182 lines
4.3 KiB
Markdown
182 lines
4.3 KiB
Markdown
# Math416 Lecture 13
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## Review on Cauchy's Theorem
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Cauchy's Theorem states that if a function is holomorphic (complex differentiable) on a simply connected domain, then the integral of that function over any closed contour within that domain is zero.
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Last lecture we proved the case for convex regions.
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### Cauchy's Formula for a Circle
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Let $C$ be a counterclockwise oriented circle, and let $f$ be a holomorphic
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function defined in an open set containing $C$ and its interior. Then,
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$$
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f(z_0)=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-z_0}dz
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$$
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for all points $z$ in the interior of $C$.
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## New materials
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### Mean value property
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#### Theorem 7.6: Mean value property
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Special case: Suppose $f$ is holomorphic on some $\mathbb{D}(z_0,R)\subset \mathbb{C}$, by cauchy's formula,
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$$
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f(z_0)=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{z-z_0}dz
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$$
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Parameterizing $C_r$, we get $\gamma(t)=z_0+re^{it}$, $0\leq t\leq 2\pi$
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$$
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\int f(z)dz=\int f(\gamma) \gamma'(t) d t
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$$
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So,
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$$
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f(z_0)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(z_0+re^{it})}{re^{it}} ire^{it} dt=\frac{1}{2\pi}\int_{0}^{2\pi} f(z_0+re^{it}) dt
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$$
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This concludes the mean value property for the holomorphic function
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If $f$ is holomorphic, $f(z_0)$ is the mean value of $f$ on any circle centered at $z_0$
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#### Area representation of mean value property
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Area of $f$ on $\mathbb{D}(z_0,r)$
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$$
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\frac{1}{\pi r^2}\int_{0}^{2\pi}\int_0^r f(z_0+re^{it})
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$$
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/*Track lost*/
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### Cauchy Integral
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#### Definition 7.7
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Let $\gamma:[a,b]\to \mathbb{C}$ be piecewise $\mathbb{C}^1$, let $\phi$ be condition on $\gamma$. Then the Cauchy interval of $\phi$ along $\gamma$ is
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$$
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F(z)=\int_{\gamma}\frac{\phi(\zeta)}{\zeta-z}d \zeta
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$$
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#### Theorem
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Suppose $F(z)=\int_{\gamma}\frac{\phi(z)}{\zeta-z}d z$. Then $F$ has a local power series representation at all points $z_0$ not in $\gamma$.
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Proof:
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Let $R=B(z_0,\gamma)>0$, let $z\in \mathbb{D}(z_0,R)$
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So
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$$
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\frac{1}{\zeta-z}=\frac{1}{(\zeta-z_0)-(z-z_0)}=\frac{1}{1-z_0}\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}
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$$
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Since $|z-z_0|<R$ and $|\zeta-z_0|>R$, $|\frac{z-z_0}{\zeta-z_0}|<1$.
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Converting it to geometric series
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$$
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\frac{1}{1-z_0}\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}=\sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n
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$$
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So,
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$$
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\begin{aligned}
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F(z)&=\int_\gamma \frac{\phi(\zeta)}{\zeta - z} d\zeta\\
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&=\int_\gamma \frac{\phi(\zeta)}{z-z_0} \sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n dz\\
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&=\sum_{n=0}^\infty (z-z_0)^n \int_\gamma \frac{\phi(\zeta)}{(\zeta-z_0)^{n+1}}
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\end{aligned}
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$$
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Which gives us an power series representation if we set $a_n=\int_\gamma \frac{\phi(\zeta)}{(\zeta-z_0)^{n+1}}$
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QED
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#### Corollary 7.7
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Suppose $F(z)=\int_\gamma \frac{\phi(\zeta)}{\zeta-z_0} dz$,
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Then,
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$$
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f^{(n)}(z)=n!\int_\gamma \frac{\phi(z)}{(\zeta-z_0)^{n+1}} d\zeta
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$$
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where $z\in \mathbb{C}\setminus \gamma$.
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Combine with Cauchy integral formula:
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If $f$ is in $O(\Omega)$, then $\forall z\in \mathbb{D}(z_0,r)$.
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$$
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f(z)=\frac{1}{2\pi i}\int_{C_r}\frac{f(\zeta)}{\zeta-z} d\zeta
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$$
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We have proved that If $f\in O(\Omega)$, then $f$ is locally given by a convergent power series
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power series has radius of convergence at $z_0$ that is $\geq$ dist($z_0$,boundary $\Omega$)
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### Liouville's Theorem
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#### Definition 7.11
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A function that is holomorphic in all of $\mathbb{C}$ is called an entire function.
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#### Theorem 7.11 Liouville's Theorem
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Any bounded entire function is constant.
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> Basic Estimate of integral
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>
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> $$\left|\int_\gamma f(z) dz\right|\leq L(\gamma) \max\left|f(z)\right|$$
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Since,
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$$
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f'(z)=\frac{1}{2\pi i} \int_{C_r} \frac{f(z)}{(\zeta-z)^2} dz
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$$
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So the modulus of the integral is bounded by
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$$
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\frac{1}{2\pi} |M|\cdot \frac{1}{R^2}=2\pi R\cdot M \frac{1}{R^2}=\frac{M}{R}
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$$
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### Fundamental Theorem of Algebra
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#### Theorem 7.12 Fundamental Theorem of ALgebra
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Every nonconstant polynomial with complex coefficients can be factored over
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$\mathbb{C}$ into linear factors.
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#### Corollary
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For every polynomial with complex coefficients.
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$$
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p(z)=c\prod_{j=i}^n(z-z_0)^{t_j}
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$$
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where the degree of polynomial is $\sum_{j=0}^n t_j$
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Proof:
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Let $p(z)=a_0+a_1z+\cdots+a_nz^n$, where $a_n\neq 0$ and $n\geq 1$.
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So
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$$
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|p(z)|=|a_nz_n|\left[\left|1+\frac{a_{n-1}}{a_nz}+\cdots+\frac{a_0}{a_nz^n}\right|\right]
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$$
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If $|z|\geq R$, $\left|1+\frac{a_{n-1}}{a_nz}+\cdots+\frac{a_0}{a_nz^n}\right|<\frac{1}{2}$ |