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Morera’s theorem gives a sufficient condition for \(f\) being holomorphic and is often taken as the converse of Cauchy’s theorem. There are several versions of this theorem. For proofs of the versions below we refer the reader to [41], p. 156.
**Theorem 138**.: _Version 1: Suppose \(f\) is continuous on an open set \(G\) and that \(\int_{\gamma}f(z)\,dz=0\) for all_ **triangles**_\(\gamma\) in \(G\). Then \(f\) is holomorphic in \(G\). Version 2: Suppose \(f\) is continuous in a simply connected region \(G\) and that \(\int_{\gamma}f(z)\,dz=0\) for all_ **closed contour**_\(\gamma\). Then \(f\) is holomorphic in \(G\)._
### Cauchy’s Formula for Derivatives and its Consequences
**Theorem 139**.: _Let \(f\) be holomorphic inside and on a positively oriented contour \(\gamma\) and let \(a\) lie inside \(\gamma\). Then \(f^{n}(a)\) exists for \(n=1,2,\dots\) and_
\[f^{n}(a)=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-a)^{n+1}}\ dw.\]
The proof of this theorem is in most complex analysis books, for example consult [41], p. 157. We observe that if we can interchange \(\int\) and the \(\frac{d}{dz}\), one can obtain an expression for \(f^{\prime}(z)\) by taking the derivative under the integral sign in Cauchy’s integral formula \(f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}\,dw\);
\[f^{\prime}(z)=\frac{d}{dz}f(z)=\frac{d}{dz}\left(\frac{1}{2\pi i}\int_{\gamma }\frac{f(w)}{w-z}\right)=\frac{1}{2\pi i}\int_{\gamma}\frac{d}{dz}\left( \frac{f(w)}{w-z}\right)\,dw\]and therefore
\[f^{\prime}(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^{2}}dw.\]
We may repeat this procedure by differentiating \(\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^{2}}dw\) with respect to \(w\), and hope to obtain \( f^{\prime\prime}(z)=\frac{2}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^{3}}dw\) where \(z\in I(\gamma)\). There is a difference between this second differentiation and the first one. Since \(f\) is holomorphic \(f^{\prime}(z)\) exists, but we do not know that \( f^{\prime\prime}(z)\) exists. However we can use Cauchy’s integral formula to examine the quotient \(\frac{f^{\prime}(z+h)-f^{\prime}(z)}{h}\) and then show that as \(h\to 0\) this limit exists and it is given by the right-hand side of \(f^{\prime}(z)\) above. Then the process can be repeated to show successively, the existence of \(f^{n}(z)\) for \(n=3,4,\dots\) all by differentiating under the integral sign to obtain
\[f^{n}(a)=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-a)^{n+1}}\ dw.\]
**Example 111**.: Suppose we want to evaluate:
\[\int_{\gamma}\frac{z+1}{z^{4}+2iz^{3}}\ dz\]
where \(\gamma:|z|=1\).
Since \(z^{4}+2iz^{3}=z^{3}(z+2i)\), \(\frac{z+1}{z^{4}+2iz^{3}}\) is not holomorphic at \(z=0\) and \(z=-2i\). However only \(z=0\) lies inside the contour \(\gamma\). By rewriting:
\[\frac{z+1}{z^{4}+2iz^{3}}=\frac{\frac{z+1}{z+2i}}{z^{3}}\]
we can identify \(a=0\), \(n=2\) and \( f(z)=\frac{z+1}{z+2i}\). The quotient rule for derivatives gives \( f^{\prime\prime}(z)=\frac{2-4i}{(z+2i)^{3}}\) and so \( f^{\prime\prime}(0)=\frac{2i-1}{4i}\). Hence the given integral
\[\int_{\gamma}\frac{z+1}{z^{4}+2iz^{3}}\ dz=\frac{2\pi i}{2!}f^{\prime\prime}( 0)=-\frac{\pi}{4}+\frac{\pi}{2}i.\]
### Cauchy’s Inequality
**Theorem 140**.: _Let \(f\) be holomorphic in a simply connected region \(G\) and \(\gamma\) be the simple closed curve \(\gamma(a,R)\) which lies entirely in \(G\). If \(|f(z)|\leq M\) for all points \(z\) on \(\gamma\), then_
Proof
: From Cauchy’s integral formula for derivatives, we have
\[f^{(n)}(a)=\frac{n!}{2\pi i}\,\int_{\gamma}\frac{f(z)}{(z-a)^{n+1}}\,dz.\]
Applying the estimation lemma, for \(\gamma\), the circle with radius \(R\), centered at \(a\), we obtain
\[\big{|}f^{(n)}(a)\big{|} =\frac{n!}{2\pi}\,\bigg{|}\int_{\gamma}\frac{f(z)}{(z-a)^{n+1}}\, dz\bigg{|}\] \[\leq\frac{n!}{2\pi}\,\int_{\gamma}\frac{\big{|}f(z)\big{|}}{\big{|} z-a\big{|}^{n+1}}\,\big{|}dz\big{|}\] \[\leq\frac{n!}{2\pi}\,\frac{M}{R^{n+1}}\,2\pi R=\frac{n!}{R^{n}}\,M.\qed\]
### Taylor’s Theorem
We have seen in Section 3.3, Theorem 125 that if \(f(z)=\sum_{n=0}^{\infty}c_{n}z^{n}\) and if \(\sum_{n=0}^{\infty}c_{n}z^{n}\) has a radius of convergence \(R\neq 0\), then
* \(f\in H(D(0;R))\). This means that convergent complex power series define a holomorphic function. How about the converse of this theorem? The answer to this question is called Taylor’s theorem. By using Cauchy’s integral formula one can prove that any function holomorphic in a disc has a power series expansion. **Theorem 141** (Taylor’s theorem).: _Let \(f\) be holomorphic in \(D(a;R)\) with \(R>0\). Then there exists unique constants \(c_{n}\) such that_ \[f(z)=\sum_{n=0}^{\infty}c_{n}(z-a)^{n}\quad z\in D(a;R),\] _where the constant \(c_{n}\) is given by the formula_ \[c_{n}=\frac{f^{(n)}(a)}{n!}=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-a)^{n +1}}\,dw\] _and \(\gamma\) is a circle \(\gamma(a;r)\) for \(0 : Fix a point \(z\in D(a;R)\) and choose \(r\) such that \(|z-a| \[f(z)=\frac{1}{2\pi i}\,\,\int_{\gamma}\frac{f(w)}{w-z}dw.\] Since \(|z-a|<|w-a|\) for all \(w\) on \(\gamma\) as shown in Figure 3.39,\[\frac{1}{w-z}=\frac{1}{(w-a)\left(1-\frac{z-a}{w-a}\right)}=\sum_{n=0}^{\infty} \frac{(z-a)^{n}}{(w-a)^{n+1}}.\] Therefore, Cauchy’s integral formula above takes the form \[f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}\,\,\,dw=\frac{1}{2\pi i}\int _{\gamma}\sum_{n=0}^{\infty}\frac{(z-a)^{n}}{(w-a)^{n+1}}\,f(w)\,dw.\] Now the question is can we interchange summation and integration? The answer to this question is yes if one has uniform convergence. We appeal to the Weierstrass M-test to see that we can do this. Then \[f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-a)^{ n+1}}\,dw\right)(z-a)^{n}.\] Now we appeal to Cauchy’s integral formula for derivatives to finish the proof. It is left to the reader to prove the uniqueness of \(c_{n}\). There are a wealth of interesting results whose proofs rest on Cauchy’s theorem and the Cauchy integral formulas. In the following we mention one more without proof. Recall that if \(f\) is continuous on a closed and bounded region \(R\), then \(f\) is bounded; that is there is some constant \(M\) such that \(|f(z)|\leq M\) for all \(z\in R\). If the boundary of \(R\) is a simple closed curve \(\gamma\), then the next theorem tells us that \(|f(z)|\) assumes its maximum value at some point on the boundary of \(\gamma\). ### Maximum Modulus Theorem **Theorem 142** ([5], p. 134).: _Suppose \(f\) is holomorphic and non-constant on a closed simple curve \(\gamma\). Then the modulus \(|f(z)|\) attains its maximum on \(\gamma\) (on the boundary)._ Figure 3.39: Proof of Taylor’s theorem **Remark 76**.: **Example 112**.: Find the maximum of \(|f(z)|=|2z+5i|\) on \(|z|\leq 2\). Since \(|z|^{2}=z\overline{z}\), we have \[|f(z)|^{2} =|2z+5i|^{2}=(2z+5i)\overline{(2z+5i)}=(2z+5i)(2\overline{z}-5i)\] \[=4z\overline{z}-10i(z-\overline{z})+25\] Since \(z-\overline{z}=2i\operatorname{Im}(z)\), the above expression takes the form \[|f(z)|^{2}=|2z+5i|^{2}=4|z|^{2}+20\text{Im}(z)+25.\]
* \(f^{\prime}(z)=\sum_{n=1}^{\infty}nc_{n}z^{n-1}\) when \(|z|
Proof
* If we assume \(f(z)\neq 0\) in the closed region bounded by \(\gamma\), then by the hypotheses of the maximum modulus theorem \(|f(z)|\) also attains its minimum on \(\gamma\).
* In real analysis there are many infinitely differentiable functions that are bounded on a closed interval \([a,b]\) but do not attain their maximum at the boundary of \([a,b]\). For example \(\sin x\) is infinitely differentiable and bounded on \([-\pi,\pi]\) and \(|\sin(\pi)|=|\sin(-\pi)|=0\) but \(|\sin x|\leq 1\) for all \(x\). The situation is different in complex analysis, where maximum modulus is attained on the boundary.
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