[1]Mémoire sur les intégrales definies prises entre des limites imaginaires (1825). The proof here given is that due to Goursat, Trans. American Math. Soc. i. (1900), p. 14. ↩

5.2 Cauchy’s theorem[1] on the integral of a function round a contour.

A simple closed curve \(C\) in the plane of the variable \(z\) is often called a contour; if \(A\), \(B\), \(D\) be points taken in order in the counter-clockwise sense along the arc of the contour, and if \(f(z)\) be a one-valued continuous[2] function of \(z\) (not necessarily analytic) at all points on the arc, then the integral \[\int_{ABDA}\!\!\!f(z)\,dz\quad \text{or} \quad \int_{(C)} \!f(z)\,dz\] taken round the contour, starting from the point \(A\) and returning to \(A\) again, is called the integral of \(f(z)\) taken along the contour. Clearly the value of the integral taken along the contour is unaltered if some point in the contour other than \(A\) is taken as the starting-point.

[2]It is sufficient for \(f(z)\) to be continuous when variations of \(z\) along the arc only are considered. ↩

We shall now prove a result due to Cauchy, which may be stated as follows. If \(f(z)\) is a function of \(z\), analytic at all points on[3] and inside a contour \(C\), then \[\int_{(C)} \!f(z)\,dz = 0.\]

[3]It is not necessary that \(f(z)\) should be analytic on \(C\) (it is sufficient that it be continuous on and inside \(C\)), but if \(f(z)\) is not analytic on \(C\), the theorem is much harder to prove. This proof merely assumes that \(f'(z)\) exists at all points on and inside \(C\). Earlier proofs made more extended assumptions; thus Cauchy’s proof assumed the continuity of \(f'(z)\). Riemann’s proof made an equivalent assumption. Goursat’s first proof assumed that \(f(z)\) was uniformly differentiable throughout \(C\).

Editor’s Note: For those like me who find the first sentence of the above note ambiguously worded, I am sure it is intended to claim that it is sufficient for \(f(z)\) to be analytic inside \(C\) and continuous on and inside \(C\). ↩

For divide up the interior of \(C\) by lines parallel to the real and imaginary axes in the manner of §5.13; then the interior of \(C\) is divided into a number of regions whose boundaries are squares \(C_1\), \(C_2, \dots \), \(C_M\) and other regions whose boundaries \(D_1\), \(D_2, \dots \), \(D_N\) are portions of sides of squares and parts of \(C\); consider \[\sum_{n=1}^M \int_{(C_n)} \! f(z)\,dz + \sum_{n=1}^N \int_{(D_n)} \! f(z)\,dz,\] each of the paths of integration being taken counter-clockwise; in the complete sum each side of each square appears twice as a path of integration, and the integrals along it are taken in opposite directions and consequently cancel;[4] the only parts of the sum which survive are the integrals of \(f(z)\) taken along a number of arcs which together make up \(C\), each arc being taken in the same sense as in \(\displaystyle \int_{(C)} \! f(z)\,dz\); these integrals therefore just make up \(\displaystyle\int_{(C)}\! f(z)\,dz\).

[4]See §4.6, example.  ↩

Now consider \(\displaystyle\int_{(C_n)} \! f(z)\,dz\). With the notation of §5.12 (and \(z_n\) from §5.13), \[\begin{align*} \int_{(C_n)}\!f(z) \,dz &= \int_{(C_n)}\!\left\{f(z_n)+(z-z_n)\;\! f'(z_n)+(z-z_n)\:\! v\right\} \,dz\\ &=\left\{f(z_n)-z_n f'(z_n)\right\}\int_{(C_n)}\!dz + f'(z_n)\int_{(C_n)}\!z\,dz + \int_{(C_n)}\!(z-z_n)\:\! v\,dz. \end{align*}\] But \[\int_{(C_n)}\!dz=\left[z\right]_{C_n}=0, \quad \int_{(C_n)}\!z\,dz=\left[\frac{1}{2}z^2\right]_{C_n}=0,\] by the examples of §4.6, since the end points of \(C_n\) coincide.

Now let \(l_n\) be the side of \(C_n\) and \(A_n\) the area of \(C_n\).

Then, using §4.62,

\[\begin{align*} \left|\,\int_{(C_n)}\!f(z) \,dz \,\right| &= \left|\,\int_{(C_n)} \!(z-z_n)\:\! v\, dz\, \right| \leq \int_{(C_n)} \! \left|\,(z-z_n)\:\! v\, dz \, \right| \\ &< \epsilon l_n \sqrt{2} \int_{(C_n)} \! \left|\, dz\,\right| = \epsilon l_n \sqrt{2} \cdot 4l_n = 4\epsilon A_n \sqrt{2} . \end{align*}\]

In like manner \[\begin{align*} \left|\,\int_{(D_n)}\!f(z) \,dz \,\right| & \leq \int_{(D_n)} \! \left|\,(z-z_n)\:\! v \, dz \, \right| \\ & \leq 4\epsilon (A'_n+ l'_n\lambda_n )\sqrt{2} , \end{align*}\] where \(A'_n\) is the area of the complete square of which \(D n\) is part, \(l'_n\) is the side of this square and \(\lambda_n\) is the length of the part of \(C\) which lies inside this square. Hence, if \(\lambda\) be the whole length of \(C\), while \(l\) is the side of a square which encloses all the squares \(C_n\) and \(D_n\), \[\begin{align*} \left|\,\int_{(C)}\!f(z) \,dz \,\right|& \leq\sum_{n=1}^M\left|\,\int_{(C_n)}\!f(z) \,dz \,\right|+\sum_{n=1}^N\left|\,\int_{(D_n)}\!f(z) \,dz \,\right|\\ &<4\epsilon \sqrt{2}\left\{\sum_{n=1}^M A_n + \sum_{n=1}^N A'_n +l\sum_{n=1}^N \lambda_n \right\}\\ &< 4\epsilon \sqrt{2}(l^2 + l\lambda). \end{align*}\] Now \(\epsilon\) is arbitrarily small, and \(l\), \(\lambda\), and \(\int_{(C)} f(z)\,dz\) are independent of \(\epsilon\). It therefore follows from this inequality that the only value which \(\int_{(C)} f(z)\,dz\) can have is zero; and this is Cauchy’s result.

Corollary 1. If there are two paths \(z_0AZ\) and \(z_0BZ\) from \(z_0\) to \(Z\), and if \(f(z)\) is. a function of \(z\) analytic at all points on these curves and throughout the domain enclosed by these two paths, then \(\int_{z_0}^Z f(z) \,dz\) has the same value whether the path of integration is \(z_0AZ\) or \(z_0BZ\). This follows from the fact that \(z_0AZBz_0\) is a contour, and so the integral taken round it (which is the difference of the integrals along \(z_0AZ\) and \(z_0BZ\) ) is zero. Thus, if \(f(z)\) be an analytic function of \(z\), the value of \(\int_{AB} f(z) \,dz\) is to a certain extent independent of the choice of the arc \(AB\), and depends only on the terminal points \(A\) and \(B\). It must be borne in mind that this is only the case when \(f(z)\) is an analytic function in the sense of §5.12.

[5]The phrase ‘analytic throughout a region’ implies one-valuedness (§5.12); that is to say that after \(z\) has described a closed path surrounding \(C_0\), \(f(z)\) has returned to its initial value. A many-valued function such as log \(z\) considered in the region \(1 \leq \left|\, z\,\right| \leq 2\) will be said to be ‘analytic at all points of the region’ (but not ‘analytic throughout the region’ –Editor).  ↩
Corollary 2. Suppose that two simple closed curves \(C_0\) and \(C_1\) are given, such that \(C_0\) completely encloses \(C_1\), as e.g. would be the case if \(C_0\) and \(C_1\) were confocal ellipses.

Suppose moreover that \(f(z)\) is a function which is analytic[5] at all points on \(C_0\) and \(C_1\) and throughout the ring-shaped region contained between \(C_0\) and \(C_1\). Then by drawing a network of intersecting lines in this ring-shaped space, we can shew, exactly as in the theorem just proved, that the integral \[\int \! f(z)\,dz\] is zero, where the integration is taken round the whole boundary of the ring-shaped space; this boundary consisting of two curves \(C_0\) and \(C_1\), the one described in the counter-clockwise direction and the other described in the clockwise direction.

Corollary 3. In general, if any connected region be given in the \(z\)-plane, bounded by any number of simple closed curves \(C_0\), \(C_1\), \(C_2, \dots \), and if \(f(z)\) be any function of \(z\) which is analytic and one-valued everywhere in this region, then \[\int f(z)\,dz\] is zero, where the integral is taken round the whole boundary of the region; this boundary consisting of the curves \(C_0\), \(C_1\), \(C_2, \dots \), each described in such a sense that the region is kept either always on the right or always on the left of a person walking in the sense in question round the boundary.

An extension of Cauchy’s theorem \(\int f(z)\, dz=0\), to curves lying on a cone whose vertex is at the origin, has been made by Ravut {Nouv. Annales de Math. (3) xvi. (1897), pp. 365–7). Also, Morera (Rend, del Ist. Lombardo, xxii. (1889), p. 191) and Osgood (Bull. Amer. Math. Soc. ii. (1896), pp. 296–302), have shewn that the property \(\int f(z)\, dz=0\) may be taken as the property defining an analytic function, the other properties being deducible from it. (See Miscellaneous Examples, example 16.)

Example. A ring-shaped region is bounded by the two circles \(\left|\, z\,\right| = 1\) and \(\left|\, z\,\right| = 2\) in the \(z\)-plane. Verify that the value of \(\displaystyle \int \frac{dz}{z}\), where the integral is taken round the boundary of this region, is zero.

For the boundary consists of the circumference \(\left|\, z\,\right| = 1\), described in the clockwise direction, together with the circumference \(\left|\, z\,\right| = 2\), described in the counter-clockwise direction. Thus, if for points on the first circumference we write \(z = e^{i\theta}\), and for points on the second circumference we write \(z = 2e^{i\phi}\), then \(\theta\) and \(\phi\) are real, and the integral becomes \[\int_0^{-2\pi}\!\frac{ie^{i\theta}\,d\theta}{e^{i\theta}}+\int_0^{2\pi}\!\frac{ie^{i\phi}\,d\phi}{e^{i\phi}}=-2\pi i+2\pi i=0.\]

5.21 The value of an analytic function at a point, expressed as an integral taken round a contour enclosing the point.

Let \(C\) be a contour within and on which \(f(z)\) is an analytic function of \(z\). Then, if \(a\) be any point within the contour, \[\frac{f(z)}{z-a}\] is a function of \(z\), which is analytic at all points within the contour \(C\) except the point \(z = a\).

Now, given \(\epsilon\), we can find \(\delta\) such that \[\left|\,f(z)-f(a)-(z-a)\;\! f'(a)\,\right| \leq \epsilon\left|\,z-a\,\right|\] whenever \(\left|\, z - a \,\right| < \delta\); with the point \(a\) as centre describe a circle \(\gamma\) of radius \(r < \delta\), \(r\) being so small that lies wholly inside \(C\).

Then in the space between \(\gamma\) and \(C\) \(\left. f(z)\middle/(z - a)\right.\) is analytic, and so, by §5.2 corollary 2, we have \[\int_C \!\frac{f(z)\,dz}{z-a}=\int_\gamma \!\frac{f(z)\,dz}{z-a}.\]

where \(\int_C\) and \(\int_\gamma\) denote integrals taken counter-clockwise along the curves \(C\) and \(\gamma\) respectively.

But, since \(\left|\, z - a\,\right| < \delta\) on \(\gamma\), we have \[\int_\gamma \!\frac{f(z)\,dz}{z-a}=\int_\gamma \!\frac{f(a)+(z-a)\;\! f'(a)+v(z-a)}{z-a}dz,\] where \(\left|\, v\,\right| < \epsilon\); and so \[\int_C \!\frac{f(z)\,dz}{z-a}=f(a)\!\int_\gamma \!\frac{dz}{z-a} + f'(a)\!\int_\gamma \!dz +\int_\gamma v\,dz.\]

Now, if \(z\) be on \(\gamma\), we may write \[z-a=re^{i\theta},\] where \(r\) is the radius of the circle \(\gamma\), and consequently \[\int_\gamma \!\frac{dz}{z-a}=\int_0^{2\pi} \! \frac{ire^{i\theta} \, d\theta}{re^{i\theta}} = i\!\int_0^{2\pi} \!d\theta = 2\pi i,\] and \[\int_\gamma \! dz =\int_0^{2\pi} \! ire^{i\theta}\, d\theta = 0;\] also, by §4.62, \[\left|\,\int_\gamma \! v\,dz \, \right| \leq \epsilon\,2\pi r.\] Thus \[\left|\,\int_\gamma \! \frac{f(z)\,dz}{z-a} - 2\pi i f(a)\, \right| = \left|\,\int_\gamma \! v\,dz \, \right| \leq 2\pi r \epsilon.\]

But the left-hand side is independent of \(\epsilon\), and so it must be zero, since \(\epsilon\) is arbitrary; that is to say \[f(a) = \frac{1}{2\pi i }\! \int_C \! \frac{f(z)\,dz}{z-a}.\]

This remarkable result expresses the value of a function \(f(z)\), (which is analytic on and inside \(C\)) at any point \(a\) within a contour \(C\), in terms of an integral which depends only on the value of \(f(z)\) at points on the contour itself.

Corollary. If \(f(z)\) is an analytic one-valued function of \(z\) in a ring-shaped region bounded by two curves \(C\) and \(C'\), and \(a\) is a point in the region, then \[f(a) = \frac{1}{2\pi i }\! \int_C \! \frac{f(z)\,dz}{z-a}-\frac{1}{2\pi i }\! \int_{C'} \! \frac{f(z)\,dz}{z-a},\] where \(C\) is the outer of the curves and the integrals are taken counter-clockwise.

5.22 The derivates of an analytic function \(f(z)\).

The function \(f'(z)\), which is the limit of \[\frac{f(z + h)-f(z)}{h} \] as \(h\) tends to zero, is called the derivate of \(f(z)\). We shall now shew that \(f'(z)\) is itself an analytic function of \(z\), and consequently itself possesses a derivate.

For if \(C\) be a contour surrounding the point \(a\), and situated entirely within the region in which \(f(z)\) is analytic, we have \[\begin{align*} f'(a) &= \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\\ \\ \\ &=\lim_{h \rightarrow 0} \frac{1}{2\pi i h}\left\{\int_C \! \frac{f(z) \, dz}{z-a-h} - \int_C \! \frac{f(z) \, dz}{z-a} \right\}\\ \\ \\ &=\lim_{h \rightarrow 0} \frac{1}{2\pi i}\int_C \frac{f(z) \, dz}{(z-a)(z-a-h)} \\ \\ \\ &=\frac{1}{2\pi i}\int_C \frac{f(z) \, dz}{(z-a)^2} + \lim_{h \rightarrow 0} \frac{h}{2\pi i}\int_C \frac{f(z) \, dz}{(z-a)^2(z-a-h)}. \end{align*}\]

Now, on \(C\), \(f(z)\) is continuous and therefore bounded, and so is \((z - a)^{-2}\); while we can take \(\left|\, h\,\right|\) less than the upper bound of \(\frac{1}{2}\left|\,z-a\,\right| \).

Therefore \(\displaystyle \left|\,\frac{f(z)}{(z-a)^2(z-a-h)}\,\right|\) is bounded; let its upper bound be \(K\). Then, if \(l\) be the length of \(C\), \[\left|\,\lim_{h \rightarrow 0} \frac{h}{2\pi i}\int_C \frac{f(z)\,dz}{(z-a)^2(z-a-h)}\,\right|\leq \lim_{h \rightarrow 0 } \:\left|\,h\,\right|\,(2\pi)^{-1} K \,l = 0,\] and consequently \[f'(z)=\frac{1}{2\pi i }\! \int_C \frac{f(z)\,dz}{(z-a)^2},\] a formula which expresses the value of the derivate of a function at a point as an integral taken along a contour enclosing the point.

From this formula we have, if the points \(a\) and \(a + h\) are inside \(C\), \[\begin{align*} \frac{f'(a+h)-f'(a)}{h}&=\frac{1}{2\pi i }\int_C \!\frac{f(z)\, dz}{h}\left\{ \frac{1}{(z-a-h)^2} - \frac{1}{(z-a)^2} \right\}\\ \\ \\ &=\frac{1}{2\pi i }\int_C f(z)\, dz \frac{2\left(z-a-\frac{1}{2}h \right)}{(z-a-h)^2(z-a)^2}\\ \\ \\ &=\frac{2}{2\pi i }\! \int_C \frac{f(z)\,dz}{(z-a)^3}+h A_h, \end{align*}\] and it is easily seen that \(A_h\) is a bounded function of \(z\) when \(\left|\, h \,\right| < \frac{1}{2}\left|\, z -a\, \right|\).

Therefore, as \(h\) tends to zero, \(h^{-1}\{f'(a + h) - f'(a)\}\) tends to a limit, namely \[\frac{2}{2\pi i }\! \int_C \frac{f(z)\,dz}{(z-a)^3}.\]

Since \(f'(a)\) has a unique differential coefficient, it is an analytic function of \(a\); its derivate, which is represented by the expression just given, is denoted by \(f''(a)\), and is called the second derivate of \(f(a)\).

Similarly it can be shewn that \(f''(a)\) is an analytic function of \(a\), possessing a derivate equal to \[\frac{2\cdot 3}{2\pi i }\! \int_C \frac{f(z)\,dz}{(z-a)^4};\]

this is denoted by \(f'''(a)\), and is called the third derivate of \(f(a)\). And in general an \(n\)th derivate \(f^{(n)}(a)\) of \(f(a)\) exists, expressible by the integral \[\frac{n!}{2\pi i }\! \int_C \frac{f(z)\,dz}{(z-a)^{n+1}},\] and having itself a derivate of the form \[\frac{(n+1)!}{2\pi i }\! \int_C \frac{f(z)\,dz}{(z-a)^{n+2}};\] the reader will see that this can be proved by induction without difficulty.

A function which possesses a first derivate with respect to the complex variable \(z\) at all points of a closed two-dimensional region in the \(z\)-plane therefore possesses derivates of all orders at all points inside the region.

5.23 Cauchy’s inequality for \(f^{(n)}(a)\).

Let \(f(z)\) be analytic on and inside a circle with centre \(a\) and radius \(r\). Let \(M\) be the upper bound of \(f(z)\) on the circle. Then, by §4.62, \[\begin{align*} \left|\, f^{(n)}(a)\,\right| &\leq \frac{n!}{2\pi }\! \int_C \! \frac{M}{r^{n+1}}\left|\,dz\,\right|\\ \\ &\leq \frac{M \cdot n!}{r^n}. \end{align*}\]

Example. If \(f(z)\) is analytic, \(z=x +iy\) and \( \nabla^2=\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}\), shew that
\[\nabla^2 \log \left|\,f(z)\,\right| = 0; \text{ and } \nabla^2 \left|\,f(z)\,\right| > 0\] unless \(f(z) =0 \) or \(f'(z) = 0\).[6] \(\vphantom{\\ 3\\}\)
(Trinity, 1910.)

[6]Editor’s Note: Either part of this example can be used to prove the maximum modulus principle.  ↩