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Introduction to complex analysis
Introduction to complex analysis

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Session 2: Integration

Introduction to integration

This session introduces complex integration, an important concept which gives complex analysis its special flavour. We spend most of this session setting up the complex integral, deriving its main properties, and illustrating various techniques for evaluating it.

To define the integral of a complex function, it is instructive to first consider real integrals, such as

integral over a under b x squared d x equals one divided by three times left parenthesis b cubed minus a cubed right parenthesis comma

where a less than b, which represents the area of the shaded part of Figure 1 (for a greater than zero). We can express this equation in words by saying that

the integral of the function f of x equals x squared over the interval left square bracket a comma b right square bracket is one divided by three times left parenthesis b cubed minus a cubed right parenthesis.

Described image
Figure 1 Area under the graph of y equals x squared between a and b

Suppose now that we wish to integrate the complex function f of z equals z squared between two points alpha and beta in the complex plane. To do this, we first need to specify exactly how to get from alpha to beta. We could, for example, choose the line segment normal cap gamma from alpha to beta, as shown in Figure 2. It turns out (as you will see later) that if we make this choice, then

the integral of the function f of z equals z squared along the line segment from alpha to beta is one divided by three times left parenthesis beta times cubed minus alpha cubed right parenthesis.

We write this as

integral over normal cap gamma z squared d z equals one divided by three times left parenthesis beta times cubed minus alpha cubed right parenthesis full stop
Described image
Figure 2 Line segment normal cap gamma from alpha to beta

But there are many other paths in the complex plane from alpha to beta, which raises the following question. Do we get the same answer if we integrate the function f of z equals z squared along a different path from alpha to beta?

In order to address this question, we first need to explain exactly what it means to ‘integrate a function along a path’. This is one of the objectives of Section 1, where we briefly review the Riemann integral from real analysis, and then use similar ideas to construct the integral of a complex function along a path in the complex plane. We will see that if f is a complex function that is continuous on a smooth path normal cap gamma colon gamma of t left parenthesis t element of left square bracket a comma b right square bracket right parenthesis in the complex plane, then the integral of f along normal cap gamma, denoted by integral over normal cap gamma f of z d z, is given by the formula

integral over normal cap gamma f of z d z equals integral over a under b f of gamma of t times gamma times super prime times left parenthesis t right parenthesis d t full stop

We can evaluate this integral by splitting f of gamma of t times gamma times super prime times left parenthesis t right parenthesis into its real and imaginary parts u of t and v of t, and evaluating the resulting pair of real integrals:

integral over a under b f of gamma of t times gamma times super prime times left parenthesis t right parenthesis d t equals integral over a under b u of t d t plus i times integral over a under b v of t d t full stop

Section 2 begins with this definition of the integral of a complex function along a smooth path, and then extends the idea to allow integration along a contour – a finite sequence of smooth paths laid end to end.

In Section 3 we prove the Fundamental Theorem of Calculus, which shows that integration and differentiation are essentially inverse processes. From this result it follows that the integral of f of z equals z squared along any contour from alpha to beta is one divided by three times left parenthesis beta times cubed minus alpha cubed right parenthesis.

We will need to be careful about how we apply results such as the Fundamental Theorem of Calculus. For example, suppose that the endpoints alpha and beta of normal cap gamma coincide, as illustrated in Figure 3. Then

equation sequence part 1 integral over normal cap gamma z squared d z equals part 2 one divided by three times left parenthesis beta times cubed minus alpha cubed right parenthesis equals part 3 zero full stop

In this case, the integral of the function f of z equals z squared along normal cap gamma is zero.

Described image
Figure 3 Contour with initial point alpha and final point beta coinciding

Consider now the function f of z equals one solidus z. We will see later in Example 4 that if we integrate f along the smooth paths normal cap gamma sub one and normal cap gamma sub two shown in Figure 4, where normal cap gamma sub one and normal cap gamma sub two are circles traversed once anticlockwise, then

integral over normal cap gamma sub one one divided by z d z equals zero comma but integral over normal cap gamma sub two one divided by z d z equals two times pi times i full stop

The reason for this difference will become apparent in Section 3.

Described image
Figure 4 Circular paths normal cap gamma sub one and normal cap gamma sub two

This OpenLearn course is an extract from the Open University course M337 Complex analysis [Tip: hold Ctrl and click a link to open it in a new tab. (Hide tip)] .