Lecture 6: Complex Integrals

Complex integration

We now introduce contour integrals (also called path integrals), a concept that is built on one-dimensional Riemann integrals but on the complex plane.

Definition 1 (Parameterized Curves). A curve on $\mathbb C$ parameterized by $t$ , $a\le t\le b$ , is the set of points $\{z \in\mathbb C\ |\ z(t)=x(t) + i y(t),\ a\le t\le b \}$ . A curve/arc is said to be simple it it does not intersect itself, that is $z(t_1)\neq z(t_2)$ $\forall t_1\neq t_2$ except possibly $z(a)=z(b)$ . The curve is said to be closed if $z(a)=z(b)$ . A simple closed curve is called a Jordan curve.

Definition 2 (Contour). A parameterized curve on $\mathbb C$ , with parameter $t$ , is said to be continuous if $z(t)$ (and therefore $x(t)$ and $y(t)$ ) is continuous functions of $t$ . The curve is said to be smooth if $z'(t)$ is continuous. A contour is an arc that consists of a finite number of connected smooth arcs.

Example 3. The unit circle centered at the origin, $\{z\in\mathbb C\ |\ |z|=1\}$ , can be written as a parameterized curve as $z(\theta) = \cos(\theta) + i\sin\theta, \ \ 0\le \theta\le 2\pi.$

Remark 4. It is a convention that when a closed contour is considered, the parameterization is taken in the way such that we are traveling in the anit-clock direction when the parameter value increases.

Definition 5 (Continuity on a Contour). Let $C$ be a parameterized curve with parameter $t$ . A function $f(z)$ is said to be (piecewise) continuous on $C$ if $f(z(t))$ is (piecewise) continuous with respect to $t$ .

Remark 6. Note that being continuous on a curve $C$ is weaker than being continuous (i.e. being continuous as a function of $z$ ).

Definition 7. A domain $\Omega\in\mathbb C$ is said to be simply connected if every simple closed contour in it encloses only points of $\Omega$ .

Example 8. It is easy to see that disks, rectangles and general polygons are all simply connected. However, the annulus $\Omega=\{z\ |\ 2\le |z|\le 4\}$ is not simply connected. For instance, the simple closed contour $\{ z\ |\ |z|=3\}$ enclose all points inside the unit disk $\{z\ |\ |z|<1\}$ which are all outside of $\Omega$ .

Definition 9 (Complex Integration). (i) Let $C$ be a smooth contour parameterized by $t\in[a, b]$ and $f(z)$ a piecewise smooth function on $C$ . We define the integral of $f(z)$ on $C$ , often called contour integral of $f$ on $C$ , as $\int_C f(z) dz := \int_a^b f(z(t))z'(t) dt$ provided that both the real part and the imaginary part of the integral with respect to $t$ exist.

(ii) Let $-C$ be the arc C traversed the opposite direction, i.e. from $t=b$ to $t=a$ . We define $\int_{-C} f(z) dz := -\int_C f(z) dz .$

(iii) If $C$ is a piecewise smooth contour that consists of $K$ smooth arcs $C_1, \cdots, C_K$ , then we define $\int_{C} f(z) dz := \sum_{k=1}^K \int_{C_k} f(z) dz .$

Example 10. Take contour $C=\{z=(\cos\theta, \sin\theta)\ |\ 0\le \theta\le 2\pi\}$ and $f(z)=z$ , $g(z)=\bar z$ , $h(z) = |z|$ , $p(z)=\dfrac{1}{z^2}$ . Then on $C$ , we have that $z=e^{i\theta}$ , and $z'(\theta) = ie^{i\theta}$ . Therefore $\int_C f(z) dz = \int_0^{2\pi} e^{i\theta} i e^{i\theta} d\theta = i \int_0^{2\pi} e^{i 2\theta} d\theta = \dfrac{i}{2i}e^{i 2\theta}|_0^{2\pi} =0.$ $\int_C g(z) dz = \int_0^{2\pi} e^{-i\theta} i e^{i\theta} d\theta = i \int_0^{2\pi} d\theta = 2\pi i.$ $\int_C h(z) dz = \int_0^{2\pi} |e^{i\theta}| i e^{i\theta} d\theta =e^{i \theta}|_0^{2\pi} =0.$ $\int_C p(z) dz = \int_0^{2\pi} e^{-2i\theta} i e^{i\theta} d\theta =-e^{-i \theta}|_0^{2\pi} =0.$

Example 11. Take contour $C=C_1\cup C_2\cup C_3\cup C_3$ with $C_1=\{z\in\mathbb C\ |\ 0\le x\le 2,\ y=0\}$ , $C_2=\{ z\in\mathbb C\ |\ x=2,\ 0\le y\le 2\}$ , $C_3=\{ z\in\mathbb C\ |\ 0\le x\le 2,\ y=2\}$ and $C_4=\{ z\in\mathbb C\ |\ x=0,\ 0\le y\le 2\}$ . Take again $f(z)=z$ , $g(z)=\bar z$ , $h(z) = |z|$ .

The natural parameterizations for the contour segments are: $x$ for $C_1$ (on which $z=x+i0$ , $dz=dx$ ) and $C_3$ (on which $z=x+2i$ , $dz=dx$ ), and $y$ for $C_2$ (on which $z=2+iy$ , $dz=idy$ ) and $C_4$ (on which $z=0+yi$ , $dz=idy$ ) and remember the ANTI-CLOCK rule. We have $\begin{gathered} \int_C f(z) dz =\sum_{k=1}^4 \int_{C_k} f(z) dz \\ = \int_0^2 (x+i0) dx + \int_0^2 (2+iy) i dy+\int_2^0 (x+2i) dx + \int_2^0 (0+iy) idy= 0.\end{gathered}$ $\begin{gathered} \int_C g(z) dz =\sum_{k=1}^4 \int_{C_k} g(z) dz \\ = \int_0^2 (x-i0) dx + \int_0^2 (2-iy) i dy+\int_2^0 (x-2i) dx + \int_2^0 (0-iy) idy= 8i.\end{gathered}$ and $\begin{gathered} \int_C h(z) dz = \int_0^2 x dx + \int_0^2 \sqrt{4+y^2} i dy+\int_2^0 \sqrt{4+x^2} dx + \int_2^0 y idy \\= 2(1-\sqrt{2}-\log(1+\sqrt{2}))(1-i).\end{gathered}$

We now prove one of the most important result about contour integration.

Theorem 12 (Fundamental Theorem of (Complex) Calculus). Let $F(z)$ be analytic such that $F'(z)=f(z)$ and assume that $f$ is continuous on domain $\Omega$ . Let $C$ be a contour in $\Omega$ parameterized by $t$ , $a\le t\le b$ . Then $\int_C f(z) d z = F(z(b))-F(z(a)).$

Proof. Let’s first assume that the contour $C$ is smooth (that is, $z'(t)$ is continuous on $C$ ). Then we have $\begin{gathered} \int_C f(z) dz=\int_C F'(z) dz=\int_a^b F'(z) z'(t) dt = \int_a^b \dfrac{d F(z(t))}{dt} dt = F(z(b))-F(z(a)). \end{gathered}$ If $C$ is only piecewise continuous, we can do the same calculation on each piece of $C$ and add the results together. This completes the proof. ◻

This theorem says that if $F(z)$ is analytic, then the value of the complex integral of $f(z)$ on contours that connect two points in the domain is independent of the contour taken. Moreover, the integral over any closed contour is $0$ for a well-defined function $F(z)$ .

Remark 13. Here are a few quick remarks regarding the Fundamental Theorem of (Complex) Calculus.

We are NOT asking $f(z)$ to be analytic but that it is the derivative of an analytic function (although later on we will see that the two actually mean the same thing).
The continuity of $f(z)$ is NOT enough to guarantee the existence of such a $F(z)$ as we will see later in the semester.
$\int_C f(z) dz =0$ for any closed contour only if $F(z)$ is well-defined, that is, single-valued. If $F(z)$ is multivalued, $F(z(b))-F(z(a))$ is not necessarily $0$ even if $b=a$ . We will see an example in the next section.

Example 14. Take contour $C=\{z=(\cos\theta, \sin\theta)\ |\ 0\le \theta\le 2\pi\}$ and $f(z)=z$ . Then $\int_C f(z) dz=0$ . This is because $f(z)=F'(z)$ with $F(z)=z^2/2$ analytic everywhere and $f(z)$ continuous everywhere.

The method of contour deformation

Many times, we need to evaluate contour integrals for a function $f(z)$ for whom we can find a $F(z)$ such that $F'(z)$ exist everywhere but on a point (or a finite number of points) inside the domain. If the contour is open or is closed but does not enclose any point where $F'(z)$ does not exist, then we can use Theorem Theorem 12 to calculate the integral. Otherwise, if the contour is closed and enclose one or more points of non-analyticity of $F$ , then the method of contour deformation is very effective in evaluate the integral. We illustrate the idea in the following two examples.

Example 15. Let $f(z)=\dfrac{1}{z}$ . Then $F(z)=\log z$ has the property that $F'(z)=f(z)$ besides at $z=0$ . Therefore $F(z)$ is not analytic at $z=0$ . Let $C$ be a closed contour that enclose $z=0$ . If $C$ has a nice parameterization (such as the unit circle as we did before), we could perform some explicit calculations to find the integral. It turns out that even when we don’t know explicitly the parameterization for $C$ we could still figure out the integral. We deformed the contour as follows: (i) we introduce a circle around the origin that is totally enclosed in $C$ ; (ii) we then introduce a crosscut $(L_1, L_2)$ that connect the circle and $C$ in the manner as shown in Figure 2. Let $\varepsilon$ be the distance between line segments $L_1$ and $L_2$ , and $\widetilde C=C_2\cup C_1\cup L_1\cup L_2$ . It is then clear that $\int_{\widetilde C} f(z) dz=0=\int_{-C_1} f(z) dz+ \int_{L_1} f(z) dz +\int_{L_2} f(z) dz+ \int_{C_2} f(z) dz$ and $\int_C f(z) dz = \lim_{\varepsilon\to 0} \int_{C_2} f(z) dz$ It is clear that as $\varepsilon\to 0$ , $\int_{L_1} f(z) dz \to -\int_{L_2} f(z) dz$ since $L_1 \to - L_2$ . We therefore have that $\int_C f(z) dz = \lim_{\varepsilon\to 0} \int_{C_2} f(z) dz = - \lim_{\varepsilon\to 0}\int_{-C_1} f(z) dz= \lim_{\varepsilon\to 0}\int_{C_1} \frac{1}{z} dz= 2\pi i.$

$\Longrightarrow$

Remark 16. We can use the Fundamental Theorem of Calculus (Theorem Theorem 12) to find the integral as follows. Let $D=\{ z\ |\ |z|\le \varepsilon\}$ be a small disk that is enclosed in the closed contour $C$ . Then $F(z)=\log z$ is analytic in $D^c$ (complement of $D$ ). By the Fundamental Theorem of Calculus (Theorem Theorem 12), the integral of $f(z)$ over the contour $C$ is simply $\log(z(b))-\log(z(a))$ with $z(a)$ and $z(b)$ the starting and ending points of the contour respectively: $z(a)=z(b)$ since the contour is closed. When we travel anti-clockwise along $C$ , the argument of $z$ increases by $2\pi$ when we return to the starting point. Therefore $\log(z(b))-\log(z(a))=2\pi i$ using the fact that $\log z = \log|z|+i(\theta_0+2k\pi)$ (taking any branch would work).

Example 17. We can use the same trick to evaluate the integral of $f(z)=\dfrac{1}{z^2}$ over a closed contour that enclose the origin. In exactly the same way, we can verify that $\int_C f(z) dz = \lim_{r\to 0} \int_{C_2} f(z) dz = - \lim_{\varepsilon\to 0}\int_{-C_1} f(z) dz = \lim_{\varepsilon\to 0}\int_{C_1} \dfrac{1}{z^2} dz = \int_0^{2\pi} e^{-i2\theta} i e^{i\theta} d\theta = 0.$

Cauchy’s Theorem

From the application point of view, the Fundamental Theorem of Calculus is extremely useful since it provides easy ways to evaluate many contour integrals when the anti-derivatives of the integrands are known (and analytic). In the same spirit, the Cauchy’s Theorem is another important theorem that provides (sometimes even more) convenient ways to evaluate contour integrals.

Theorem 18 (Cauchy’s Theorem). Let $f(z)$ be analytic in a simply connected domain $\Omega$ . Then for any simple closed contour $C$ in $\Omega$ , we have $\int_C f(z) dz =0.$

Proof. (Based on the assumption that $f'(z)$ is continuous on $\Omega$ so that we can use Green’s theorem.) Let $f(z) = u + i v$ , $z=x+iy$ and $dz = dx + i dy$ . Then $\int_C f(z) dz = \int_C \left(u dx - v dy\right)+ i\int_C \left(u dy + v dx\right)=\int_C \left(u dx + (- v) dy\right)+ i\int_C \left (u dy + v dx\right)$ Let $\Omega'$ be the region enclosed in the contour $C$ . By the Green’s Theorem, which states $\int_C \left(u dx + v dy\right) =-\int_{\Omega'} (u_y-v_x) dx dy$ , we have $\int_C f(z) dz = \int_{\Omega'} (u_y + v_x) dx dy+ i\int_C (u_x - v_y) dx dy= 0.$ The last step follows from the Cauchy-Riemann equations since $f(z)$ is analytic. ◻

Remark 19. Cauchy’s Theorem requires that $f(z)$ be analytic. This automatically ensures that $f'(z)$ is continuous as we will see later. Therefore, our assumption on the continuity of $f'(z)$ in the proof is legimate.

Remark 20. For a non-simple closed contour with finite number of, say $M$ , self-intersections, we can decompose the contour into $M$ simple closed contours; see Figure 3 for an example with $M=2$ self-intersections. Cauchy’s Theorem can then be applied to each simple closed contour in the decomposition. This shows that Cauchy’s Theorem can be applied to non-simple closed contour of finite number of self-intersections.

Self-intersecting contour C decomposed into the three simple loops C1, minus C2, and C3. — Example of a non-simple closed contour $C$ that can be decomposed into $3$ simple closed contours: $C_1$ , $-C_2$ and $C_3$ .

Cauchy’s Theorem can be used to check whether or not an anti-derivative exists for a continuous function, as is illustrated in the following theorem

Theorem 21 (Morera’s Theorem). Let $f(z)$ be continuous in a simply connected domain $\Omega$ . If $\int_C f(z) dz=0$ for every simple closed contour $C$ in $\Omega$ , then there exists analytic function $F(z)$ in $\Omega$ such that $F'(z)=f(z)$ .

Proof. We construct $F(z)$ explicitly as we did in the case of real variable. Let $z_0$ , $z$ and $z+h$ be three points in $\Omega$ . We integrate $f(z)$ along a simple closed contour that is formed by there arcs: $z_0\to z+h$ , $z+h\to z$ and $z\to z_0$ . This gives $\int_{z_0}^{z+h} f(z') dz' + \int_{z+h}^{z} f(z') dz' + \int_{z}^{z_0} f(z') dz' =0$ by the assumption of the theorem. We now define $F(z) = \int_{z_0}^z f(z') d z'.$ Then the previous equation implies that $F(z+h)-F(z) = \int_{z_0}^{z+h} f(z') dz' - \int_{z_0}^{z} f(z') dz' = \int_{z}^{z+h} f(z') dz'.$ Therefore $\dfrac{F(z+h)-F(z)}{h} = \dfrac{ \int_{z}^{z+h} f(z') dz'}{h}.$ Taking $h\to 0$ and utilizing the fact that $f(z)$ is continuous, we find that $F'(z) = f(z).$ This concludes the proof. ◻

It is clear that since $F'(z)$ exists and is continuous, $F$ is analytic. As we will see very soon, if $F$ is analytic, then its derivatives are also analytic. This means that $f(z)$ is analytic. Therefore, Morera’s Theorem says that if $f(z)$ is continuous in a simply connected domain $\Omega$ and $\int_C f(z) dz=0$ for every simple closed contour $C$ in $\Omega$ , then $f$ is analytic.

Applications of Cauchy’s Theorem.

Cauchy’s Theorem can be used to find contour integrals of analytic functions. As a comparison between the Fundamental Theorem of Calculus and Cauchy’s Theorem, we emphasize the following.

Fundamental Theorem of Calculus: $f(z)$ being continuous + anti-derivative $F(z)$ being analytic.
Cauchy’s Theorem: $f(z)$ being analytic + $\Omega$ being simply connected.

We now give some examples to see how to utilize Cauchy’s Theorem.

Example 22. The integral of $z^n$ ( $n\ge 1$ ), $e^z$ , $\sin z$ , $\cos z$ and $ze^z$ etc along any closed contour on $\mathbb C$ is $0$ since these functions are all analytic on $\mathbb C$ .

Example 23. Let $f(z)=\dfrac{1}{z^n}$ ( $n\ge 1$ ). Then $f(z)$ is analytic everywhere but at $z=0$ .

(i) For any simple closed contour $C$ that does not enclose the origin, $\int_C f(z) dz=0$ by Cauchy’s Theorem.

(ii) When $C$ is a closed contour around the origin, we use the Method of Contour Deformation to find that $\int_C f(z) dz=\left\{ \begin{array}{cc} 2\pi i, & n=1\\ 0, & n\ge 2 \end{array} \right.$

Example 24. Let $f(z)=\dfrac{1}{(z-z_0)^n}$ ( $n\ge 1$ ). Then $f(z)$ is analytic everywhere but at $z=z_0$ .

(i) For any simple closed contour $C$ that does not enclose $z_0$ , $\int_C f(z) dz=0$ by Cauchy’s Theorem.

(ii) When $C$ is a closed contour around $z_0$ , we use the Method of Contour Deformation to find that $\int_C f(z) dz=\left\{ \begin{array}{cc} 2\pi i, & n=1\\ 0, & n\ge 2 \end{array} \right.$

Example 25. Let $f(z)=\dfrac{1}{(z-z_1)(z-z_2)}$ ( $z_1\neq z_2$ ). Then $f(z)$ is analytic everywhere but at $z=z_1$ and $z=z_2$ .

(i) For any simple closed contour $C$ that does not enclose $z_1$ and $z_2$ , $\int_C f(z) dz=0$ by Cauchy’s Theorem.

(ii) When $C$ is a closed contour enclosing $z_1$ , but not $z_2$ , we use the Method of Contour Deformation to find that $\int_C f(z) dz=\int_C \dfrac{1}{z_1-z_2} \left[ \dfrac{1}{z-z_1}-\frac{1}{z-z_2}\right] dz= \dfrac{2\pi i}{z_1-z_2},$ since $\dfrac{1}{z-z_2}$ is analytic in the region enclosed in the contour.

(iii) When $C$ is a closed contour enclosing both $z_1$ and $z_2$ , we use the Method of Contour Deformation to find that $\int_C f(z) dz=\dfrac{1}{z_1-z_2} \int_C \left[ \dfrac{1}{z-z_1}-\frac{1}{z-z_2}\right] dz = \dfrac{2\pi i-2\pi i}{z_1-z_2}=0.$

Guided review and additional examples

Learning goals

Parameterize piecewise smooth contours with the correct orientation.
Evaluate contour integrals from the definition.
Use antiderivatives when they are available.
Estimate integrals with the ML inequality.

Concept connection: the path contributes through $dz$

For a parameterization $z=z(t)$ , the contour integral becomes $\int f(z(t))z'(t)\,dt$ . Both the values of $f$ on the path and the direction encoded in $z'(t)$ matter. Reversing orientation multiplies the integral by $-1$ .

Worked example 1: integrate along a line segment

Let $C$ be the line from $0$ to $1+i$ , parameterized by $z(t)=t(1+i)$ for $0\le t\le1$ . Then

$\int_C z^2\,dz =\int_0^1 t^2(1+i)^3\,dt =\frac{(1+i)^3}{3} =\frac{-2+2i}{3}.$

The same answer follows from the antiderivative $z^3/3$ .

Worked example 2: a path-dependent integrand

On the positively oriented unit circle, $z=e^{it}$ and $dz=ie^{it}dt$ . Since $\overline z=e^{-it}$ on this circle,

$\int_C \overline z\,dz =\int_0^{2\pi}e^{-it}ie^{it}\,dt =2\pi i.$

The integrand $\overline z$ has no complex antiderivative on the plane, so endpoint evaluation is not available.

Check your understanding

If $|f(z)|\le M$ on a contour of length $L$ , what does the ML inequality say?

Show the answer

It gives $\left|\int_C f(z)\,dz\right|\le ML$ . The estimate is often more useful than an exact parameterization when one only needs to show that an integral tends to zero.