]> 20.6 Integration Over Curves in the Complex Plane

## 20.6 Integration Over Curves in the Complex Plane

The definite integral considered so far represents area in the $x y$ plane. We defined it however by dividing the interval in $x$ into small subintervals (say of width $d$ ) and taking the sum of an estimate of the area of each subinterval, namely $f ( x ' ) d$ where $x '$ is a point in that subinterval.

Suppose now we let $C$ be a curve in the complex plane, and let $f$ be a function of the variable $z$ with $z = x + i y$ . We can define integration along that curve of an integrand $f ( z )$ by dividing the curve into small pieces, and summing $f ( z ' ) ( z i − z i − 1 )$ where $z '$ is a point in the i-th interval whose endpoints are $z i$ and $z i − 1$ , over all the pieces.

This integral along a curve will no longer represent area, since neither $f$ nor the difference between $z$ 's will be real numbers. But we can multiply complex numbers together, so that this definition makes perfect sense.

Such an entity is called a contour integral in the complex plane. Though this integral no longer has the interpretation of area, it still has the property that it is well defined if the path $C$ is finite and the function $f$ is bounded and continuous on it. We denote it as follows

$∫ C f ( z ) d z$

Integrals of this kind are enormously valuable mathematical tools, as we may soon see.

Do not be put off by the strangeness of integrating in the complex plane and dealing with functions that have complex values. Almost everything you can say about ordinary integrals apply to these.

Exercises:

20.3 Integrate the function $sin ⁡ z$ up the imaginary axis, from 0 to $i$ using what you know about the integral of the sine function.

20.4 Do the same thing numerically on a spreadsheet, writing $sin ⁡ z = e i z − e − i z 2 i$ and $z = x + i y$ , with $x = 0$ (and hence $z = i y$ ), $y$ going from 0 to 1.