Cauchy integral formula

Section 1.15 Cauchy integral formula

Lemma 1.15.1.

Assume \(f\) is continuous on an open set \(U\) containing \(z_0\text{.}\) For each \(r> 0\) let \(\gamma_r=z_0+re^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) We have

\begin{equation} \lim\limits_{r\to 0}\int_{\gamma_r}\frac{f(z)}{z-z_0}\, dz=2\pi i f(z_0)\text{.}\tag{1.60} \end{equation}

Proof.

First observe that for all \(r\) we have

\begin{align*} \int_{\gamma_r} \frac{f(z)}{z-z_0}\, dz \amp = \int_{\gamma_r} \frac{f(z)-f(z_0)+f(z_0)}{z-z_0}\, dz\\ \amp = \int_{\gamma_r} \frac{f(z)-f(z_0)}{z-z_0}\, dz +\int_{\gamma_r} \frac{f(z_0)}{z-z_0}\, dz\\ \amp =\int_{\gamma_r} \frac{f(z)-f(z_0)}{z-z_0}\, dz+2\pi i f(z_0)\text{,} \end{align*}

and thus

\begin{equation*} \abs{\int_{\gamma_r} \frac{f(z)}{z-z_0}\, dz-2\pi i f(z_0)}=\abs{\int_{\gamma_r} \frac{f(z)-f(z_0)}{z-z_0}\, dz}\text{.} \end{equation*}

Let \(\epsilon> 0\text{.}\) We will show that there is a \(\delta> 0\) such that \(r< \delta\) implies

\begin{equation*} \abs{\int_{\gamma_r} \frac{f(z)}{z-z_0}\, dz-2\pi i f(z_0)}=\abs{\int_{\gamma_r} \frac{f(z)-f(z_0)}{z-z_0}\, dz}\leq \epsilon\text{,} \end{equation*}

and thus that the limit statement holds. Using continuity at \(z_0\text{,}\) for any \(\epsilon > 0\text{,}\) we can find a \(\delta> 0\) such that \(\abs{z-z_0}< \delta\) implies \(\abs{f(z)-f(z_0)}< \epsilon/(2\pi)\text{.}\) It follows that for any \(0< r< \delta\) we have

\begin{align*} \abs{\int_{\gamma_r} \frac{f(z)}{z-z_0}\, dz-2\pi i f(z_0)}\amp=\abs{\int_{\gamma_r} \frac{f(z)-f(z_0)}{z-z_0}\, dz}\amp \\ \amp \leq \frac{\epsilon}{2\pi r}\cdot 2\pi r \amp (ML-\text{ineq.})\\ \amp =\epsilon \text{,} \end{align*}

as desired.

Theorem 1.15.2. Cauchy-integral formula.

Assume \(f\) is holomorphic on an open set \(U\) containing the \(z_0\text{,}\) suppose \(\overline{B}_r(z_0)\subseteq U\) for some \(r> 0\text{,}\) and let \(\gamma=z_0+re^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) For any \(w\in B_{r}(z_0)\) we have

\begin{equation} f(w)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-w}\, dz\tag{1.61} \end{equation}

Proof.

Given any \(w\in B_{r}(z_0)\text{,}\) for any \(s> 0\) with \(B_s(z_0)\subseteq B_r(z_0)\text{,}\) the principle of deformation theorem implies that

\begin{equation*} \int_\gamma\frac{f(z)}{z-w}\, dz=\int_{\gamma_s}\frac{f(z)}{z-w}\, dz\text{,} \end{equation*}

where \(\gamma_s=w+se^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) (Note that \(f(z)/(z-w)\) is holomorphic at all points lying within \(\gamma\) and outside of \(\gamma_s\text{.}\)) Since this is true for all (sufficiently small) \(s\text{,}\) it follows from Lemma 1.15.1 that

\begin{align*} f(w) \amp = \lim\limits_{s\to 0}\int_{\gamma_s}\frac{f(z)}{z-w}\, dz\\ \amp = \lim\limits_{s\to 0}\int_{\gamma}\frac{f(z)}{z-w}\, dz \\ \amp = \int_{\gamma}\frac{f(z)}{z-w}\, dz\text{,} \end{align*}

as claimed.

Example 1.15.3. Cauchy Integral formula.

Compute \(\int_\gamma \frac{e^{z^2-1}}{z^2+6z+5}\, dz\text{,}\) where \(\gamma\) is the triangular closed path \(\gamma_{(z_0,z_1,z_2)}\text{,}\) where \(z_0=2-i\text{,}\) to \(z_1=-1+1\text{,}\) to \(z_2=-3-i\)

Solution.

We have

\begin{align*} \int_\gamma \frac{e^{z^2-1}}{z^2+6z+5}\, dz \amp = \int_\gamma \frac{e^{z^2-1}}{(z+1)(z+5)}\, dz \\ \amp = \int_\gamma \frac{g(z)}{z+1}\, dz\text{,} \end{align*}

where \(g(z)=e^{z^2-1}/(z+5)\) is holomorphic away from \(-5\text{.}\) In particular, \(g\) is holomorphic in the open ball \(U=B_5(0)\text{.}\) Consider the path \(\alpha=-1+4e^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) (See Figure 1.15.4.) Since \(\gamma\) lies within \(\alpha\text{,}\) and since \(g(z)/(z+1)\) is holomorphic at all points within \(\alpha\) and outside of \(\gamma\text{,}\) we have

\begin{equation*} \int_\alpha \frac{g(z)}{z+1}\, dz=\int_\gamma \frac{g(z)}{z+1}\, dz\text{.} \end{equation*}

Now using the Cauchy integral formula, applied to \(U=B_5(0)\) and \(\alpha\text{,}\) we have

\begin{align*} \int_\gamma \frac{e^{z^2-1}}{z^2+6z+5}\, dz \amp = \int_\alpha \frac{g(z)}{z+1}\, dz \\ \amp = 2\pi i g(-1)\\ \amp = 2\pi i \cdot \frac{e^0}{4}\\ \amp = \frac{\pi i}{2}\text{.} \end{align*}

Figure 1.15.4. Cauchy integral formula on triangular path

Example 1.15.5. Cauchy integral formula: cosine.

Compute \(\int_0^{2\pi}\frac{1}{2-\cos t}\, dt\text{.}\)

Solution.

Writing \(\cos t=(e^{it}+e^{-it})/2=(e^{it}+1/e^{it})/2\text{.}\) We see that the integral in question is almost the result of integrating \(f(z)=1/(2-(z+1/z)/2)\) over the path \(\gamma(t)=e^{it}\text{,}\) \(t\in [0,1]\text{;}\) we are only missing the \(\gamma'(t)=ie^{it}\) term in the numerator. Let’s explore further, using a little algebra:

\begin{align*} \int_0^{2\pi}\frac{1}{2-\cos t}\, dt \amp= \int_0^{2\pi}\frac{1}{2-(e^{it}+e^{-it})/2}\, dt\\ \amp = \int_0^{2\pi}\frac{1}{2-(e^{it}+e^{-it})/2}\cdot \frac{ie^{it}}{ie^{it}}\, dt \\ \amp = \int_\gamma\frac{1}{2-(z+1/z)/2}\cdot \frac{1}{iz}\, dz\\ \amp = \int_{\gamma}\frac{2}{i(4z-z^2-1)}\, dz\\ \amp = \int_{\gamma}\frac{2i}{z^2-4z+1}\, dz\\ \amp = \int_{\gamma}\frac{2i}{(z-\alpha)(z-\beta)}\, dz\text{,} \end{align*}

where \(\gamma(t)=e^{it}\text{,}\) \(t\in [0, 2\pi]\text{,}\) \(\alpha=2-\sqrt{3}\text{,}\) \(\beta=2+\sqrt{3}\text{.}\) Since \(\alpha\) lies within \(\gamma\text{,}\) and \(\beta\) lies without \(\gamma\text{,}\) can apply the Cauchy integral formula to the function \(f(z)=2i/(z-\beta)\) (which is holomorphic on \(\C-\{\beta\}\)) to conclude that

\begin{align*} \int_0^{2\pi}\frac{1}{2-\cos t}\, dt \amp= \int_{\gamma}\frac{f(z)}{(z-\alpha)}\, dz \\ \amp = 2\pi i f(\alpha)\\ \amp = 2\pi i\cdot \frac{2i}{\alpha-\beta}\\ \amp = \frac{-4\pi}{-2\sqrt{3}}\\ \amp =\frac{2\pi}{\sqrt{3}}\text{.} \end{align*}

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