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# Lecture 25: November 14, 2023 ## Proof of Cauchy's Integral Formula ### Theorem Suppose $f$ is analytic on a simply connnected domain $D$. Let $\gamma$ be a simple closed curve in $D$ oriented counterclockwise. For any $z_0 \in D$ not on $\gamma$, $$ f(z_0) = \frac{1}{2\pi i} \int_\gamma \frac...

# Lecture 25: November 14, 2023 ## Proof of Cauchy's Integral Formula ### Theorem Suppose $f$ is analytic on a simply connnected domain $D$. Let $\gamma$ be a simple closed curve in $D$ oriented counterclockwise. For any $z_0 \in D$ not on $\gamma$, $$ f(z_0) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-z_0}dz $$ **Proof:** Since $z_0$ is not on $\gamma$, $f(z)/(z-z_0)$ is analytic on $D - {z_0}$. Let $C_\epsilon$ be a circle of radius $\epsilon$ centered at $z_0$ contained in $D$ oriented counterclockwise. $\frac{f(z)}{z-z_0}$ is analytic on the region between $\gamma$ and $C_\epsilon$. By the deformation invariance theorem: $$ \int_\gamma \frac{f(z)}{z - z_0} dz = \int_{C_\epsilon} \frac{f(z)}{z - z_0} dz $$ Parametrize $C_\epsilon$ by $z(t) = z_0 + \epsilon e^{it}$ for $0 \le t \le 2\pi$, $z'(t) = i\epsilon e^{it}$. $$ \int_{C_\epsilon} \frac{f(z)}{z - z_0} dz = \int_0^{2\pi} \frac{f(z_0 + \epsilon e^{it})}{\epsilon e^{it}} i\epsilon e^{it} dt = i \int_0^{2\pi} f(z_0 + \epsilon e^{it}) dt $$ Now, $$ i \int_0^{2\pi} f(z_0 + \epsilon e^{it}) dt - 2\pi i f(z_0) = i \int_0^{2\pi} f(z_0 + \epsilon e^{it}) - f(z_0) dt $$ $$ \left| i \int_0^{2\pi} f(z_0 + \epsilon e^{it}) - f(z_0) dt \right| \le \int_0^{2\pi} |f(z_0 + \epsilon e^{it}) - f(z_0)| dt $$ Since $f$ is continuous at $z_0$, for any $\varepsilon > 0$, there exists $\delta > 0$ such that if $|z - z_0| < \delta$, then $|f(z) - f(z_0)| < \varepsilon$. Choose $\epsilon < \delta$. Then $|f(z_0 + \epsilon e^{it}) - f(z_0)| < \varepsilon$. Then, $$ \int_0^{2\pi} |f(z_0 + \epsilon e^{it}) - f(z_0)| dt \le \int_0^{2\pi} \varepsilon \ dt = 2\pi \varepsilon $$ Because we can make this integral arbitrarily small, it must be zero. $$ i \int_0^{2\pi} f(z_0 + \epsilon e^{it}) dt - 2\pi i f(z_0) = 0 $$ $$ \int_{C_\epsilon} \frac{f(z)}{z - z_0} dz = 2\pi i f(z_0) $$ $$ \int_\gamma \frac{f(z)}{z - z_0} dz = 2\pi i f(z_0) $$ Thus, $$ f(z_0) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-z_0}dz $$ Q.E.D. ## Cauchy's Integral Formula for Derivatives ### Theorem Suppose $f$ is analytic on a simply connected domain $D$, let $\gamma$ be a simple closed curve in $D$ oriented counterclockwise. For any $z_0 \in D$ not on $\gamma$ and any $n \ge 0$. $$ f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)}{(z-z_0)^{n+1}} dz $$ **Proof:** We'll prove this by induction. For $n = 0$, this is Cauchy's Integral Formula. So assume the formula holds for $n = k$. $$ f^{(k)}(z_0) = \frac{k!}{2\pi i} \int_\gamma \frac{f(z)}{(z-z_0)^{k+1}} dz $$ We want to show that $$ f^{(k+1)}(z_0) = \frac{(k+1)!}{2\pi i} \int_\gamma \frac{f(z)}{(z-z_0)^{k+2}} dz $$ By definition $$ f^{(k+1)}(z_0) = \lim_{h \to 0} \frac{f^{(k)}(z_0 + h) - f^{(k)}(z_0)}{h} $$ Using the inductive hypothesis $$ f^{(k+1)}(z_0) = \lim_{h \to 0} \frac{\frac{k!}{2\pi i} \int_\gamma \frac{f(z)}{(z - (z_0 + h))^{k+1}} dz - \frac{k!}{2\pi i} \int_\gamma \frac{f(z)}{(z-z_0)^{k+1}} dz}{h} $$ $$ = \frac{k!}{2\pi i} \lim_{h \to 0} \int_\gamma \frac{1}{h} \left[ \frac{f(z)}{(z - (z_0 + h))^{k+1}} - \frac{f(z)}{(z-z_0)^{k+1}} \right] dz $$ Consider $$ \frac{1}{h} \left[ \frac{1}{(z - (z_0 + h))^{k+1}} - \frac{1}{(z-z_0)^{k+1}} \right] = \frac{1}{h} \left[ \frac{(z-z_0)^{k+1} - (z - (z_0 + h))^{k+1}}{(z - (z_0 + h))^{k+1} (z-z_0)^{k+1}}\right] $$ Using the binomial theorem $$ (z-(z_0 + h))^{k+1} = (z-z_0 - h)^{k+1} = \sum_{j=0}^{k+1} \binom{k+1}{j} (z-z_0)^{k+1-j} (-h)^j $$ $$ = (z-z_0)^{k+1} + (k+1)(z-z_0)^k (-h) + \binom{k+1}{2} (z-z_0)^{k-1} (-h)^2 + \dots + (-h)^{k+1} $$ $$ (z-z_0)^{k+1} - (z - (z_0 + h))^{k+1} = (k+1)(z-z_0)^k h - \binom{k+1}{2} (z-z_0)^{k-1} h^2 + \dots - (-h)^{k+1} $$ $$ \frac{(z-z_0)^{k+1} - (z - (z_0 + h))^{k+1}}{h} = (k+1)(z-z_0)^k - \binom{k+1}{2} (z-z_0)^{k-1} h + \dots - (-h)^{k} $$ Taking the limit as $h \to 0$ $$ \lim_{h \to 0} \frac{(z-z_0)^{k+1} - (z - (z_0 + h))^{k+1}}{h} = (k+1)(z-z_0)^k $$ So, $$ \lim_{h \to 0} \frac{1}{h} \left[ \frac{1}{(z - (z_0 + h))^{k+1}} - \frac{1}{(z-z_0)^{k+1}} \right] = \frac{(k+1)(z-z_0)^k}{(z-z_0)^{k+2}(z-z_0)^{k+1}} = \frac{(k+1)}{(z-z_0)^{k+2}} $$ Therefore, $$ f^{(k+1)}(z_0) = \frac{k!}{2\pi i} \int_\gamma f(z) \frac{(k+1)}{(z-z_0)^{k+2}} dz = \frac{(k+1)!}{2\pi i} \int_\gamma \frac{f(z)}{(z-z_0)^{k+2}} dz $$ Q.E.D.