Lecture 10: using Cauchy’s formula

Last time, we studied complex path integrals. Most directly, we saw how to compute them by parametrizing the path. More abstractly, we observed that if $f(z)$ is analytic on a domain $D$, then $f(z)dz$ is closed; so if $D$ is star-shaped, then $f(z)dz$ is exact, and so we can find a primitive $F$ for $f$, i.e. an analytic function $F$ on $D$ such that $F^{\prime }=f$, and then we can evaluate path integrals on $D$ using the fundamental theorem of calculus.

The higher-level version of this is Cauchy’s theorem: if $D$ is a bounded domain with boundary $\partial D$ piecewise smooth and $f$ is analytic on $D$, extending smoothly to the boundary $\partial D$, then

$${\int }_{\partial D}f(z)𝑑z=0.$$

As a corollary, we have Cauchy’s theorem: for any $z\in D$,

$$f(z)=\frac{1}{2\pi i}{\int }_D\frac{f(w)}{w-z}𝑑w.$$

Today, we want to give some first applications of Cauchy’s theorem and formula. The most straightforward application is the computation of integrals: if $f$ is analytic on a region $D$, then Cauchy’s theorem tells us that its integral over the boundary of $D$ is zero. Similarly, Cauchy’s formula tells us that the integral over a domain $D$ of anything of the form $\frac{f(z)}{z-z_0}$ is just $2\pi if(z)$.

For example, consider the disk , so $\partial D$ is the circle $\gamma$ of radius $3$. Then

$${\int }_{\gamma }\frac{z^2}{z+2}𝑑z=2\pi i\cdot {(-2)}^2=8\pi i.$$

An interesting simpler example is, for $\gamma$ any simple closed curve bounding a domain containing the origin,

$${\int }_{\gamma }\frac{1}{z}𝑑z=2\pi i.$$

When $\gamma$ is a circle, this matches our calculation last time.

In fact, we can push this method a little further: recall we had the more general formula

$$f^{(n)}(z)=\frac{n!}{2\pi i}{\int }_{\partial D}\frac{f(w)}{{(w-z)}^{n+1}}𝑑w,$$

so we can evaluate anything of this form as well. For example, for $\gamma$ the circle of radius $3$ as above,

$${\int }_{\gamma }\frac{e^z}{{(z+2)}^3}𝑑z$$

comes from the $n=2$ case, with $f(z)=e^z$. Since $f^{(2)}(z)=e^z$, this gives

$${\int }_{\gamma }\frac{e^z}{{(z+2)}^3}𝑑z=\frac{2\pi i}{2!}{e}^{-2}=\frac{\pi i}{e^2}.$$

This would otherwise be a very difficult integral to compute!

For a slightly harder version, consider an integral like

$${\int }_{\gamma }\frac{e^z}{z^2(z+2)}𝑑z.$$

This is not of the form we can evaluate using Cauchy’s formula, even the more general version: we would need to use either $f(z)=\frac{e^z}{z^2}$ or $g(z)=\frac{e^z}{z+2}$, neither of which is analytic on the whole disk.

However, we can use the following strategy. Let $D_0$ be a disk of some very small radius $r$ centered at $0$, and $D_{-2}$ another disk of radius $r$ centered at $-2$ (with $r$ small enough that they don’t intersect). On $D^{\prime }=D\setminus D_0\setminus D_{-2}$, the function $h(z)=\frac{e^z}{z^2(z+2)}$ is analytic, so

$${\int }_{\partial D^{\prime }}h(z)𝑑z=0.$$

Now, we can write $\partial D^{\prime }$ (keeping in mind orientations) as $\gamma$, in the positive direction, together with two circles of radii $r$ centered at $0$ and $-2$ in the negative direction, which we call ${\gamma }_0$ and ${\gamma }_{-2}$ respectively. Therefore

$$0={\int }_{\partial D^{\prime }}h(z)={\int }_{\gamma }h(z)𝑑z-{\int }_{{\gamma }_0}h(z)𝑑z-{\int }_{{\gamma }_{-2}}h(z)𝑑z,$$

i.e.

$${\int }_{\gamma }h(z)𝑑z={\int }_{{\gamma }_0}h(z)𝑑z+{\int }_{{\gamma }_{-2}}h(z)𝑑z.$$

For the first term, note that on the disk $D_0$ of sufficiently small radius $r$ near $0$, $g(z)=\frac{e^z}{z+2}$ is analytic, and so we can use Cauchy’s formula (for derivatives):

$${\int }_{{\gamma }_0}\frac{e^z}{z^2(z+2)}𝑑z=\frac{2\pi i}{1!}{g}^{\prime }(0)=\frac{\pi i}{2}$$

(evaluating the derivative is left as an exercise for the reader). For the second term, similarly $f(z)=\frac{e^z}{z^2}$ is analytic on $D_{-2}$ for $r$ sufficiently small, so

$${\int }_{{\gamma }_{-2}}\frac{e^z}{z^2(z+2)}𝑑z=2\pi if(-2)=\frac{\pi i}{2e^2}.$$

Therefore the total integral is

$${\int }_{\gamma }\frac{e^z}{z^2(z+2)}𝑑z=\frac{\pi i}{2}(1+e^{-2}).$$

We can also use Cauchy’s formula to bound the derivatives of $f$. For a fixed point $z_0$, choose $r$ small enough that $f$ is defined on the closed disk $\{z:|z|\le r\}$. By Cauchy’s formula (for derivatives), we have

$f^{(n)}(z_0)$ $={\displaystyle \frac{n!}{2\pi i}}{\displaystyle {\int }_{|z|=r}}{\displaystyle \frac{f(z)}{{(z-z_0)}^{n+1}}}𝑑z$

$={\displaystyle \frac{n!}{2\pi i}}{\displaystyle {\int }_0^{2\pi }}{\displaystyle \frac{f(z_0+r{e}^{i\theta })}{r^{n+1}{e}^{i\theta (n+1)}}}ri{e}^{i\theta }𝑑\theta$

$={\displaystyle \frac{n!}{2\pi r^n}}{\displaystyle {\int }_0^{2\pi }}f(z_0+r{e}^{i\theta })e^{-in\theta }𝑑z.$

If $|f(z)|\le M$ for $|z-z_0|=r$, the $ML$ bound gives

$$|f^{(n)}(z_0)|\le \frac{n!}{2\pi r^n}\cdot 2\pi M=\frac{n!}{r^n}M.$$

This is the Cauchy estimate for higher derivatives. That is:

Proposition (Cauchy estimate).

Let $D$ be a disk of radius $r$ centered at a point $z_0\in ℂ$, and let $f:D\to ℂ$ be an analytic function extending continuously to $\overline{D}$. If $|f(z)|\le M$ for all $z\in \partial D$, then

$$|f^{(n)}(z_0)|\le \frac{n!}{r^n}M.$$

As a corollary, we deduce the following extremely nice result. We say a function is entire if it is analytic on the whole complex plane. Recall also that a function $f$ is bounded if there exists a real number $M$ such that for any $z$ in the domain of $f$, $|f(z)|\le M$.

Theorem (Liouville’s theorem).

Any bounded entire function is constant.

This is very surprising: in the real case, we have plenty of very nicely behaved non-constant analytic functions which are bounded, e.g. $\sin (x)$ and $\cos (x)$. We have seen that these functions have nice extensions to the complex plane, and are analytic everywhere, hence entire; so Liouville’s theorem tells us that, somewhat counter-intuitively, $\sin z$ and $\cos z$ are not bounded in the complex setting (since they are entire and not constant). In particular the bounds $|\sin (z)|\le 1$, $|\cos (z)|\le 1$ are false in general for $z$ complex.

Given the Cauchy estimate, the proof is now straightforward. If $f$ is bounded and entire, we fix some $M$ such that $|f(z)|\le M$ for all $z\in ℂ$. The Cauchy estimate tells us that $|f^{(n)}(z)|\le \frac{n!}{r^n}M$ for any $r>|z|$. Since we can take $r$ arbitrarily large (since $f$ is entire), we can take the right-hand side arbitrarily close to zero, so in fact we must have $|f^{(n)}(z)|=0$ for all $n$ and all $z$. In fact the case $n=1$ is enough: this shows that $f^{\prime }(z)=0$ for all $z$, hence $f$ is constant.

Liouville’s theorem gives yet another proof of the fundamental theorem of algebra: suppose that $P(z)=a_n{z}^n+\mathrm{\cdots }+a_{1}z+a_0$ is a complex polynomial with no roots in $ℂ$. Then $\frac{1}{P(z)}$ is an entire function. If $n\ge 1$, then $|P(z)|\to \mathrm{\infty }$ as $|z|\to \mathrm{\infty }$, so $\left|\frac{1}{P(z)}\right|\to 0$ as $|z|\to \mathrm{\infty }$; therefore $\frac{1}{P}$ is bounded. (In fact, if $n=0$ then $P$ is constant so again bounded.) Therefore by Liouville’s theorem $\frac{1}{P}$ must be constant, i.e. $P$ is constant, so the only polynomials with no complex zeros are the constant ones, i.e. the fundamental theorem of algebra holds.