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\[\lim_{\epsilon\to 0}\frac{1}{2\pi i}\int_{\gamma_{\epsilon}}\frac{f(z)}{z-z_{0}}\, dz=f(z_{0}).\]
10. Let \(f(z)\) be a holomorphic function defined for \(|z|\leq 1\) and \(u(x,y)=\mbox{Re}f(z)\), where \(z=x+iy\). Show that
\[\int_{\gamma}\frac{\partial u}{\partial y}dx-\frac{\partial u}{\partial x}dy=0,\]
where \(\gamma\) is the unit circle.
11. Show that
\[\int_{0}^{\infty}\cos x^{2}\,dx=\int_{0}^{\infty}\sin x^{2}\,dx=\sqrt{\pi/8}.\]
Hint: Integrate \(e^{iz^{2}}\) round the contour \(\gamma\) shown in Figure 3.38. Use Cauchy’s theorem and the inequality \(\sin 2\theta\geq\frac{4\theta}{\pi}\) for \(\theta\in\left[0,\frac{\pi}{4}\right]\).
### 3.8 Cauchy’s Formulae
**Cauchy’s Integral Formula**
**Theorem 135**.: _Let \(f\) be holomorphic inside and on a positively oriented contour \(\gamma\). Then if \(a\) is inside \(\gamma\),_
\[f(a)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-a}\ dw.\]
The above formula for \(f(a)\) is remarkable in the sense that it states the values of \(f\) on \(\gamma\) completely determine the values of \(f\) inside \(\gamma\). In other words, the value of \(f\) is determined by its “boundary values.”
Proof
: First note that since \(f(a)\) is constant, we have
\[\int_{\gamma}\frac{f(a)}{w-a}\ dw=f(a)\int_{\gamma}\frac{1}{w-a}\ dw=2\pi if(a)\]
from the fundamental integral. We also know by the Jordan Curve Theorem that there exists \(r\) such that \(D(a;r)\) lies inside \(\gamma\). For any \(\epsilon \[\int_{\gamma}\frac{f(w)}{w-a}\ dw=\int_{\gamma(a;\epsilon)}\frac{f(w)}{w-a}\ dw,\] by the deformation theorem. Now consider the difference \[\left|\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-a}\ dw-f(a)\right|=\left| \frac{1}{2\pi i}\int_{\gamma(a;\epsilon)}\frac{f(w)-f(a)}{w-a}\ dw\right|\] and use the estimation lemma to obtain \[\left|\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-a}\ dw-f(a)\right|\leq\frac{ 1}{2\pi}\,2\pi\sup_{t\in[0,2\pi]}|f(a+\epsilon e^{it})-f(a)|.\] Figure 3.38: Contour for Exercise 11Now all we have to do is to remember that \(f\) is continuous at \(a\), so the supremum above tends to zero as \(\epsilon\to 0\). Thus the left-hand side must be zero. **Example 108**.: ### Liouville’s Theorem **Theorem 136** (Liouville).: _Let \(f\) be holomorphic and bounded in the complex plane \(\mathbb{C}\). Then \(f\) is constant._ : Suppose \(|f(w)|\leq M\) for all \(w\in\mathbb{C}\). Fix \(a,b\) in \(\mathbb{C}\), and take \(R\) such that \[|w-a|\geq\frac{1}{2}R\quad\text{and}\quad|w-b|\geq\frac{1}{2}R\quad\text{ whenever}\quad|w|=R.\] Cauchy’s integral formula applied to \(f(a)\) and \(f(b)\) with \(\gamma(0,R)\) and then applying the estimation lemma yields \[|f(a)-f(b)|\leq\frac{1}{2\pi}\,(2\pi R)M\frac{|a-b|}{(1/2R)^{2}}.\] Thus \(f(a)=f(b)\) when \(R\) is sufficiently large. **Remark 75**.: In real analysis there are many infinitely differentiable functions that are bounded, i.e., \(|f(x)|\leq M\) for all \(x\in\mathbb{R}\). For example, \(f(x)=\sin x\) is infinitely differentiable and \(|\sin x|\leq 1\) for all \(x\), but it is **not** a constant function. In complex analysis, Liouville’s theorem asserts that the only bounded entire functions (entire means holomorphic in the whole complex plane \(\mathbb{C}\)) are constant functions. **Example 109**.: Let \(M\geq 0\) and \(f\) be an entire function such that \(\operatorname{Re}f(z)\leq M\) for all \(z\in\mathbb{C}\). Show that \(f\) is a constant function. Consider the function \[F(z)=e^{f(z)}\quad\text{ for all }\quad z\in\mathbb{C}.\] Then, \(F(z)\) is an entire function and \(|F(z)|=|e^{f(z)}|=e^{\operatorname{Re}f(z)}\) and thus \[|F(z)|\leq e^{M}.\] By Liouville’s theorem \(F(z)\) is constant and therefore \(f\) must be a constant function. ### Fundamental Theorem of Algebra **Theorem 137**.: _Let \(p(z)\) be a non-constant polynomial with complex coefficients. Then, there exists a complex number \(\xi\) such that \(p(\xi)=0\). Consequently a complex polynomial of degree \(n>1\) has \(n\) roots (not necessarily distinct) in \(\mathbb{C}\)._ : The proof is done by contradiction. Suppose we have a non-constant polynomial \(p(z)\) and \(p(z)\neq 0\) for every \(z\in\mathbb{C}\). Then, \(\dfrac{1}{p(z)}\) is holomorphic. Next, we claim that \(\dfrac{1}{p(z)}\) is bounded on \(\mathbb{C}\). Since \(|p(z)|\to\infty\) as \(|z|\to\infty\), there exists \(R\) such that \[\left|\dfrac{1}{p(z)}\right|<1\quad\text{for}\quad|z|>R.\] On the other hand, on the compact set \(|z|\leq R\), the function \(\dfrac{1}{p(z)}\) is continuous and hence bounded (it is bounded because it is a continuous functions on a compact set). By Liouville’s theorem \(\dfrac{1}{p(z)}\) must be constant, a contradiction. The last statement in the theorem is proved by induction on \(n\) (the degree of the polynomial). **Example 110**.: Show that every polynomial \(p(z)=a_{n}z^{n}+\cdots+a_{0}\) of degree \(n\geq 1\) has precisely \(n\) roots in \(\mathbb{C}\). If these roots are denoted by \(w_{1},w_{2},\ldots,w_{n}\) then \(p\) can be factored as \[p(z)=a_{n}(z-w_{1})(z-w_{2})\cdots(z-w_{n}).\]By the fundamental theorem of algebra \(p\) has a root, say \(w_{1}\). Then writing \(z=(z-w_{1})+w_{1}\) and inserting this for \(z\) in \(p\), using the binomial formula we get \[p(z)=b_{n}(z-w_{1})^{n}+\dots+b_{1}(z-w_{1})+b_{0},\] where \(b_{0},b_{1},\dots,b_{n}\) are new coefficients and \(b_{n}=a_{n}\). Since \(p(w_{1})=0\) we know \(b_{0}=0\), therefore we can factor out \((z-w_{1})\) and write \[p(z)=(z-w_{1})q(z),\] where \(q(z)\) is a polynomial of degree \(n-1\). By induction on the degree of the polynomial we conclude that \[p(z)=c(z-w_{1})(z-w_{2})\dots(z-w_{n})\] for some \(c\in\mathbb{C}\). Expanding the right-hand side shows that the coefficient of \(z^{n}\) is \(c\) and therefore \(c=a_{n}\) as desired. ### Morera’s Theorem
* Suppose \(\gamma\) is the unit circle \(|z|=1\), then using Cauchy’s integral formula: \[\int_{\gamma}\frac{e^{z}+\sin z}{z}\ dz=2\pi if(0)=2\pi i.\] Here the denominator of the integrand is \(z=z-0\) and \(f(w)=e^{w}+\sin w\), which is holomorphic inside and on \(\gamma\). The value \(0\) is in \(\gamma\), so we can apply Cauchy’s Integral Theorem with \(f(0)=1\) to obtain the claim.
* Evaluating \[\int_{\gamma}\frac{z^{2}}{z^{2}+1}\ dz,\] where \(\gamma=\gamma(i,1)\), we first observe \(z^{2}+1=(z+i)(z-i)=0\). For \(z=i\) and \(z=-i\), \(\frac{z^{2}}{z^{2}+1}\) is not holomorphic, but only \(z=i\) is inside \(\gamma\). By rewriting \[\int_{\gamma}\frac{z^{2}}{z^{2}+1}\ dz=\int_{\gamma}\frac{z^{2}}{z-i}\ dx=2\pi if(i)=-\pi.\] Because \(f(w)=\frac{w^{2}}{w+i}\), which is holomorphic inside and on \(\gamma\), by substituting \(f(i)=\frac{i^{2}}{i+i}=-\frac{1}{2i}\) into Cauchy’s integral formula we prove the claim.
Proof
Proof
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