Proof. Sufficiency: Let \(C\) be any curve joining \(z_{0}\) to \(z_{1}\) . We have \[\int_{z_{0}}^{z_{1}} f(z) \,dz = \lim_{\Delta z \rightarrow 0} \sum F'(\zeta_{\nu})(z_{\nu+1}-z_{\nu}).\] Since \(f(z)\) is continuous, the sum may be written in the form \[\sum\left(\frac{F(z_{\nu+1})-F(z_{\nu})}{z_{\nu+1}-z_{\nu}}+\varepsilon_{\nu}\right)(z_{\nu+1}-z_{\nu})\] where \(\left|\varepsilon_{\nu}\right|\) is less than some fixed \(\varepsilon\) , (3.11) , Chapter 2. Thus we obtain \begin{align} &\sum\big[F(z_{\nu+1})-F(z_{\nu})\big]+\sum \varepsilon_{\nu}(z_{\nu+1}-z_{\nu}) \\ &= F(z)-F(z_{0})+\sum \varepsilon_{\nu} \Delta z_{\nu}. \end{align} Since \(\big|\sum \varepsilon_{\nu} \Delta z_{\nu}\big|<\varepsilon\left|z_{1}-z_{0}\right|\) we have in the limit

\[\int_{z_{0}}^{z_{1}} f(z) \,d z=F(z)-F(z_{0}).\] 

Necessity: Suppose the integral

\[\int_{z_{0}}^{z_{1}} f(z) \,d z\] is independent of the curve of integration. We define a function \[F(z)=\int_{z_{0}}^{z} f(\zeta) \,d\zeta.\] Form the difference quotient \[\frac{F(z+\Delta z)-F(z)}{\Delta z}=\int_{z}^{z+\Delta z} f(\zeta) \,d\zeta \frac{1}{\Delta z}.\] The integral from \(z\) to \(z+\Delta z\) is independent of the curve. We take the straight line path. By taking \(|\Delta z|\) sufficiently small we may be sure that \(|f(\zeta)-f(z)|<\varepsilon\) whenever \(|\zeta-z|<|\Delta z|\) . Hence \begin{align} \left|f(z)-\frac{F(z+\Delta z)-F(z)}{\Delta z}\right|&=\left|\frac{1}{\Delta z} \int_{z}^{z+\Delta z}\big[f(z)-f(\zeta)\big]d\zeta\right|\\ &<\varepsilon. \end{align} We conclude that \(F'(z)=f(z)\) . ◻

Proof. We give the famous Goursat proof.

A. The theorem is true if \({C}\) is a rectangle. For let \({R}_{0}\) be any rectangle in \(D\) . We partition \({R}_{0}\) into four equal rectangles by means of parallels to the sides. If \({R}_{1}^{*}\) , \({R}_{2}^{*}\) , \({R}_{3}^{*}\) , \({R}_{4}^{*}\) denote the rectangles of the partition then by (1.34) and (1.35) it follows that \[\int_{{R}_{0}} f({z})\,{dz}=\int_{{R}_{1}^{*}}+\int_{{R}_{2}^{*}}+\int_{{R}_{3}^{*}}+\int_{{R}_{4}^{*}}.\] Therefore if we set \[I_{0}=\left|\int_{R_{0}} f(z) \,d z\right|\] It follows that \[|I_{0}| \leq\left|\int_{{R}_{1}^{*}}\right|+\cdots+\left|\int_{{R}_{4}^{*}}\right|\] Consequently for at least one of the \(R_{1}^{*}\) , denote it by \(R_{1}\) , we have \[I_{1}=\left|\int_{R_{1}} f(z) \,d z\right| \geq \frac{I_{0}}{4}.\] If we repeat the process for the rectangle \(R_{1}\) we obtain a rectangle \(R_{2}\) with \[I_{2}=\left|\int_{R_{1}} f(z) \,d z\right| \geq \frac{I_{1}}{4} \geq \frac{I_{0}}{4^{2}}\] and in \(n\) steps we obtain an \(R_{n}\) with \[I_{n}=\left|\int_{R_{n}} f(z) \,d z\right| \geq \frac{I_{0}}{4^{n}}.\] 

In this manner we determine a sequence of rectangles \(R_{0}, R_{1}, R_{2}, \ldots\) where each rectangle is contained in the preceding and the maximum diameter (i.e. the diagonal) tends to zero. We refer to such a sequence as a "nested set". It is not difficult to show that there is one and only one point common to all the rectangles of a nested set. The proof is left as an exercise.

Let \(\zeta\) denote the point common to all the \(R_{i}\) . Since \(f(z)\) is analytic we may approximate \(f(z)\) by a linear function at \(\zeta\) . Namely, \[f(z)-f(\zeta)=\big[f'(\zeta)+\eta(z, \zeta)\big](z-\zeta)\] where \(\eta(z, \zeta)\) can be made arbitrarily small by taking \(z\) close to \(\zeta\) . ¹ That is, given any \(\varepsilon>0\) we can find a \(\delta\) such that \(|z-\zeta|<\delta\) implies \(|\eta(z, \zeta)|<\varepsilon\) .

Choose \(n\) so large that \(R_{n}\) is contained in the circular domain \[|z-\zeta|<\delta.\] On \(R_{n}\) we have \[f(z)=f(\zeta)+z f'(\zeta)-\zeta f'(\zeta)+(z-\zeta) \eta.\] Consequently \begin{align} \int_{R_{n}} f(z) \,d z&=f(\zeta) \int_{R_{n}} \,d z+f'(\zeta) \int_{R_{n}} z \,d z\\ & \quad -\zeta f'(\zeta) \int_{R_{n}} \,d z+\int_{R_{n}}(z-\zeta) \eta \,d z. \end{align} 

By equations (1.37) , (1.38) we know that the first three integrals are independent of the path and hence vanish. Thus \[\int_{R_{n}} f(z) \,d z=\int_{R_{n}}(z-\zeta) \eta \,d z.\] On \(R_{n}\) we have \(|\eta|<\varepsilon\) since the entire rectangle is within a \(\delta\) -radius of \(\zeta\) . Furthermore, \(|z-\zeta| \leq \frac{L_{n}}{2}\) where \(L_{n}\) is the perimeter of \(R_{n}\) . Using the estimate (1.33) we obtain \[I_{n}=\left|\int_{R_{n}} f(z) \,d z\right|<\varepsilon \cdot \frac{L_{n}}{2} \cdot L_{n}=\frac{\varepsilon}{2} \frac{L_{0}^{2}}{4^{n}}.\] Combining this with our previous result we obtain \[\frac{\varepsilon}{2} \frac{L_{0}^{2}}{4^{n}}>I_{n} \geq \frac{I_{0}}{4^{n}}\] Consequently \[I_{0}<\frac{\varepsilon L_{0}^{2}}{2}\] Thus \(I_{0}\) must be less than any positive number and therefore \(I_{0}=0\) .

The remainder of the proof consists of an extension of the theorem to more general curves.

B. The theorem is true for "step" polygons, i.e. polygons consisting of a finite number of segments parallel to the coordinate axes. For proof let us suppose first that \(C\) is a step polygon and simple. We partition \(C\) into rectangles as follows:

Form the rectangular lattice obtained by extending the sides of the polygon. The interior of \({C}\) is certainly contained in the rectangles of this lattice since it is bounded four lines obtained from the highest side, the lowest side, and the sides farthest to the left and right. Now, the interior of every rectangle of the lattice must be completely in the interior of \({C}\) or completely in the exterior since no lattice rectangle contains a point of \(C\) in its interior. By our orientation convention ² \(C\) is the sum of the rectangles inside \(C\) . We conclude by (1.34) that the theorem is true for simple step polygons.

If \({C}\) is a step polygon and not simple, then we may decompose it into simple sections as follows: Let \(A\) be any point on \(C\) . Beginning from \(A\) follow \(C\) in any direction until the path meets itself for the first time at some point \(B\) .

The part of the path running from the first encounter with \(B\) to the second is a closed polygon, as is also the remainder of \({C}\) ; each of these two polygons can be further decomposed, if necessary, in a similar manner, and so forth. Since the number of self intersections of \({C}\) is finite, a decomposition into simple closed polygons is obtained in a finite number of steps, and the theorem then holds for \(C\) by (1.34) .

C. We are now in a position to prove the theorem for arbitrary rectifiable curves. It is only necessary to show that any rectifiable curve \(C\) can be approximated as closely as we please by step polygons. Since \(C\) is rectifiable it can be sub-divided into arcs of arbitrarily small length. Let \({z}_{0}, {z}_{1}, \ldots\) denote the successive points of a sub-division. In each interval we approximate \(C\) by means of a "step".

For the interval from \(z_{\nu-1}\) to \(z_{\nu}\) we take the step running from \(z_{\nu-1}\) to \(x_{\nu-1}+i y_{\nu}\) to \(z_{\nu}\) . Clearly the length of such a step is less than \(2|z_{\nu}-z_{\nu-1}|\) . Hence the length of such an approximating polygon is less than twice the length of \({C}\) . By taking the subdivision sufficiently fine we can bring the step polygon arbitrarily close to \(C\) . For if the length of the arc \(\overset{\Huge \frown}{z_{\nu}z_{\nu-1}} < \varepsilon\) , no point on the step between \(z_{\nu}\) and \(z_{\nu-1}\) is as far away as \(2 \varepsilon\) . The theorem follows by Lemma 3.1.10 .◻

Proof. This is proved by constructing a simply-connected domain. We join each of the \(C_{i}\) to \(C\) by means of non-intersecting paths. If we restrict ourselves to curves which do not cross these paths the resultant domain is simply connected. Hence the integral over the new boundary curve vanishes. But the integrals over the connecting paths are taken in both directions and cancel Lemma 3.1.6 . We conclude that \[\oint_{C} f(z) \,d z+\oint_{C_{1}} f(z) \,d z+\cdots+\oint_{C_{n}} f(z) \,d z=0\] which proves the theorem.◻