### 1 Complex integrals

If $f\left(z\right)$ is a single-valued, continuous function in some region $R$ in the complex plane then we define the integral of $f\left(z\right)$ along a path $C$ in $R$ (see Figure 7) as

$\phantom{\rule{2em}{0ex}}{\int }_{C}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\int }_{C}\left(u+iv\right)\left(dx+i\phantom{\rule{0.3em}{0ex}}dy\right)$ .

Figure 7

Here we have written $f\left(z\right)$ and $dz$ in real and imaginary parts:

$\phantom{\rule{2em}{0ex}}f\left(z\right)=u+iv\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}dz=dx+i\phantom{\rule{0.3em}{0ex}}dy.$

Then we can separate the integral into real and imaginary parts as

$\phantom{\rule{2em}{0ex}}{\int }_{C}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\int }_{C}\left(u\phantom{\rule{0.3em}{0ex}}dx-v\phantom{\rule{0.3em}{0ex}}dy\right)+i{\int }_{C}\left(v\phantom{\rule{0.3em}{0ex}}dx+u\phantom{\rule{0.3em}{0ex}}dy\right).$

We often interpret real integrals in terms of area; now we define complex integrals in terms of line integrals over paths in the complex plane. The line integrals are evaluated as described in HELM booklet  29.

##### Example 10

Obtain the complex integral:

${\int }_{C}z\phantom{\rule{0.3em}{0ex}}dz$

where $C$ is the straight line path from $z=1+i$ to $z=3+i$ . See Figure 8.

Figure 8

##### Solution

Here, since $y$ is constant ( $y=1$ ) along the given path then $z=x+i$ , implying that $u=x$ and $v=1$ . Also, as $y$ is constant, $dy=0$ .

Therefore,

$\phantom{\rule{2em}{0ex}}\begin{array}{ccc}\hfill {\int }_{C}z\phantom{\rule{0.3em}{0ex}}dz& \hfill =\hfill & {\int }_{C}\left(u\phantom{\rule{0.3em}{0ex}}dx-v\phantom{\rule{0.3em}{0ex}}dy\right)+i{\int }_{C}\left(v\phantom{\rule{0.3em}{0ex}}dx+u\phantom{\rule{0.3em}{0ex}}dy\right)\hfill \\ & & \\ \hfill & \hfill =\hfill & {\int }_{1}^{3}x\phantom{\rule{0.3em}{0ex}}dx+i{\int }_{1}^{3}1\phantom{\rule{0.3em}{0ex}}dx\hfill \\ & & \\ \hfill & \hfill =\hfill & {\left[\frac{{x}^{2}}{2}\right]}_{1}^{3}+i\left[\rightx{\left]\right}_{1}^{3}\hfill \\ & & \\ \hfill & \hfill =\hfill & \left(\frac{9}{2}-\frac{1}{2}\right)+i\left(3-1\right)=4+2i.\hfill \\ \hfill \end{array}$

Evaluate ${\int }_{{C}_{1}}z\phantom{\rule{0.3em}{0ex}}dz$ where ${C}_{1}$ is the straight line path from $z=3+i$ to $z=3+3i$ .

First obtain expressions for $u,v,dx$ and $dy$ by finding an appropriate expression for $z$ along the path:

Along the path $z=3+iy$ , implying that $u=3$ and $v=y$ . Also $dz=0+idy$ . Now find limits on $y$ :

The limits on $y$ are: $\phantom{\rule{1em}{0ex}}y=1$ to $y=3.$ Now evaluate the integral:

$\phantom{\rule{2em}{0ex}}\begin{array}{ccc}\hfill {\int }_{{C}_{1}}z\phantom{\rule{0.3em}{0ex}}dz& \hfill =\hfill & {\int }_{{C}_{1}}\left(u\phantom{\rule{0.3em}{0ex}}dx-v\phantom{\rule{0.3em}{0ex}}dy\right)+i{\int }_{{C}_{1}}\left(v\phantom{\rule{0.3em}{0ex}}dx+u\phantom{\rule{0.3em}{0ex}}dy\right)\hfill \\ & & \\ \hfill & \hfill =\hfill & {\int }_{1}^{3}-y\phantom{\rule{0.3em}{0ex}}dy+i{\int }_{1}^{3}3\phantom{\rule{0.3em}{0ex}}dy\hfill \\ & & \\ \hfill & \hfill =\hfill & {\left[\frac{-{y}^{2}}{2}\right]}_{1}^{3}+i\left[\right3y{\left]\right}_{1}^{3}=\left(-\frac{9}{2}+\frac{1}{2}\right)+i\left(9-3\right)\hfill \\ & & \\ \hfill =-4+6i.\end{array}$

Evaluate ${\int }_{{C}_{2}}z\phantom{\rule{0.3em}{0ex}}dz$ where ${C}_{2}$ is the straight line path from $z=1+i$ to $z=3+3i$ .

We first need to find the equation of the line ${C}_{2}$ in the Argand plane.

We note that both points lie on the line $y=x$ so the complex equation of the straight line is $z=x+ix$ giving $u=x$ and $v=x$ . Also $dz=dx+idx=\left(1+i\right)dx$ .

$\therefore \phantom{\rule{2em}{0ex}}{\int }_{{C}_{2}}z\phantom{\rule{0.3em}{0ex}}dz={\int }_{{C}_{2}}\left(x\phantom{\rule{0.3em}{0ex}}dx-x\phantom{\rule{0.3em}{0ex}}dx\right)+i{\int }_{{C}_{2}}\left(x\phantom{\rule{0.3em}{0ex}}dx+x\phantom{\rule{0.3em}{0ex}}dx\right)$ .

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=i{\int }_{{C}_{2}}\left(2x\phantom{\rule{0.3em}{0ex}}dx\right)$

Next, we see that the limits on $x$ are $x=1$ to $x=3$ . We are now in a position to evaluate the integral:

$\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{\int }_{{C}_{2}}z\phantom{\rule{0.3em}{0ex}}dz=i{\int }_{1}^{3}2x\phantom{\rule{0.3em}{0ex}}dx=i\left[\right{x}^{2}{\left]\right}_{1}^{3}=i\left(9-1\right)=8i.$

Note that this result is the sum of the integrals along $C$ and ${C}_{1}$ . You might have expected this.

A more intricate example now follows.

##### Example 11

Evaluate ${\int }_{{C}_{1}}{z}^{2}\phantom{\rule{0.3em}{0ex}}dz$ where ${C}_{1}$ is that part of the unit circle going anticlockwise from the point $z=1$ to the point $z=i$ . See Figure 9.

Figure 9

##### Solution

First, note that $\phantom{\rule{2em}{0ex}}{z}^{2}={\left(x+iy\right)}^{2}={x}^{2}-{y}^{2}+2xyi$ and $dz=dx+i\phantom{\rule{1em}{0ex}}dy$ giving

$\phantom{\rule{2em}{0ex}}{\int }_{{C}_{1}}{z}^{2}\phantom{\rule{0.3em}{0ex}}dx={\int }_{{C}_{1}}\left\{\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0.3em}{0ex}}dx-2xy\phantom{\rule{0.3em}{0ex}}dy\right\}+i{\int }_{{C}_{1}}\left\{2xy\phantom{\rule{0.3em}{0ex}}dx+\left({x}^{2}-{y}^{2}\right)dy\right\}.$

This is obtained by simply expressing the integral in real and imaginary parts. These integrals cannot be evaluated in this form since $y$ and $x$ are related. Instead we re-write them in terms of the single variable $\theta$ .

Note that on the unit circle: $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}x=cos\theta ,\phantom{\rule{1em}{0ex}}y=sin\theta$  so that   $dx=-sin\theta \phantom{\rule{0.3em}{0ex}}d\theta$   and   $dy=cos\theta \phantom{\rule{0.3em}{0ex}}d\theta$ .

The expressions $\left({x}^{2}-{y}^{2}\right)$ and $2xy$ can be expressed in terms of $2\theta$ since

$\phantom{\rule{2em}{0ex}}{x}^{2}-{y}^{2}={cos}^{2}\theta -{sin}^{2}\theta \equiv cos2\theta \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}2xy=2cos\theta sin\theta \equiv sin2\theta$ .

Now as the point $z$ moves from $z=1$ to $z=i$ along the path ${C}_{1}$ the parameter $\theta$ changes from $\theta =0$ to $\theta =\frac{\pi }{2}$ . Hence,

${\int }_{{C}_{1}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz={\int }_{0}^{\frac{\pi }{2}}\left\{-cos2\theta sin\theta \phantom{\rule{0.3em}{0ex}}d\theta -sin2\theta cos\theta \phantom{\rule{0.3em}{0ex}}d\theta \right\}+i{\int }_{0}^{\frac{\pi }{2}}\left\{-sin2\theta sin\theta \phantom{\rule{0.3em}{0ex}}d\theta +cos2\theta cos\theta \phantom{\rule{0.3em}{0ex}}d\theta \right\}.$

We can simplify these daunting-looking integrals by using the trigonometric identities:

$\phantom{\rule{2em}{0ex}}sin\left(A+B\right)\equiv sinAcosB+cosAsinB\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}cos\left(A+B\right)\equiv cosAcosB-sinAsinB$ .

We obtain (choosing $A=2\theta$ and $B=\theta$ in both expressions):

$\phantom{\rule{2em}{0ex}}-cos2\theta sin\theta -sin2\theta cos\theta \equiv -\left(sin\theta cos2\theta +cos\theta sin2\theta \right)\equiv -sin3\theta$ .

Also $\phantom{\rule{2em}{0ex}}-sin2\theta sin\theta +cos2\theta cos\theta \equiv cos3\theta .$

Now we can complete the evaluation of our integral:

$\begin{array}{ccc}\hfill {\int }_{{C}_{1}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dx& \hfill =\hfill & {\int }_{0}^{\frac{\pi }{2}}\left(-sin3\theta \right)d\theta +i{\int }_{0}^{\frac{\pi }{2}}cos3\theta \phantom{\rule{0.3em}{0ex}}d\theta \hfill \\ & & \\ \hfill & \hfill =\hfill & {\left[\frac{1}{3}cos3\theta \right]}_{0}^{\frac{\pi }{2}}+i{\left[\frac{1}{3}sin3\theta \right]}_{0}^{\frac{\pi }{2}}=\left(0-\frac{1}{3}\right)+i\left(-\frac{1}{3}-0\right)=-\frac{1}{3}-\frac{1}{3}i\equiv -\frac{1}{3}\left(1+i\right).\hfill \end{array}$

In the last Task we integrated ${z}^{2}$ over a given path. We had to perform some intricate mathematics to get the value. It would be convenient if there was a simpler way to obtain the value of such complex integrals. This is explored in the following Tasks.

Evaluate ${\left[\frac{1}{3}{z}^{3}\right]}_{1}^{i}$

We obtain $-\frac{1}{3}\left(1+i\right)$ again, which is the same result as from the previous Task.

It would seem that, by carrying out an analogue of real integration (simply integrating the function and substituting in the limits) we can obtain the answer much more easily. Is this coincidence?

$\phantom{\rule{2em}{0ex}}\begin{array}{ccc}\hfill {\left[\frac{1}{2}{z}^{2}\right]}_{1+i}^{3+3i}& \hfill =\hfill & \frac{1}{2}\left\{{\left(3+3i\right)}^{2}-{\left(1+i\right)}^{2}\right\}\hfill \\ & & \\ \hfill & \hfill =\hfill & \frac{1}{2}\left\{9+18i-9-1-2i+1\right\}\hfill \\ & & \\ \hfill & \hfill =\hfill & \frac{1}{2}\left(16i\right)=8i,\hfill \\ \hfill \end{array}$

the result we obtained earlier.

We shall investigate these ‘coincidences’ in Section 26.5.

As a variation on this example, suppose that the path ${C}_{1}$ is the entire circumference of the unit circle travelled in an anti-clockwise direction. The limits are $\theta =0$ and $\theta =2\pi$ . Hence

$\begin{array}{rcll}{\int }_{{C}_{1}}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz& =& {\int }_{0}^{2\pi }\left(-sin3\theta \right)d\theta +i{\int }_{0}^{2\pi }cos3\theta \phantom{\rule{0.3em}{0ex}}d\theta \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& {\left[\frac{1}{3}cos3\theta \right]}_{0}^{2\pi }+i{\left[\frac{1}{3}sin3\theta \right]}_{0}^{2\pi }& \text{}\\ & =& \left(\frac{1}{3}-\frac{1}{3}\right)+i\left(0-0\right)=0.& \text{}\\ & & & \text{}\end{array}$

Is there an underlying reason for this result? (We shall see in Section 26.5.)

Another technique for evaluating integrals taken around the unit circle is shown in the next example, in which we need to evaluate

$\phantom{\rule{2em}{0ex}}{∮}_{C}\frac{1}{z}dz$ where $C$ is the unit circle.

Note the use of $∮$ since we have a closed path; we could have used this notation earlier.

Evaluate ${∮}_{C}\frac{1}{z}dz$ where $C$ is the unit circle.

First show that a point $z$ on the unit circle can be written $z={\text{e}}^{i\theta }$ and hence find $dz$ in terms of $\theta$ :

On the unit circle a point $\left(x,y\right)$ is such that $x=cos\theta ,\phantom{\rule{1em}{0ex}}y=sin\theta$ and hence $z=cos\theta +isin\theta$ which, using De Moivre’s theorem, can be seen to be $z={\text{e}}^{i\theta }.$

. Now evaluate the integral ${∮}_{C}\frac{1}{z}dz$ .

$\phantom{\rule{2em}{0ex}}{∮}_{C}\frac{1}{z}dz={\int }_{0}^{2\pi }\frac{1}{{\text{e}}^{i\theta }}i{\text{e}}^{i\theta }d\theta ={\int }_{0}^{2\pi }id\theta =2\pi i.$

We now quote one of the most important results in complex integration which incorporates the last result.

##### Key Point 1

If $n$ is an integer and $C$ is the circle centre $z={z}_{0}$ and radius $r$ , that is, it has equation $\phantom{\rule{1em}{0ex}}\left|z-{z}_{0}\right|=r$ then

$\phantom{\rule{2em}{0ex}}{∮}_{C}\frac{dz}{{\left(z-{z}_{0}\right)}^{n}}=\left\{\begin{array}{cc}\hfill 0,\hfill & \hfill n\ne 1;\hfill \\ \hfill 2\pi i,\hfill & \hfill n=1.\hfill \end{array}\right\$

Note that the result is independent of the value of $r$ .