I am reading various reviews of Ablowitz and Fokas' Complex Variables: Introduction and Applications. In one review, the reviewer wrote that the authors' proof (which will be replicated below) of Cauchy-Goursat theorem is simply wrong.
However, to my knowledge, the more or less identical proofs have been used in many books, such as Brown and Churchill's Complex Variables and Applications, 8/e (2004, pp.152-156), Moore and Hadlock's Complex Analysis (1991, p.68), James Kelly's Graduate Mathematical Physics (2006, pp. 29-31) as well as Krishna's Complex Analysis (2010, pp.293-295) written by Vasishtha et al.. Here are my questions:
- Is the proof really wrong?
- If this proof is correct, is it a good one? It seems that this proof is fairly shorter than many others (such as Priestley's or Bak and Newman's).
Below are the statement and proof of the theorem that appear on pp.105-108 of the second edition of Ablowitz and Fokas' text. (A more detailed proof in the same spirit with very similar wordings can be found in Brown and Churchill's textbook.)
Theorem 2.7.1 (Cauchy-Goursat) If a function $f(z)$ is analytic at all points interior to and on a simple closed contour, then $$\oint_C f(z)dz=0.\tag{2.7.1}$$
Proof Consider a finite region $R$ consisting of points on and within a simple closed contour $C$. We form a square mesh over the region $R$ by drawing lines parallel to the $x$ and $y$ axes such that we have a finite number of square subregions in which each point of $R$ lies in at least one subregion. If a particular square contains points not in $R$, we delete these points. Such partial squares will occur at the boundary (see Figure 2.7.1).
We can refine this mesh by dividing each square in half again and again and redefine partial squares as above. We do this until the length of the diagonal of each square is sufficiently small.
We note that the integral around the contour $C$ can be replaced by a sum of integrals around the boundary of each square or partial square $$\oint_C f(z)dz = \sum_{j=1}^n \oint_{C_j} f(z)dz\tag{2.7.2}$$ where it is noted that all interior contours will mutually cancel because each inner side of a square is covered twice in opposite directions.
Introduce the following equality: $$f(z)=f(z_j)+(z-z_j)f'(z_j)+(z-z_j)\tilde{f}_j(z)\tag{2.7.3a}$$ where $$\tilde{f}(z)=\left(\frac{f(z)-f(z_j)}{z-z_j}\right)-f'(z_j)\tag{2.7.3b}$$ We remark that $$\oint_{C_j}dz=0, \quad \oint_{C_j}(z-z_j)dz=0\tag{2.7.4}$$ which can be established either by direct integration or from the known anti-derivatives: $$1=\frac{d}{dz}z, \quad (z-z_j)=\frac{d^2}{dz^2}\frac{(z-z_j)^2}{2}, \quad\ldots$$ then using the results of Theorem 2.4.1...
(Remark by question asker: theorem 2.4.1 here refers to the contour integral analogue of the fundamental theorem of calculus, i.e. $\int_C f(z)dz = F(z_2)-F(z_1)$ when $F$ is an analytic function with a continuous derivative $f$ in a domain $D$ and $C$ is a contour lying inside $D$ with endpoints $z_1$ and $z_2$.)
... Then, it follows that \begin{align} \left|\oint_C f(z)dz\right| &\leq \sum_{j=1}^n \left|\oint_{C_j}f(z)dz\right|\\ &= \sum_{j=1}^n \left|\oint_{C_j}(z-z_j)\tilde{f}_j(z)dz\right|\\ &\leq \sum_{j=1}^n \oint_{C_j}|z-z_j|\left|\tilde{f}_j(z)\right|dz\tag{2.7.5} \end{align} It can be established (remark by question asker: see Brown and Churchill for details) the mesh can be refined sufficiently such that $$\left|\tilde{f}_j(z)\right|=\left|\frac{f(z)-f(z_j)}{z-z_j}-f'(z_j)\right|<\epsilon\tag{2.7.6}$$ Calling the area of each square $A_j$, we observe the geometric fact that $$|z-z_j|\leq\sqrt{2A_j}\tag{2.7.7}$$ Thus, using Theorem 2.4.2 (remark by question asker: i.e. the M-L formula) for all interior squares, we have $$\oint_{C_j}|z-z_j|\left|\tilde{f}_j(z)\right|dz \leq (\sqrt{2A_j})\epsilon(4\sqrt{A_j})=4\sqrt{2}\epsilon A_j\tag{2.7.8}$$ and for all boundary squares, the following upper bound holds: $$\oint_{C_j}|z-z_j|\left|\tilde{f}_j(z)\right|dz \leq (\sqrt{2A_j})\epsilon(4\sqrt{A_j}+L_j)\tag{2.7.9}$$ where $L_j$ is the length of the portion of the contour in the partial square $C_j$. Then $\oint_C f(z)dz$ is obtained by adding over all such contributions Eqs. $(2.7.8)$ and $(2.7.9)$. Calling $A=\sum A_j,\ L=\sum L_j$, quantity $A$ being the area of the square mesh bounded by the contour $C$ and $L$ the length of the contour $C$, we have $$\oint_C f(z)dz \leq \left(4\sqrt{2}A+\sqrt{2AL}\right)\epsilon\tag{2.7.10}$$ We can refine our mesh indefinitely so as to be able to choose $\epsilon$ as small as we wish. Hence the integral $\oint_C f(z)dz$ must be zero.$\ {}_\blacksquare$