \(z_k\), \((k=1,\cdots,n)\) we have Riemann summation

\[ \sum_{k=1}^n f(\xi_k)(z_k-z_{k-1}). \]

Denote \(s_k\) to be the length of arc from \(z_{k-1}\) to \(z_k\). If \(\lambda=\max\{s_k: 1\leq k\leq n\}\rightarrow 0\), the above summation has a unique limit, then we call complex function \(f\) Riemann-integrable. And we denote the integral of \(f\) along \(\gamma\) as

\[ \int_\gamma f(z)dz=\lim_{\lambda\rightarrow 0}\sum_{k=1}^n f(\xi_k)(z_k-z_{k-1}). \]

Sufficient condition for Riemann integrability

If complex function \(f=u+iv\) is continuous on \(\gamma\) which is length-integrable, then it is Riemann-integrable on \(\gamma\), and

\[ \int_\gamma f(z)dz=\int_\gamma (u dx-vdy)+i\int_\gamma (vdx+udy). \]

Just compose complex number into two real parts. Denote \(z_k=x_k+iy_k\), \(\xi_k=\eta_k+i\zeta_k\), \(\Delta x_k=x_k-x_{k-1}, \Delta y_k=y_k-y_{k-1}\) then \(f(\xi_k)=u(\eta_k,\zeta_k)+iv(\eta_k,\zeta_k)\), so Riemann summation

\[ \begin{align*} \sum_{k=1}^n f(\xi_k)(z_k-z_{k-1})=&\sum_{k-1}^n [u(\eta_k,\zeta_k)+iv(\eta_k,\zeta_k)]\cdot [(x_k-x_{k-1})+i(y_k-y_{k-1})]\\ =&\sum_{k=1}^n\{ [u(\eta_k,\zeta_k)\Delta x_k-v(\eta_k,\zeta_k)\Delta y_k]+\\ &i [u(\eta_k,\zeta_k)\Delta y_k+v(\eta_k,\zeta_k)\Delta x_k]\} \end{align*} \]

when \(u, v\) is continuous on \(\gamma\), the above summation converges to the needed formula.

The above formula is easy to know by

\[ f(z)dz=(u+iv)(dx+idy)=udx-vdy+i(udy+vdx). \]

Integral along a smooth curve

Assume smooth curve \(\gamma\subset \mathbb{C}\) parametrized by \(z: [a,b]\mapsto \mathbb{C}\), and \(f\) is a continuous function on \(\gamma\). Then the integral of \(f\) along \(\gamma\)

\[ \int_\gamma f(z)dz=\int_a^b f(z(t))z'(t)dt. \]

The logic is simple, even a little rude: Parametrize the four real integral along \(\gamma\) and then sum them up. That is, \(f=u+iv\), \(z=x+iy\), then \(z'(t)=x'(t)+iv'(t)\).

Since the real and imaginary parts of integral are

\[ \int_\gamma udx-vdy=\int_a^b [u(x(t),y(t))x'(t)-v(x(t),y(t))y'(t)]dt \]
\[ \int_\gamma vdx+udy=\int_a^b [v(x(t),y(t))x'(t)+u(x(t),y(t))y'(t)]dt \]

we have

\[ \int_\gamma f(z)dz=\int_a^b [u(x(t),y(t))+iv(x(t),y(t))][x'(t)+iy(t)]dt=\int_a^b f(z(t))z'(t)dt. \]

Example. Calculate

(i)

\[ \int_\gamma dz=z(b)-z(a)\neq \int_a^b |z'(t)|dt=\text{length}(\gamma) \]

(ii)

\[ \int_\gamma \frac{dz}{(z-z_0)^n}, \]

where \(n\in \mathbb{Z}\), \(\gamma=C_r(z_0)\).

(i) \(n=1\).

(ii) \(n\neq 1\).

Properties for Integral of complex function

If complex function \(f\) and \(g\) is continuous on \(\gamma\) which is length-calculatable, then

(i) \(\gamma^-\) is the reverse of \(\gamma\),

\[ \int_{\gamma^-}f(z)dz=-\int_{\gamma}f(z)dz. \]

(ii) For all \(\alpha, \beta\in \mathbb{C}\),

\[ \int_{\gamma} (\alpha f(z)+\beta g(z))dz=\alpha\int_{\gamma}f(z)dz+\beta \int_{\gamma}g(z)dz. \]

(iii) Inequation for absolute estimation

\[ \left|\int_\gamma f(z)dz\right|\leq \sup_{z\in \gamma}|f(z)|\cdot \text{length}(\gamma) \]

Cauchy's Theorem

Definition of Primitive function

Assume \(f\) is a function on the open set \(\Omega\). A Primitive for \(f\) on \(\Omega\) is a function \(F\) that is holomorphic on \(\Omega\), such that \(F'(z)=f(z),\forall z\in \Omega\).

Cauchy's Theorem

If a continuous function \(f\) has a primitive \(F\) on open set \(\Omega\), and \(\gamma\) is a piecewise-smooth curve begins at \(\omega_1\) and ends at \(\omega_2\), then

\[ \int_\gamma f(z)dz=F(\omega_2)-F(\omega_1). \]

(i) For \(\gamma\) is smooth.

Choose a parametrization \(z: [a,b]\mapsto \mathbb{C}\) of \(\gamma\), such that \(z(a)=\omega_1, z(b)=\omega_2\).

\[ \begin{align*} \int_\gamma f(z)dz&=\int_a^b f(z(t))z'(t)dt\\ &=\int_a^b F'(z(t))z'(t)dt\\ &=\int_a^b \frac{d}{dt}F(z(t))dt\\ &=F(z(b))-F(z(a))=F(\omega_2)-F(\omega_1). \end{align*} \]

where the third "\(=\)" is by chain rule, and the fouth "\(=\)" is by fundamental theorem of calculas.

(ii) For \(\gamma\) is piecewise-smooth, just add these smooth part up.

The condition and result of the above theorem could be replaced by the following statement.

\(F\) is holomorphic and \(F'\) is continuous on \(\Omega\), then

\[ \int_\gamma F'(z)dz=F(\omega_2)-F(\omega_1). \]

Corollary: Cauchy's closed curve theorem

If \(\gamma\) is a closed smooth curve in an open set \(\Omega\), and \(f\) is continuous and has a primitive in \(\Omega\), then

\[ \int_\gamma f(z)dz=0. \]

The condition and result of the above theorem could be replaced by the following statement.

\(F\) is holomorphic and \(F'\) is continuous on \(\Omega\), then

\[ \int_\gamma F'(z)dz=0. \]

Corollary 2: zero-derivatives means constant function

If \(f\) is holomorphic in a region \(\Omega\), and \(f'=0\), then \(f\) is a constant.

\(\forall \omega_0\in \Omega\), \(\forall \omega\),

\[ f(\omega)-f(\omega_0)=\int_\gamma f'(z)dz=0. \]

so \(f(\omega)=f(\omega_0)\).

Actually the above theorems also hold for general curves.

Goursat Theorem

Goursat's Theorem

Assume \(\Omega\) is an open set in \(\mathbb{C}\). If \(f\in H(\Omega)\), then for any a triangle \(T\subset \Omega\) whose interior is also contained in \(\Omega\),

\[ \int_T f(z)dz=0. \]

Bisect \(\mathcal{T}^0\), we have a sequence of triangles

\[ \mathcal{T}^0\supset \mathcal{T}^1\supset\cdots \]

use nested compact set theorem to get a point \(z_0\) in these subtriangles.

Use properties of holomorphic function,

\[ f(z)=f(z_0)+f'(z_0)(z-z_0)+\psi(z)(z-z_0) \]

where \(\lim\limits_{z\rightarrow z_0}\psi(z)=0\). The first and second items of the rightside equation equals zero after integrating over a closed curve, while last item is bounded by a small \(\varepsilon_n=\sup_{z\in T^{n}}|\psi(z)|\). That is,

\[ \left|\int_{T^n}fdz\right|=\left|\int_{T^n}\psi (z-z_0) dz\right|\leq \varepsilon_n d^{(n)}p^{(n)}. \]

where \(d^{(n)}, p^{(n)}\) denote the diameter and perimeter of the \(n\)th triangles. Each time after bisection, we have

\[ d^{(n+1)}\leq \frac{1}{2}d^{(n)},\quad p^{(n+1)}\leq \frac{1}{2}p^{(n)} \]

and the integral along \(T^{(n)}\), must hold for some \(j\in \{0,1,2,3\}\)

\[ \left|\int_{T^{(n)}} f(z)dz\right|\leq 4\left|\int_{T^{(n+1)}_j} f(z)dz\right| \]

So

\[ \begin{align*} \left|\int_{T^{(0)}} f(z)dz\right|&\leq 4^n \left|\int_{T^{(n)}_j} f(z)dz\right|\\ &\leq 4^n \varepsilon_n d^{(n)}p^{(n)}\\ &\leq 4^n \varepsilon_n 4^{-n} d^{(0)}p^{(0)}=\varepsilon_n\rightarrow 0(n\rightarrow \infty) \end{align*} \]

Corollary: integral in Rectangle curve

Assume \(f\in H(\Omega)\). For any rectangle \(R\subset \Omega\) whose interior is also contained in \(\Omega\),

\[ \int_T f(z)dz=0. \]

Local existence of Primitive

By constructing method.

A holomorphic function has a primitive

Assume \(f\) is a holomorphic function in an open disc, then \(f\) has a primitive in the disc.

Actually we use Goursat's Theorem and continuity of \(f\), without utilizing Holomorphism of \(f\). That is, a continuous function on an open set must have a primitive, if it satisfies that its integral on every triangle on \(\Omega\) equals zero. In the following parts, we will show that this equals to holomorphic.

Define

\[ F(z)=\int_{\gamma_z} f(\omega)d\omega, \]

where \(\gamma_z\) is a polygomal line from \(0\) to \(z\).

by Goursat's Theorem, we can show that

\[ \begin{align} F(z+h)-F(z)=\int_\eta f(\omega)d\omega,\label{primitive} \end{align} \]

where \(\eta\) is a straight line from \(z\) to \(z+h\).

Then use the continuity of \(f\) in \(\Omega\), i.e at point \(z\),

\[ f(\omega)=f(z)+\psi(\omega) \]

where \(\lim\limits_{\omega\rightarrow z}\psi(\omega)=0\). Substitute in equation \(\ref{primitive}\), we have

\[ \begin{align*} F(z+h)-F(z)&=f(z)\int_\eta d\omega + \int_\eta \psi(\omega)d\omega.\\ \frac{F(z+h)-F(z)}{h}&=f(z)+\frac{\int_\eta \psi(\omega)d\omega}{h} \end{align*} \]

the latter integral goes to \(0\) as \(h\) tend to \(0\) because

\[ \left|\frac{\int_\eta \psi(\omega)d\omega}{h} \right|\leq \frac{\sup\limits_{\omega\in \eta}|\psi(\omega)|\cdot |h|}{|h|} \]

So

\[ \lim_{h\rightarrow 0}\frac{F(z+h)-F(z)}{h}=f(z). \]

Corollary: Cauchy's Theorem for a disc

Assume \(D\) is an open disc, and \(f\in H(D)\). Then for closed curve \(\gamma\),

\[ \int_\gamma f(z)dz=0 \]

By using Corollary: Cauchy's closed curve theorem.

Actually, Cauchy's Theorem could apply to any closed curve with unambiguious interior, not just a disc, whose interior is clear.

Cauchy's Integral Formula

Using the following integral formula, we could show that holomorphic function has derivatives of arbitrary order \(n\), and could be expanded by power series.

Cauchy's integral formula

Assume \(f\) is holomorphic in an open set that contains the closure of a disc \(D\). If \(C\) denotes the boundary circle of this disc with the positive orientation, then for any point \(z\in D\),

\[ f(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta-z}d\zeta. \]

We could use a "key hole" to prove. Look the following image.

Let \(F(z)=\frac{f(\zeta)}{\zeta-z}\), then it is holomorphic away from point \(z\), so

\[ \int_{\gamma} F(\zeta)d\zeta=0=\int_{C_\varepsilon^+} F(\zeta)d\zeta + \int_{\Gamma^+}F(\zeta)d\zeta +\int_{\Gamma^-}F(\zeta)d\zeta +\int_{C_r^-}F(\zeta)d\zeta. \]

Let \(\delta\rightarrow 0\), we have \(\int_{\Gamma^+}F(\zeta)d\zeta +\int_{\Gamma^-}F(\zeta)d\zeta=0\). If we rewrite

\[ F(z)=\frac{f(\zeta)-f(z)}{\zeta-z}+\frac{f(z)}{\zeta-z}, \]

and by the example commen integral, we know

\[ \int_{C_\varepsilon^+}\frac{f(z)}{\zeta-z}d\zeta=2\pi i f(z), \]

and \(\frac{f(\zeta)-f(z)}{\zeta-z}\) is bounded, whose integral on \(C_\varepsilon^+\) converges to \(0\). Actually \(\frac{f(\zeta)-f(z)}{\zeta-z}\) has a removable singularity at \(z\), we could extend it into a holomorphic function

\[ g(\zeta)=\begin{cases} \frac{f(\zeta)-f(z)}{\zeta-z},\quad &\zeta\neq z\\ f'(z),\quad &\zeta=z \end{cases}. \]

so its integral vanishes. So

\[ \int_{C_\varepsilon^+}F(\zeta)d\zeta = 2\pi i f(z). \]

and we are done.

The following theorem is amazing, called regularity.

Corollary: Cauchy integral formula with Infinitely many derivatives

If \(f\) is holomorphic in an open set \(\Omega\), then \(f\) has infinitely many complex derivatives in \(\Omega\).

Moreover, if \(C\subset \Omega\) is a circle whose interier is also contained in \(\Omega\), then

\[ f^{(n)}(z)=\frac{n!}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta. \]

for all \(z\) in the interior of \(C\)

By induction on \(n\). The theorem holds for \(n=0\) by Cauchy's integral formula.

Assume it holds for \(n=k\), i.e.

\[ f^{(k)}(z)=\frac{k!}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta-z)^{k+1}}d\zeta. \]

Then we have its quotient

\[ \begin{align} \frac{f^{(k)}(z+h)-f^{(k)}(z)}{h}&=\frac{k!}{2\pi i h}\int_C f(\zeta) \left[\frac{1}{(\zeta-z-h)^{k+1}} -\frac{1}{(\zeta-z)^{k+1}}\right] d\zeta.\label{quotient} \end{align} \]

Notice that \(A^{k+1}-B^{k+1}=(A-B)\left(\sum_{i=0}^{k} A^i B^{k-i}\right)\), with

\[ A-B = \frac{1}{\zeta-z-h} -\frac{1}{\zeta-z}=\frac{h}{(\zeta-z-h)(\zeta-z)} \]

so if we choose \(h\) small enough, then equation \(\ref{quotient}\) becomes

\[ \begin{align*} \lim_{h\rightarrow 0}\frac{f^{(k)}(z+h)-f^{(k)}(z)}{h}&=\lim_{h\rightarrow 0}\frac{k!}{2\pi i h}\int_C f(\zeta) \frac{h}{(\zeta-z-h)(\zeta-z)} \left[\sum_{i=0}^{k} \left(\frac{1}{\zeta-z-h}\right)^i \left(\frac{1}{\zeta-z}\right)^{k-i}\right] d\zeta\\ &=\lim_{h\rightarrow 0}\frac{k!}{2\pi i}\int_C f(\zeta) \frac{1}{(\zeta-z-h)(\zeta-z)} \left[\sum_{i=0}^{k} \left(\frac{1}{\zeta-z-h}\right)^i \left(\frac{1}{\zeta-z}\right)^{k-i}\right] d\zeta\\ &=\frac{k!}{2\pi i}\int_C \lim_{h\rightarrow 0}f(\zeta) \frac{1}{(\zeta-z-h)(\zeta-z)} \left[\sum_{i=0}^{k} \left(\frac{1}{\zeta-z-h}\right)^i \left(\frac{1}{\zeta-z}\right)^{k-i}\right] d\zeta\\ &=\frac{k!}{2\pi i}\int_C \frac{1}{(\zeta-z)^2} \frac{k+1}{(\zeta-z)^{k}}d\zeta\\ &=\frac{(k+1)!}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta-z)^{k+2}}d\zeta. \end{align*} \]

The third "\(=\)" holds because when \(h\rightarrow 0\), the integrated function converges uniformly whenever \(\zeta\in C\).

Corollary: Cauchy Inequalities

If \(f\) is holomorphic in an open set that contains the closure of a disc \(D\) centered at \(z\) and of radius \(R\) (or \(D_R(z)\)), then

\[ |f^{(n)}(z)|\leq \frac{n!\|f\|_{\partial D}}{R^n}, \]

where \(\|f\|_{\partial D}=\sup\limits_{z\in \partial D}|f(z)|\) denotes the supremum of \(|f|\) on the boundary circle \(\partial D\).

Using Cauchy integral formula to estimate

\[ \begin{align*} |f^{(n)}(z)|&\leq \frac{n!}{2\pi}\left|\int_{\partial D} \frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta\right|\\ &=\frac{n!}{2\pi} \left| \int_{\partial D} \frac{f(\zeta)}{R^{n+1}}d\zeta\right|\\ &\leq \frac{n!}{2\pi} \frac{\sup\limits_{\zeta\in \partial D}|f(\zeta)|}{R^{n+1}} \left|\int_{\partial D} d\zeta \right|\\ &=\frac{n!\|f\|_{\partial D}}{2\pi R^{n+1}} \cdot 2\pi R\\ &=\frac{n!\|f\|_{\partial D}}{R^n}. \end{align*} \]

Power Series Expansion

Theorem of holomorphic function with power series expansion

Assume \(f\) is holomorphic in an open set \(\Omega\). If \(D_R(z_0)\) is a disc centered at \(z_0\), whose closure is contained in \(\Omega\), then \(f\) has a power series expansion at \(z_0\)

\[ f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n,\quad \forall z\in D_R(z_0) \]

with coefficients given by

\[ a_n=\frac{f^{(n)}(z_0)}{n!},\quad n\geq 0. \]

Fix \(z\in D\). From Cauchy's integral formula, we have

\[ f(z)=\frac{1}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{\zeta-z}d\zeta. \]

The following deduction is classic.

\[ \frac{1}{\zeta-z}=\frac{1}{(\zeta-z_0)-(z-z_0)}=\frac{1}{(\zeta-z_0)}\frac{1}{1-\frac{z-z_0}{\zeta-z_0}} \]

the last item could be expanded by geometric series expansion, which is uniformly convengent for \(\zeta\in \partial D\) because for fixed \(z\) and \(z_0\), \(\exists r\in (0,1)\), s.t. \(\frac{|z-z_0|}{|\zeta-z_0|}<r\). Therefore

\[ \frac{1}{1-\frac{z-z_0}{\zeta-z_0}}=\sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n \]

so

\[ \begin{align*} f(z)&=\frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{(\zeta-z_0)} \sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n d\zeta\\ &=\sum_{n=0}^\infty \frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{(\zeta-z_0)} \left(\frac{z-z_0}{\zeta-z_0}\right)^n d\zeta\quad \text{by series convergence}\\ &=\sum_{n=0}^\infty\left( \frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta \right) (z-z_0)^n, \end{align*} \]

and we are done.

This gives another perspective to see that holomorphic function has infinitely many derivatives in terms of power series.

The above proof also tells us the power series converges in where \(f\) is holomorphic. If \(f\) is holomorphic on \(\mathbb{C}\), then the power series converges on whole \(\mathbb{C}\).

Liouville's Theorem

If \(f\) is entire and bounded, then \(f\) is constant.

For arbitrary \(z\in \mathbb{C}\) and arbitrary \(R>0\) (D_R(z)), using Cauchy Inequalities

\[ |f'(z)|\leq \frac{B}{R}\rightarrow 0,\quad R\rightarrow\infty. \]

So \(f'=0\), then by zero-derivatives means constant function, we are done.

Analytic Continuation

The genetic code of a holomorphic function is determined if we know its values on appropriate arbitrarily small subsets.

Theorem for zero accumulation

Assume \(f\) is a holomorphic function in a region \(\Omega\), and \(f\) vanishes on a sequence of distinct points with a limit point in \(\Omega\), then \(f\) is identically zero.

Suppose \(z_0\in \Omega\) is a limit point for sequence \(w_n\) which satisfies \(f(w_n)=0\).

  • We first show that \(f(z)=0\) for \(z\in B_r(z_0)\), where \(r\) is a small enough number.

Easy to have choose \(r\) such that \(B_r(z_0)\subset \Omega\), and we could use power series expansion of \(f\)

\[ f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n,\quad \forall z\in D_r(z_0). \]

If \(f(z)\) does not vanish in \(D_r(z_0)\), then there must exists a smallest integer \(m\) such that \(a_m\neq 0\). So expansion becomes

\[ f(z)=\sum_{n=m}^\infty a_n (z-z_0)^n=a_m(z-z_0)^m(1+g(z-z_0)). \]

where \(g(z-z_0)\rightarrow 0 (z\rightarrow z_0)\). Since \(z_0\) is a limit point, then no matter how small \(r\) is, \(\exists w_K\in B_r(z_0)\), such that \(w_K\neq z_0\), \(f(w_k)=0\), but \(w_K-z_0\neq 0\) and \(1+g(w_K-z_0)\neq 0\), contradicting!

Actually, \(f(z)=(z-z_0)^m h(z)\), here \(h(z_0)=a_m\neq 0\). By continuity of \(f\) in \(D_r(z_0)\), we have \(f(z)\neq 0\) for \(z\in D_\delta(z_0)\). But we could always find a point of \(w_K\in D_\delta(z_0)\) such that \(f(w_K)=0\) which contradicts!

  • Extend to the whole region \(\Omega\) using connectedness. One method is by iterating the above process, since after one round, we have more points which satisfies \(f(z)=0\). We continue this extension until the boundary of the region. Another method is by the following trick.

We define \(U\) to be the interior of sets of points \(z\) which satisfies \(f(z)=0\). It is open by definition. It is also closed, which is not easy to see, since choose \(z_n\rightarrow z\), and \(f(z_n)=0\), then \(f(z)=0\) by continuity. So \(f\) vanishes in the neighborhood of \(z\) as we have discussed above. So \(z\in U\). Define \(V=\Omega-U\) is open, i.e. we find two disjoint open set to compose \(\Omega\). However since \(\Omega\) is connected, it contradicts!

Corollary

Assume \(f\) and \(g\) are holomorphic in a region \(\Omega\) and \(f(z)=g(z)\) for all \(z\) in some non-empty open subset of \(\Omega\), then \(f(z)=g(z)\) throughout \(\Omega\).

Definition of Analytic Continuation

Given a pair of functions \(f\) and \(f\), which are holomorphic in region \(\Omega\) and \(\Omega'\), respectively. If \(f=F\) on \(\Omega\), then we say \(F\) is a analytic continuation of \(f\) into the region \(\Omega'\).

Applications

Morera's Theorem

Morera's Theorem

Suppose \(f\) is a continuous function on a open disc \(D\) such that for any triangle \(T\) contained in \(D\),

\[ \int_T f(z)dz=0, \]

then \(f\) is holomorphic on \(D\).

Just use the proof process in Local existence of Primitive.

Sequence of holomorphic functions

Sequence of holomorphic functions

Assume \(\{f_n\}\) is a sequence of holomorphic functions that converges uniformly to a function \(f\) in every compact subset of \(\Omega\), then \(f\) is holomorphic in \(\Omega\).

Moreover, sequence \(\{f'_n\}\) converges uniformly to \(f'\) on every compact subset of \(\Omega\).

Notice "compact" guarantee \(f\) is also in \(\Omega\).

Use Goursat's Theorem, we have for any triangle \(T\) in a closed disc \(D\) contained in \(\Omega\)

\[ \int_T f_n(z)dz=0. \]

By mathematical analysis, since \(f_n\rightrightarrows f\), \(f\) is continuous, we could change the order of limit and integral

\[ \int_T f(z)dz =\lim_{n\rightarrow \infty}\int_T f_n(z)dz=0. \]

By Morera's Theorem, \(f\) is holomorphic.

For additional parts, we have to show that

\[ \sup_{z\in \Omega_\delta}|F'(z)|\leq \frac{1}{\delta}\sup_{\xi\in\Omega}|F(\xi)| \]

where \(\Omega_\delta=\{z\in \Omega: \overline{D}_\delta(z)\in \Omega\}\).

We usually construct a holomorphic function using

\[ F(z)=\sum_{i=1}^\infty f_i(n). \]

If \(f_i\) is holomorphic, and the above series converges uniformly to \(F\), then \(F\) is holomorphic.

In terms of Integrals

Holomorphic function in terms of integrals

Assume \(F(z,s)\) defined on \(\Omega\times [0,1]\), satisfies

(i) \(F(z,s)\) is holomorphic with respect to \(z\) for each fixed \(s\);

(ii) \(F(z,s)\) is continuous in \(\Omega\times [0,1]\);

then a function \(f\) defined by

\[ f=\int_0^1 F(z,s)ds \]

is holomorphic on \(\Omega\).

We could use Sequence of holomorphic functions.

Define Riemann summation

\[ f_n(z):=\frac{1}{n}\sum_{i=1}^n F(z,\frac{i}{n}) \]

which is holomorphic on \(\Omega\). We claim \(f\) is a uniform convergence of \(f_n\). We could use continuous function in a compact set is uniformly continuous, i.e. \(\forall \varepsilon>0\), \(\exists \delta>0\), \(\forall |s_1-s_2|\leq \delta\), \(\forall\) disc \(D\subset \Omega\),

\[ |F(z, s_1)-F(z,s_2)|<\varepsilon,\quad \forall z\in D. \]

Actually

\[ \begin{align*} \left|f_n-f\right|&=\left|\sum_{i=1}^{n}\int_\frac{i-1}{n}^\frac{i}{n} \left[F\left(z,\frac{i}{n}\right)-F(z,s)\right]ds\right|\\ &\leq \sum_{i=1}^{n}\int_\frac{i-1}{n}^\frac{i}{n} \left|F\left(z,\frac{i}{n}\right)-F(z,s)\right|ds\\ &\leq \sum_{i=1}^{n} \frac{\varepsilon}{n}=\varepsilon. \end{align*} \]

Homotopies and simply connected domains

Definition of Homotopies

Assume two continuous curves \(\gamma_0\), \(\gamma_1\) defined on an open set \(\Omega\), are given by a parametrization \(\gamma_0(t)\) and \(\gamma_1(t)\), \(t\in [a,b]\), with common end-points

\[ \gamma_0(a)=\gamma_1(a)=\alpha,\quad \gamma_0(b)=\gamma_1(b)=\beta \]

then the above curve are said to be homotopic if \(\forall s\in [0,1]\), there exsits another curve \(\gamma_s\subset \Omega\), parametrized with \(\gamma_s(t)\), \(t\in [a,b]\), such that

\[ \gamma_s(a)=\alpha, \quad \gamma_s(b)=\beta, \]

and \(\gamma_s(t)|_{s=0}=\gamma_0(t)\), and \(\gamma_s(t)|_{s=1}=\gamma_1(t)\). Moreover, \(\gamma_s(t)\) is continuous jointly on \(s\in [0,1]\) and \(t\in [a,b]\).

Loosely speaking, two curves are homotopic if one curve could be continuously deformed into the other without leaving \(\Omega\).

Theorem for homotopic curves

Assume two curves \(\gamma_0\) and \(\gamma_1\) with same end-points are homotopic in an open set \(\Omega\), \(f\) is holomorphic on \(\Omega\), then

\[ \int_{\gamma_0}f(z)dz=\int_{\gamma_1}f(z)dz. \]

Definition of Simply connected region

A region (connected) \(\Omega\) is said to be simply connected, if for all two curves \(\gamma_0,\gamma_1\subset \Omega\) with same end-points, they are homotopic.

Theorem for primitives on simply connected region

Every function on a simply connected region has a primitive.

just the same logic as our proof in Local existence of Primitive, where we use rectangular to formulate a primitive. Here we just use a continuous curve.

The punctured plane is not simply connected, since on which \(f=\frac{1}{z}\) is holomorphic, however \(\int_{C_\varepsilon}\frac{1}{z}dz=2\pi i\neq 0\). This means \(f\) does not have a primitive on \(D_\delta(0)-\{0\}\), which implies it is not simply connected.

The complex Logarithm

Simply connected region guarantees the uniqueness of complex logarithm,

\[ \log z=\log r+ i\theta. \]

with \(z=re^{i\theta}\).

The following theorem makes use of the inspiration that \(\log z\) is a primitive of \(\frac{1}{z}\) on a simply connected region. So we could construct a logarithm function.

Theorem for complex logarithm

Assume \(\Omega\) is a simply connected region with \(1\in \Omega\) and \(0\not\in \Omega\). Then there exsits a holomorphic function \(F(z)=\log (z)\), such that

(i) \(F\) is holomorphic on \(\Omega\);

(ii) \(e^{F(z)}=z\) for all \(z\in \Omega\),

(iii) \(F(r)=\log r, \forall r\) whenever \(r\) is a real number and near \(1\).

(i) Assume \(\gamma\subset \Omega\) starts from \(1\) and ends at \(z\), define

\[ F(z):=\int_\gamma \frac{1}{w}dw. \]

arguing as Theorem for primitives on simply connected region, since \(\frac{1}{z}\) is holomorphic on \(\Omega\), we have \(F\) is a primitive of \(\frac{1}{z}\) and

\[ F'=\frac{1}{z}. \]

(ii) Equivalently, we have to prove

\[ e^{-F(z)}z=1. \]

Actually, \(\frac{d}{dz}e^{-F(z)}z=e^{-F(z)}(1-F'(z)z)=0\), so by zero-derivatives means constant function, we have \(e^{-F(z)}z=C\). Given the initial value \(F(1)=0\), we have \(c=1\) and we prove (ii).

(iii) Still by definition \(F(r)=\int_1^r\frac{1}{w}dw=\log r\).

Choose \(\Omega=\mathbb{C}-(-\infty, 0]\), then we have the principle branch of logarithm

\[ \log z=\log r+i\theta \]

with \(|\theta|<\pi\). We define it by a specific curve, which starts from \(1\), firstly goes straight to \(r\) and then rotate about \(\theta\) positively.

\[ \log z=\int_1^r\frac{1}{w}dw+\int_0^\theta \frac{1}{re^{it}}rie^{it}dt=\log r+o\theta. \]

If \(g=\log f\), then \(g'=\frac{f'}{f}\). So we could construct a primitive using integral of \(\frac{f'}{f}\) to get \(g\) from a arbitrary function which is holomorphic on a simply connected region.

Theorem for generating logarithm of a function

Assume \(\Omega\) is a simply connected region, \(f\) is holomorphic on \(\Omega\) and does not vanish, then there exists a holomorphic function such that \(e^{g(z)}=f(z)\).

Still the same logic of constructing a primitive of \(\frac{f'}{f}\).

Define

\[ g(z)=\int_\gamma \frac{f'(w)}{f(w)}dw+c_0 \]

where \(\gamma\subset \Omega\) starts from \(z_0\in \Omega\) and ends at \(z\), and \(c_0\) is used for compensation such that \(e^g(z_0)=f(z_0)=e^{c_0}\). Arguing as Theorem for primitives on simply connected region, we have \(g(z)\) is holomorphic and

\[ g'(z)=\frac{f'(z)}{f(z)}. \]

To prove \(e^{g(z)}=f(z)\) we only need to prove \(f(z)e^{-g(z)}=1\). Actually

\[ \frac{d}{dz}\left[f(z)e^{-g(z)}\right]=e^{-g(z)}(f'(z)-g'(z)f(z))=0 \]

so \(f(z)e^{-g(z)}=c\). Notice that \(g(z_0)=c_0\), so we have \(c=f(z_0)e^{-c_0}=1\). So \(e^{g(z)}=f(z)\).

Example. Assume \(f(z)=1+z\), use

\[ \frac{f'}{f}=\frac{1}{1+z}=\sum_{n=0}^\infty (-1)^n z^n. \]

near \(z=0\) to show that

\[ \log(1+z)=\sum_{n=0}^\infty \frac{(-1)^n z^{n+1}}{n+1}. \]

We know from the above theorem there exists a function \(g\) such that \(e^{g(z)}=1+z\). We have to prove that \(\log(1+z)\) is primitive of \(\frac{1}{1+z}\). Just take a derivative on both sides of the desired equaiton and show that they equal by condition. So they actually differ by a constant. Then use initial value at \(z=0\) to show that the constant is \(0\).