Re: 7.1 Contour Integrals - Help needed!

billroot wrote:

What is the line from a to b meant to represent? Why not a straight line? I tried substituting simple complex values for points a and b, doing the integration, applying the limits and I always get the same value for the integral regardless of whether I go straight from a to b or via some other artbitary points, chosen to go different ways round the origin. When I look more carefully at fig 7.3. I notice that Prof. Penrose has not drawn a circle round the origin from b which is what I would expect, the last contour in fig 7.3(d) has point b well inside it. What am I missing here?

Without an understanding of this I won't understand the solution to 7.1 or what comes after.

Quote:

What is the line from a to b meant to represent?

The line from

to

is meant to represent the path of integration. In the theory of integration for real variables you always integrate along the real x-axis, but as soon as you go into the complex domain you have a dilemma: how to go from

to

?

Quote:

Why not a straight line?

Of course you could draw a straight line but Penrose has avoided that because it doesn't look general enough - it would be

a particular path from

to

, and he's trying to talk generally here.

In diagram 7.3(b) - top right - instead of going directly to

from

the path goes round the origin once to get to

.

Now, you could say, "but point

in diagram (b) is the same point as point

in diagram (a)".

This is correct in purely geometric terms, but when you are doing integration, you are moving around the curve in small increments and

is changing continuously as you go round the origin.

So, if you think of the initial value of

as

, then after you've gone round the origin once and back to the same point the value of

is now

.

That means that whatever the value was for the integral in diagram (a), let's say

, then it's value in diagram (b) is

, because the value of

has increased by

.

In diagram (c), Penrose shows the integration continuing from

back to

. The value of this integral from

to

must be the negative of the original integral in diagram (a) since the path of integration has been reversed, and so its value must be

. So the total integral round a

closed contour enclosing the origin from

and back to

must be

.

This is true whatever values we choose for and . (Added by vasco on 25th January 2010 at 17:22 GMT).Quote:

I notice that Prof. Penrose has not drawn a circle round the origin from b which is what I would expect, the last contour in fig 7.3(d) has point b well inside it. What am I missing here?

The point here is that

and

represent two completely arbitrary points in the complex plane and therefore the squid-shaped contour in diagram (c) might as well be any shape you like and pass through any points you like, as long as it encloses the origin. You can continuously deform the contour into any shape you like without crossing over the origin and the integral will still be equal to

.

I hope this helps. If you need any further clarification then please post again.

Vasco