Chapter 25Complex Integration and Cauchy's Integral Theorem

Proof

Let $z(t)=x(t)+iy(t)\ (a\le t\le b)$.

(i), (ii): Linearity. If we split $f(z(t))\pm g(z(t))$ into real and imaginary parts, we are simply applying the linearity $\int(p\pm q)\,dt=\int p\,dt\pm\int q\,dt$ of definite integrals of real-valued functions to the real part and the imaginary part separately. Similarly, if we write $\alpha=p+iq$ ($p,q\in\R$), then the real and imaginary parts of $\alpha f(z(t))z'(t)$ are linear combinations, with coefficients $p,q$, of the real and imaginary parts of $f(z(t))z'(t)$, and (ii) follows from the linearity of real integrals.

(iii): Reversal of orientation. Since $-C$ is the curve that traces the same figure in the opposite direction, we can take $\tilde z(t):=z(a+b-t)=z(s)$, with $s=a+b-t$, as a parametrization of $-C$ (at $t=a$ we have $s=b$, and at $t=b$ we have $s=a$, so the direction is reversed). By the chain rule, $\tilde z'(t)=-z'(s)$ (differentiating $s$ with respect to $t$ contributes the factor $ds/dt=-1$). Hence

$$\int_{-C}f(z)\,dz=\int_a^b f(\tilde z(t))\tilde z'(t)\,dt=\int_a^b f(z(s))\big(-z'(s)\big)\,dt$$

Here we use the substitution $s=a+b-t,\ ds=-dt$. Since $dt=-ds$, the sign of $\big(-z'(s)\big)\,dt$ is reversed once, giving $z'(s)\,ds$, and in addition the interval of integration $t:a\to b$ corresponds to $s:b\to a$ (from $b$ to $a$), that is, to the opposite direction. Therefore,

$$\int_a^b f(z(s))\big(-z'(s)\big)\,dt=\int_{s=b}^{s=a}f(z(s))z'(s)\,ds=-\int_a^b f(z(s))z'(s)\,ds=-\int_C f(z)\,dz$$

(The last equality uses the property of real integrals $\int_b^a=-\int_a^b$: restoring the interval of integration $b\to a$ to the usual direction $a\to b$ reverses the sign.) Note that the sign is reversed twice (once through $dt=-ds$ and once through restoring the interval), and in the end only one sign change remains.

(iv): Splitting of the path. We can choose the parameter for the whole of $C$ so that $C_1$ corresponds to $t\in[a,c]$ and $C_2$ corresponds to $t\in[c,b]$ ($a\lt c\lt b$) (this is possible because the endpoint of $C_1$ coincides with the starting point of $C_2$). We need only apply the property of real integrals $\int_a^b=\int_a^c+\int_c^b$ as it stands.

(v): The $ML$ inequality. Let $I=\int_C f(z)\,dz$. If $I=0$ the inequality is trivial, so suppose $I\neq0$ and write $I=|I|e^{i\theta}$ ($\theta=\arg I$) in polar form. Then $|I|=e^{-i\theta}I$ is a real number. Now

$$|I|=e^{-i\theta}I=e^{-i\theta}\int_a^b f(z(t))z'(t)\,dt=\int_a^b e^{-i\theta}f(z(t))z'(t)\,dt$$

In this form, the left-hand side $|I|$ is a real number, so the value of the integral on the right-hand side is also real, and we may regard the right-hand side as equal to its own real part. In general, for a complex number $w=p+iq$ ($p,q\in\R$), $|w|^2=p^2+q^2\ge p^2$ gives $|w|\ge|p|\ge p$, that is, $\operatorname{Re}(w)\le|w|$ (the real part does not exceed the absolute value), so we obtain

$$|I|=\operatorname{Re}\int_a^b e^{-i\theta}f(z(t))z'(t)\,dt=\int_a^b\operatorname{Re}\big[e^{-i\theta}f(z(t))z'(t)\big]dt \le \int_a^b\big|e^{-i\theta}f(z(t))z'(t)\big|dt$$

Since $|e^{-i\theta}|=1$, we have $\big|e^{-i\theta}f(z(t))z'(t)\big|=|f(z(t))|\,|z'(t)|\le M|z'(t)|$ ($M$ is the maximum of $|f|$ on $C$). Moreover, $|z'(t)|=\sqrt{x'(t)^2+y'(t)^2}$ is precisely the arc-length element that you learned in calculus, and $L=\int_a^b|z'(t)|\,dt$ is the length of the path $C$. Therefore we obtain

$$|I|\le\int_a^b M|z'(t)|\,dt=M\int_a^b|z'(t)|\,dt=ML$$

∎

Proof

Put $z=re^{i\theta}$ ($r$ fixed, $0\le\theta\le2\pi$). Then $z^n=r^ne^{in\theta}$ and $\dfrac{dz}{d\theta}=ire^{i\theta}$ (we use the derivative of the exponential function, $\frac{d}{d\theta}e^{i\theta}=ie^{i\theta}$, just as it stands).

The case $n\neq-1$.

$$\oint_C z^n\,dz=\int_0^{2\pi}r^ne^{in\theta}\cdot ire^{i\theta}\,d\theta=ir^{n+1}\int_0^{2\pi}e^{i(n+1)\theta}\,d\theta$$

Here, since $n+1\neq0$, an antiderivative of $e^{i(n+1)\theta}$ is $\dfrac{e^{i(n+1)\theta}}{i(n+1)}$ (we can check this by differentiating with respect to $\theta$: $i(n+1)\cdot\frac{e^{i(n+1)\theta}}{i(n+1)}=e^{i(n+1)\theta}$, which returns the original function). Hence

$$ir^{n+1}\Big[\frac{e^{i(n+1)\theta}}{i(n+1)}\Big]_0^{2\pi}=\frac{r^{n+1}}{n+1}\Big(e^{i(n+1)2\pi}-e^0\Big)$$

Since $n+1$ is an integer other than $0$, $e^{i(n+1)2\pi}=\cos\big(2\pi(n+1)\big)+i\sin\big(2\pi(n+1)\big)=1$ (the angle turns through exactly an integer multiple of $2\pi$, so we return to the original position). Therefore $e^{i(n+1)2\pi}-1=0$, and

$$\oint_C z^n\,dz=\frac{r^{n+1}}{n+1}(1-1)=0$$

The case $n=-1$. Since $z^{-1}=\dfrac1z=\dfrac1{re^{i\theta}}$,

$$\oint_C z^{-1}\,dz=\int_0^{2\pi}\frac1{re^{i\theta}}\cdot ire^{i\theta}\,d\theta=\int_0^{2\pi}i\,d\theta=i\big[\theta\big]_0^{2\pi}=2\pi i$$

(The fact that $r$ and $e^{i\theta}$ cancel exactly, so that the integrand becomes the constant $i$, is the reason why only $n=-1$ is special.) ∎

Proof

In the original notes, this proof breaks off after the heading (“Evidence”) alone, so we give a self-contained proof starting from the basics of multiple integrals and line integrals learned in calculus. To keep the argument simple, we assume that $D$ is a “well-behaved” region that can be described in both of the following two ways (for a general region, the result can be shown by dividing it into such small pieces and stacking up the same argument piece by piece).

(a) Vertical-line description: $D=\{(x,y)\mid a\le x\le b,\ \varphi_1(x)\le y\le\varphi_2(x)\}$; that is, there are two curves, $y=\varphi_1(x)$ (lower) and $y=\varphi_2(x)$ (upper), that sandwich $D$ from below and above, and the boundary $C$ consists of these two curves (and, if there are any, vertical line segments at the two ends).

Then the double integral on the right-hand side can be computed by integrating first with respect to $y$ and then with respect to $x$ (an iterated integral). The inner integral is, for each fixed $x$, an integral of a single variable, so the fundamental theorem of calculus can be used as it stands.

$$\iint_D\frac{\partial f}{\partial y}\,dx\,dy=\int_a^b\left[\int_{\varphi_1(x)}^{\varphi_2(x)}\frac{\partial f}{\partial y}(x,y)\,dy\right]dx=\int_a^b\Big[f\big(x,\varphi_2(x)\big)-f\big(x,\varphi_1(x)\big)\Big]dx$$

(The rewriting of the inner $[\ ]$ is nothing more than an application of the fundamental theorem $\int_{\varphi_1}^{\varphi_2}\partial f/\partial y\,dy=f(x,\varphi_2(x))-f(x,\varphi_1(x))$, based on the fact that the antiderivative of $f(x,y)$, regarded as a function of $y$ with $x$ fixed, is $f(x,y)$ itself.)

On the other hand, we compute the line integral $\oint_C f\,dx$ on the left-hand side. Tracing the boundary $C$ counterclockwise, we go along the lower curve $y=\varphi_1(x)$ in the direction $x:a\to b$, and along the upper curve $y=\varphi_2(x)$ in the direction $x:b\to a$ (the reverse direction). (Even if there are vertical segments at the two ends, $x$ is constant on them, so $dx=0$ and they do not contribute to $\oint f\,dx$.) Hence, using property (iii) (reversal of orientation), we have

$$\oint_C f\,dx=\int_a^b f\big(x,\varphi_1(x)\big)\,dx+\int_b^a f\big(x,\varphi_2(x)\big)\,dx=\int_a^b f\big(x,\varphi_1(x)\big)\,dx-\int_a^b f\big(x,\varphi_2(x)\big)\,dx$$

This is exactly the value of the double integral computed above, with the opposite sign. That is, we have shown that

$$\oint_C f\,dx=-\int_a^b\Big[f\big(x,\varphi_2(x)\big)-f\big(x,\varphi_1(x)\big)\Big]dx=-\iint_D\frac{\partial f}{\partial y}\,dx\,dy$$

holds.

(b) Horizontal-line description: For $D=\{(x,y)\mid c\le y\le d,\ \psi_1(y)\le x\le\psi_2(y)\}$ as well (here $D$ is sandwiched from the left and the right by two curves: $x=\psi_1(y)$ on the left side and $x=\psi_2(y)$ on the right side), if we argue in exactly the same way with the roles of $x,y$ interchanged, then when we go counterclockwise we trace the right-hand curve in the direction $y:c\to d$ and the left-hand curve in the direction $y:d\to c$, and so we obtain

$$\oint_C f\,dy=\int_c^d f\big(\psi_2(y),y\big)\,dy-\int_c^d f\big(\psi_1(y),y\big)\,dy=\int_c^d\Big[f\big(\psi_2(y),y\big)-f\big(\psi_1(y),y\big)\Big]dy=\iint_D\frac{\partial f}{\partial x}\,dx\,dy$$

(Note that this time the sign is not reversed, and the two sides agree as they are. This is because the directions in which the left and right curves are traversed are the reverse of those for the upper and lower curves in the case above.) ∎

Proof

Let $f(z)=u(x,y)+iv(x,y)$. From the assumption that $f$ is holomorphic on $C,D$ and that $f'(z)$ is continuous, all the first-order partial derivatives of $u,v$ are also continuous (by the discussion of the Cauchy–Riemann equations in Chapter 24, $f'(z)=u_x+iv_x=v_y-iu_y$, so if $f'$ is continuous, the combinations $u_x,v_x,v_y,u_y$ must also be continuous). Hence Green's theorem (Theorem 25.1) can be applied to $u,v$ as it stands.

Applying Green's theorem to each of $u,v$, we obtain the following four equations.

$$\oint_C u\,dx=-\iint_D \frac{\partial u}{\partial y}\,dx\,dy,\qquad \oint_C u\,dy=\iint_D \frac{\partial u}{\partial x}\,dx\,dy$$ $$\oint_C v\,dx=-\iint_D \frac{\partial v}{\partial y}\,dx\,dy,\qquad \oint_C v\,dy=\iint_D \frac{\partial v}{\partial x}\,dx\,dy$$

Here we apply to the closed curve $C$ the decomposition of a complex integral into its real and imaginary parts, $\displaystyle\int_C f(z)\,dz=\int_C(u\,dx-v\,dy)+i\int_C(u\,dy+v\,dx)$, which we confirmed in Definition 25.1.

$$\oint_C f(z)\,dz=\Big(\oint_C u\,dx-\oint_C v\,dy\Big)+i\Big(\oint_C u\,dy+\oint_C v\,dx\Big)$$

Substituting the four equations above into each of the real and imaginary parts, we obtain

$$\oint_C f(z)\,dz=\iint_D\underbrace{\left(-\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)}_{\text{①}}dx\,dy+i\iint_D\underbrace{\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)}_{\text{②}}dx\,dy$$

Now, since $f$ is holomorphic on $C,D$, the Cauchy–Riemann equations (Chapter 24) $\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$ hold. Substituting these into ① and ②, we get

$$\text{①}=-\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}=\frac{\partial v}{\partial x}-\frac{\partial v}{\partial x}=0,\qquad \text{②}=\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}=\frac{\partial v}{\partial y}-\frac{\partial v}{\partial y}=0$$

(In ① we used $u_y=-v_x$, and in ② we used $u_x=v_y$.) Since the integrands of ① and ② both become identically $0$ on $D$, each of the double integrals is also $0$. Hence $\displaystyle\oint_C f(z)\,dz=0+i\cdot0=0$. ∎

Proof

If we join $C_1$ ($\alpha\to\beta$) and $-C_2$ ($\beta\to\alpha$; this is $C_2$ traversed in the reverse direction) in this order, we obtain a single simple closed curve that starts at $\alpha$ and returns to $\alpha$. Applying Theorem 25.2 (Cauchy's integral theorem) to it, we have, by property (iv) (splitting of the path),

$$\int_{C_1}f(z)\,dz+\int_{-C_2}f(z)\,dz=0$$

Substituting property (iii) (reversal of orientation), $\displaystyle\int_{-C_2}f(z)\,dz=-\int_{C_2}f(z)\,dz$, we obtain

$$\int_{C_1}f(z)\,dz-\int_{C_2}f(z)\,dz=0\qquad\therefore\ \int_{C_1}f(z)\,dz=\int_{C_2}f(z)\,dz$$

∎

Proof

Let the path $\alpha\to\beta$ be $C_1$ and the path $\beta\to\alpha$ be $C_2$ ($C_2$ starts from $\beta$ and runs in the direction opposite to $C_1$). Traversing $C_1$ and then $C_2$ gives a simple closed curve that returns to $\alpha$, so by Theorem 25.2 (Cauchy's integral theorem; since $D$ is simply connected, the interior is entirely contained in $D$) and property (iv),

$$\int_{C_1}f(z)\,dz+\int_{C_2}f(z)\,dz=0$$

Now let $C_3$ be the path that traverses $C_2$ in the opposite direction (that is, $C_3$ runs $\alpha\to\beta$ and $C_2=-C_3$). Then by property (iii) (reversal of orientation), $\displaystyle\int_{C_2}f(z)\,dz=\int_{-C_3}f(z)\,dz=-\int_{C_3}f(z)\,dz$. Substituting this into the equation above, we obtain

$$\int_{C_1}f(z)\,dz-\int_{C_3}f(z)\,dz=0\qquad\therefore\ \int_{C_1}f(z)\,dz=\int_{C_3}f(z)\,dz$$

Since $C_1,C_3$ are both paths $\alpha\to\beta$, this shows that “the integral of a holomorphic function does not depend on the path.” ∎

Proof

As in Figure 25.9(b), draw one bridging segment from the hole $C_1$ to the outer boundary $C$, and let $C_4$ (outward) and $-C_4$ (inward, the reverse of $C_4$) denote the path that goes back and forth along it. Similarly, draw a bridging segment from the hole $C_2$ as well, and denote the round trip by $C_3,-C_3$. Once these two cuts are made, the domain obtained from $D$ by removing the two holes becomes a simply connected domain surrounded by a single simple closed curve, which is traversed as follows: go along the outer boundary in the direction of $C$ (counterclockwise), go along the bridging segment $C_4$ to the vicinity of hole 1, go around the boundary of hole 1 in the reverse direction $-C_1$ (once the cuts are made, the boundary of a hole is clockwise as seen from the outside), return to the outer boundary along $-C_4$, continue further along $C$, go along the bridging segment $C_3$ to the vicinity of hole 2, go along $-C_2$ (the reverse direction), and return to the outer boundary along $-C_3$.

This simple closed curve is the boundary of a simply connected domain (the domain after the cuts have been made), and $f$ is holomorphic there, so by Theorem 25.2 (Cauchy's integral theorem) the integral along this closed curve is $0$. Using property (iv) (splitting of the path) and adding up the integrals in the order in which the closed curve is traversed, we obtain

$$\oint_C f\,dz+\oint_{-C_1}f\,dz+\int_{C_4}f\,dz+\int_{-C_4}f\,dz+\oint_{-C_2}f\,dz+\int_{C_3}f\,dz+\int_{-C_3}f\,dz=0$$

Here, by property (iii) (reversal of orientation), the two integrals along the round trip on a bridging segment satisfy $\displaystyle\int_{C_4}f\,dz+\int_{-C_4}f\,dz=\int_{C_4}f\,dz-\int_{C_4}f\,dz=0$, and likewise $\displaystyle\int_{C_3}f\,dz+\int_{-C_3}f\,dz=0$, so they cancel neatly (the trip there and the trip back run along the same path, with the same values of the integrand, so if the directions are opposite only the sign is reversed and the two integrals add up to zero). What remains is

$$\oint_C f\,dz+\oint_{-C_1}f\,dz+\oint_{-C_2}f\,dz=0$$

Furthermore, substituting $\displaystyle\oint_{-C_1}f\,dz=-\oint_{C_1}f\,dz$ and $\displaystyle\oint_{-C_2}f\,dz=-\oint_{C_2}f\,dz$ (property (iii)), we obtain

$$\oint_C f\,dz-\oint_{C_1}f\,dz-\oint_{C_2}f\,dz=0\qquad\therefore\ \oint_C f\,dz=\oint_{C_1}f\,dz+\oint_{C_2}f\,dz$$

The same argument works in exactly the same way when there is one hole (doubly connected) or three or more holes ($n$-fold connected); one only has to increase or decrease the number of bridging segments. ∎

Proof

The function $f(z)=\dfrac1{z-\alpha}$ is holomorphic on the whole complex plane except at $z=\alpha$. Since $C$ is a simple closed curve that encloses only $\alpha$ in its interior, if we take a circle $C_r:|z-\alpha|=r$ centered at $\alpha$ whose radius $r$ is small enough for the circle to fit entirely inside $C$, then $f$ is holomorphic on the domain between $C$ and $C_r$ (a doubly connected domain). Hence, by Corollary 25.2 (deformation of the path of integration in a doubly connected domain),

$$\oint_C \frac1{z-\alpha}\,dz=\oint_{C_r} \frac1{z-\alpha}\,dz$$

The right-hand side can be computed directly on the small circle $C_r$. Setting $z=\alpha+re^{i\theta}\ (0\le\theta\le2\pi)$, we have $z-\alpha=re^{i\theta}$ and $dz=ire^{i\theta}d\theta$, so

$$\oint_{C_r}\frac1{z-\alpha}\,dz=\int_0^{2\pi}\frac1{re^{i\theta}}\cdot ire^{i\theta}\,d\theta=\int_0^{2\pi}i\,d\theta=2\pi i$$

(This is just the case $n=-1$ of Formula 25.2, repeated with the center moved from the origin to $\alpha$.) Therefore $\displaystyle\oint_C \frac1{z-\alpha}\,dz=2\pi i$ holds regardless of the specific shape of $C$. ∎