Complex Integration

The magic and power of calculus ultimately rests on the amazing fact that differentiation and integration are mutually inverse operations. And, just as complex functions enjoy remarkable differentiability properties not shared by their real counterparts, so the sublime beauty of complex integration goes far beyond its real progenitor.

Peter J. Oliver

Contour integral

Consider a contour $C$ parametrized by $z (t) = x (t) + i y (t)$ for $a \leq t \leq b .$ We define the integral of the complex function along $C$ to be the complex number $\begin{array}{r} (1) & \int_{C} f (z) d z = \int_{a}^{b} f (z (t)) z^{'} (t) d t . \end{array}$ Here we assume that $f (z (t))$ is piecewise continuous on the interval $a \leq t \leq b$ and refer to the function $f (z)$ as being piecewise continuous on $C .$ Since $C$ is a contour, $z^{'} (t)$ is also piecewise continuous on $a \leq t \leq b$ ; and so the existence of integral ( $1$ ) is ensured.

The right hand side of ( $1$ ) is an ordinary real integral of a complex-valued function; that is, if $w (t) = u (t) + i v (t),$ then $\begin{array}{r} (2) & \int_{a}^{b} w (t) d t = \int_{a}^{b} u (t) d t + i \int_{a}^{b} v (t) d t \end{array}$

Now let us write the integrand $f (z) = u (x, y) + i v (x, y)$ in terms of its real and imaginary parts, as well as the differential $d z = \frac{d z}{d t} d t = (\frac{d x}{d t} + i \frac{d y}{d t}) d t = d x + i d y$ Then the complex integral ( $1$ ) splits up into a pair of real line integrals:

\begin{array}{r} (3) & \int_{C} f (z) d z = \int_{C} (u + i v) (d x + i d y) = \int_{C} (u d x - v d y) + i \int_{C} (v d x + u d y) . \end{array}

Example 1: Let's evaluate $\int_{C} \overset{―}{z} d z,$ where $C$ is given by $x = 3 t, y = t^{2},$ with $- 1 \leq t \leq 4.$

Piecewise smooth curve — $C$ : $z (t) = 3 t + i t^{2},$ with $- 1 \leq t \leq 4.$

Here we have that $C$ is $z (t) = 3 t + i t^{2} .$ Therefore, with the identification $f (z) = \overset{―}{z}$ we have $f (z (t)) = \overset{―}{3 t + i t^{2}} = 3 t - i t^{2} .$ Also, $z^{'} (t) = 3 + 2 i t,$ and so the integral is $\begin{array}{rcl} \int_{C} \overset{―}{z} d z & = & \int_{- 1}^{4} (3 t - i t^{2}) (3 + 2 i t) d t \\ = & \int_{- 1}^{4} (2 t^{3} + 9 t + 3 t^{2} i) d t \\ = & \int_{- 1}^{4} (2 t^{3} + 9 t) d t + i \int_{- 1}^{4} 3 t^{2} d t \\ = & (\frac{1}{2} t^{4} + \frac{9}{2} t^{2}) |_{- 1}^{4} + i t^{3} |_{- 1}^{4} \\ = & 195 + 65 i . \end{array}$

Example 2: Now let's evaluate $\int_{C} \frac{1}{z} d z,$ where $C$ is the circle $x = \cos t, y = \sin t,$ with $0 \leq t \leq 2 π .$

In this case $C$ is $z (t) = \cos t + i \sin t = e^{i t},$ $f (z (t)) = \frac{1}{e^{i t}}$ and, $z^{'} (t) = i e^{i t} .$ Thus $\begin{array}{rcl} \int_{C} \frac{1}{z} d z & = & \int_{0}^{2 π} (e^{- i t}) i e^{i t} d t = i \int_{0}^{2 π} d t = 2 π i . \end{array}$

Numerical evaluation of complex integrals

Exploration 1

Use the following applet to explore numerically the integral $\int_{C} \overset{―}{z} d z$ with different contours $C$ :

Line segments.
Semicircles.
Circles, positively and negatively oriented.

You can also change the domain coloring plotting option. Drag the points around and observe carefully what happens. Then solve Exercise 1 below.

The arrows on the contours indicate direction.

Exercise 1: Use definition ( $1$ ) to evaluate $\int_{C} \overset{―}{z} d z,$ for the following contours $C$ from $z_{0} = - 2 i$ to $z_{1} = 2 i$ :

Line segment. That is, $z (t) = - 2 i (1 - t) + 2 i t,$ with $0 \leq t \leq 1.$
Right-hand semicircle. That is, $z (θ) = 2 e^{i θ}$ with $- \frac{π}{2} \leq θ \leq \frac{π}{2} .$
Left-hand semicircle. That is, $z (θ) = - 2 i e^{- i θ}$ with $0 \leq θ \leq π .$

Use the applet to confirm your results.

What conclusions (if any) can you draw about the function $\overset{―}{z}$ from this?

Exploration 2

Now use the applet below to explore numerically the integrals $\int_{C} (z^{2} + z) d z; \int_{C} \frac{1}{z^{2}} d z$ with different contours $C$ (line segments, semicircles, and circles). Drag the points around and observe carefully what happens. You can select the functions z^2+z or 1/z^2 from the list at the left-top corner. Then solve Exercises 2 and 3.

Exercise 2: Consider the integral $I_{1} = \int_{C} (z^{2} + z) d z .$ Use the applet to analyze the value of $I_{1}$ in the following cases:

$C$ is any contour from $z_{0} = - 1 - i$ to $z_{1} = 1 + i .$
$C$ is the circle with center $z_{0}$ and radius $r > 0,$ $| z - z_{0} | = r$ ; positively or negatively oriented. In this cases select Circle ↺ or Circle ↻.

What conclusions (if any) can you draw about the value of $I_{1}$ and the function $z^{2} + z$ from this?

Exercise 3: Now considering integral $I_{2} = \int_{C} \frac{1}{z^{2}} d z .$ First, in the applet select the function f(z)=1/z^2. Then analyze the values of $I_{2}$ in the following cases:

$C$ is any contour from $z_{0} = - i$ to $z_{1} = i .$ What happens when you select Line Segment in the applet? What happens when you select Semicircles?
$C$ is the circle with center $z_{0}$ and radius $r > 0,$ $| z - z_{0} | = r$ ; positively or negatively oriented. In this case select Circle ↺ or Circle ↻. What happens if $z = 0$ is inside or outside the circle? What happens if $z = 0$ lies on the contour, e.g. when $z_{0} = 1$ and $r = 1$ ?

What conclusions (if any) can you draw about the value of $I_{2}$ and the function $\frac{1}{z^{2}}$ from this?

Properties

All the following familiar properties of integrals can be proved directly from the definition given in ( $2$ ). If $w (t) = u (t) + i v (t)$ and $s (t) = υ (t) + i ν (t)$ are complex-valued functions of a real variable $t$ continuous on an interval $[a, b],$ then

\begin{array}{rcl} (4) & \int_{a}^{b} k w (t) d t & = & k \int_{a}^{b} w (t) d t, k is a complex constant, \\ (5) & \int_{a}^{b} [w (t) + s (t)] d t & = & \int_{a}^{b} w (t) d t + \int_{a}^{b} s (t) d t, \\ (6) & \int_{a}^{b} w (t) d t & = & \int_{a}^{c} w (t) d t + \int_{c}^{b} w (t) d t c \in [a, b], \\ (7) & \int_{a}^{b} w (t) d t & = & - \int_{b}^{a} w (t) d t . \end{array}

From definition ( $1$ ), and the properties just mentioned above, it also follows immediately that

\begin{array}{rcl} (8) & \int_{C} z_{0} f (z) d z & = & z_{0} \int_{C} f (z) d z, z_{0} is a complex constant, \\ (9) & \int_{C} [f (z) + g (z)] d z & = & \int_{C} f (z) d z + \int_{C} g (z) d z \end{array}

Now consider the contour defined for the integral ( $1$ ). The contour $- C$ consists of the same set of points but with the order reversed so that the new contour extends from $z_{2}$ to $z_{1},$ as shown in Figure 3.

Reversed contour — The contour $- C$ extends from $z_{2}$ to $z_{1} .$

The contour $- C$ has a parametric representation $z = z (- t)$ with $- b \leq t \leq - a .$ Hence

\begin{array}{r} \int_{- C} f (z) d z = \int_{- b}^{- a} f [z (- t)] \frac{d}{d t} z (- t) d t = - \int_{- b}^{- a} f [z (- t)] z^{'} (- t) d t . \end{array}

If we use the substitution $τ = - t$ in the last integral, then

\begin{array}{r} \int_{- C} f (z) d z = - \int_{a}^{b} f [z (τ)] z^{'} (τ) d t . \end{array}

Hence, we obtain the following property

\begin{array}{r} (10) & \int_{- C} f (z) d z = - \int_{C} f (z) d z . \end{array}

Consider now a path $C,$ parametrized by $z (t)$ for $t \in [a, b],$ that consists of a contour $C_{1}$ from $z_{1}$ to $z_{2}$ followed by a contour $C_{2}$ from $z_{2}$ to $z_{3},$ the initial point of $C_{2}$ being the final point of $C_{1} .$ Then

\begin{array}{r} (11) & \int_{C} f (z) d z = \int_{C_{1}} f (z) d z + \int_{C_{2}} f (z) d z . \end{array}

Exercise 4: Use the properties for integrals of functions $w (t)$ to prove ( $11$ ).

Finally we introduce an inequality involving contour integrals that is extremely important in various applications. Consider $C$ a smooth curve parametrized by $z (t) = x (t) + i y (t)$ with $t \in [a, b] .$ Suppose that the components $x^{'} (t)$ and $y^{'} (t)$ of the derivative $z^{'} (t) = x^{'} (t) + i y^{'} (t)$ are continuous on the entire interval $[a, b] .$ Then the real-valued function $| z^{'} (t) | = \sqrt{{[x^{'} (t)]}^{2} + {[y^{'} (t)]}^{2}}$ is integrable on $[a, b] .$ According to the definition of arc length from calculus, the length of $C$ is the number $L = \int_{a}^{b} | z^{'} (z) | d t .$

If $C$ is a contour of length $L$ and $f$ is a piecewise continuous function on $C,$ and $M$ is a nonnegative constant such that $| f (z) | \leq M$ for all points $z$ on $C$ at which $f (z)$ is defined, then $\begin{array}{r} (12) & | \int_{C} f (z) d z | \leq M L \end{array}$

Note that since $C$ is a contour and $f$ is piecewise continuous on $C,$ a number $M$ appearing in the inequality $| f (z) | \leq M$ will always exist. The reason is that the real-valued function $| f (z (t)) |$ is continuous on the closed bounded interval $[a, b]$ when $f$ is continuous on $C;$ and this function always reaches a maximum value $M$ on that interval. Therefore $| f (z) |$ has a maximum value on $C$ when $f$ is continuous on it. This is also true when $f$ is piecewise continuous on $C .$

The property ( $12$ ) is often used in the theory of complex integration and is sometimes referred to as the ML-inequality.

Example 3: Let $C$ be the contour of the circle $| z | = 2$ from $z_{1} = 2$ to $z_{2} = 2 i$ that lies in the first quadrant, as shown in Figure 5. We can use inequality ( $12$ ) to show that $| \int_{C} \frac{d z}{z^{2} - 1} | \leq \frac{π}{3} .$ without evaluating the integral.

ML-inequality example — Circle $| z | = 2$ from $z_{1} = 2$ to $z_{2} = 2 i .$

Note that if $z$ is a point of $C,$ then

\begin{array}{r} | z^{2} - 1 | \geq | | z^{2} | - 1 | = | {| z |}^{2} - 1 | = | 4 - 1 | = 3 \end{array}

Thus $| \frac{1}{z^{2} - 1} | = \frac{1}{| z^{2} - 1 |} \leq \frac{1}{3}$ Also the length of $C$ is $\frac{1}{4} (4 π) .$ Thus, if we take $M = \frac{1}{3}$ and $L = π,$ we find that $| \int_{C} \frac{d z}{z^{2} - 1} | \leq M L = \frac{π}{3} .$

Example 4: We can also use the ML-inequality to find the value of the integral $lim_{R \to \infty} \int_{C_{R}} \frac{z^{1 / 2}}{z^{2} + 1} d z$ where $C_{R}$ is the arc $z (t) = R e^{i t},$ with $t \in [0, π]$ and $z^{1 / 2}$ denotes the branch $z^{1 / 2} = \exp (\frac{1}{2} \log z) = \sqrt{r} e^{i t / 2}, (r > 0, - \frac{π}{2} < t < \frac{3 π}{2})$ of the square root function.

Note that when $| z | = R > 1,$ then $| z^{1 / 2} | = | \sqrt{R} e^{i t / 2} | = \sqrt{R}$ and $| z^{2} + 1 | \geq | | z^{2} | - 1 | = R^{2} - 1.$ Thus, at points on $C_{R}$ , we have

| \frac{z^{1 / 2}}{z^{2} + 1} | \leq M_{R} where M_{R} = \frac{\sqrt{R}}{R^{2} - 1} .

Since the length of $C_{R}$ is $L = π R,$ using ( $12$ ), we obtain

| \int_{C_{R}} \frac{z^{1 / 2}}{z^{2} + 1} d z | \leq M_{R} L = \frac{π R \sqrt{R}}{R^{2} - 1} \cdot \frac{1 / R^{2}}{1 / R^{2}} = \frac{\frac{π}{\sqrt{R}}}{1 - \frac{1}{R^{2}}} .

The term on the right tends to zero as $R$ tends to infinity. Hence $lim_{R \to \infty} \int_{C_{R}} \frac{z^{1 / 2}}{z^{2} + 1} d z = 0.$

Antiderivatives

While the value of a contour integral of a function $f (z)$ from a fixed point $z_{0}$ to a fixed point $z_{1}$ generally depends on the chosen path, there are certain functions for which the integrals from $z_{0}$ to $z_{1}$ have values that are independent of path, as you have seen in Exercises 2 and 3. These examples also illustrate the fact that the values of integrals around closed paths are sometimes, but not always, zero. The next theorem is useful in determining when integration is independent of path and, moreover, when an integral around a closed path has value zero. This theorem is known as the complex version of the Fundamental Theorem of Calculus and involves the concept of antiderivative (or primitive) of a continuous function $f (z)$ on a domain $D,$ that is, a function $F (z)$ such that $F^{'} (z) = f (z)$ for all $z \in D .$

Suppose that

f (z)

is continuous on a domain

D .

f (z)

has an antiderivative throughout

D,

then for any contour

C

lying entirely in

D

, with initial fixed point

z_{1}

and final fixed point

z_{2},

we have that

\begin{array}{r} (13) & \int_{C} f (z) d z = F (z) |_{z_{1}}^{z_{2}} = F (z_{2}) - F (z_{1}) . \end{array}

This follows from definition (

1

) and the chain rule. That is

\begin{array}{rcl} \int_{C} f (z) d z & = & \int_{C} F^{'} (z (t)) \frac{d z}{d t} d t \\ = & \int_{a}^{b} \frac{d}{d t} F (z (t)) d t \\ = & F (z (b)) - F (z (a)) \\ = & F (z_{2}) - F (z_{1}) \end{array}

where

z_{1} = z (a)

and

z_{2} = z (b)

are the endpoints of the contour

C .

As a consequence of the previous Theorem 1, for closed curves we have $\begin{array}{r} (14) & \int_{C} f (z) d z = \int_{C} F^{'} (z) d z = 0. \end{array}$ for every closed contour $C,$ that is, the endpoints are equal.

If the function $f (z)$ satisfies the hypothesis of Theorem 1, then for all contours $C$ lying in $D$ beginning at $z_{1}$ and ending at $z_{2}$ we have expression ( $13$ ). Hence the result demonstrates that the integral is independent of path. This fact is illustrated in Figure 6.

Path independence — Independent paths forming closed curve

Considering Figure 6, we have $\int_{C_{1}} f (z) d z = \int_{C_{2}} f (z) d z$ because

\begin{array}{r} \int_{C} f (z) d z = \int_{C_{1}} f (z) d z - \int_{C_{2}} f (z) d z = 0 \end{array}

where the closed contour is

C = C_{1} - C_{2} .

In other words:

f

is continuous and has an antiderivative throughout a domain

D,

then

\int_{C} f (z) d z

is independent of path.

In addition, we have the following sufficient condition for the existence of an antiderivate:

f

is continuous and

\int_{C} f (z) d z

is independent of path, then

f

has an antiderivate everywhere in

D .

Consider the function

\begin{array}{r} (15) & F (z) = \int_{z_{0}}^{z} f (s) d s \end{array}

where

s

denotes a complex variable,

z_{0}

is a fixed points in

D,

and

z

represents any point in

D .

Assuming that

f

is continuous and

\int_{C} f (z) d z

is independent of path, we wish to prove that

F^{'} (z) = f (z),

that is, (

15

) is an antiderivate of

f

D .

Using properties of the integral we have

\begin{array}{r} (16) & F (z + Δ z) - F (z) = \int_{z_{0}}^{z + Δ z} f (s) d s - \int_{z_{0}}^{z} f (s) d s = \int_{z}^{z + Δ z} f (s) d s . \end{array}

Since $D$ is a domain, we can choose $Δ z$ so that $z + Δ z$ is in $D .$ Furthermore, $z$ and $z + Δ z$ can be joined by a straight segment, as shown un Figure 7. This is the contour we use in the last integral in ( $16$ ). For $z$ fixed, we can write

\begin{array}{r} (17) & f (z) Δ z = f (z) \int_{z}^{z + Δ z} d s = \int_{z}^{z + Δ z} f (z) d s or f (z) = \frac{1}{Δ z} \int_{z}^{z + Δ z} f (z) d s . \end{array}

Proof — The points $z$ and $z + Δ z$ can be joined by a straight segment.

From ( $16$ ) and ( $17$ ) we obtain

\begin{array}{r} \frac{F (z + Δ z) - f (z)}{Δ z} = \frac{1}{Δ z} \int_{z}^{z + Δ z} [f (s) - f (z)] d s . \end{array}

\begin{array}{rcl} | \frac{F (z + Δ z) - f (z)}{Δ z} | & = & | \frac{1}{Δ z} \int_{z}^{z + Δ z} [f (s) - f (z)] d s | \\ = & | \frac{1}{Δ z} | | \int_{z}^{z + Δ z} [f (s) - f (z)] d s | \\ \leq & | \frac{1}{Δ z} | ε | Δ z | = ε . \end{array}

Therefore, we have proved that

\begin{array}{r} lim_{Δ z \to 0} \frac{F (z + Δ z) - f (z)}{Δ z} = f (z) or F^{'} (z) = f (z) . \end{array}