## Saturday, November 23, 2013

### Solution to Contour Integration problems 1) C is the right hand half of the circle  êz ê = 2  from z = -2 i to z = 2 i. Hence we want to find the integral of:
I = òC  z* dz
For which z = 2 exp (i q)  (- p/2 <   q  <  p/2)
Then:
I=  òp/2 -p/2  (2 exp(i q)) (2 exp(- i q))  dq  =  4i   òp/2 -p/2   dq
= 4 i(p/2  - (-p/2)) = 4i   (p)  =  4 pi

Note that for such a point z on the circle êz ê  = 2 , it follows that zz* = 4 or z* = 4/z. So that the result  4 pi can also be written:

I = òC  dz/ z =  pi

2) We used the revised contour shown in the graphic which will consist of two parts: C1 and C2. For C1 we will have:

òC1  f(z) dz  =   òOA  f(z) dz  +    òAB  f(z) dz

And for C2:

òC2  f(z) dz  =   òOC  f(z) dz  +   ò CB  f(z) dz

Where f(z) =  y – x   -i3x2   (z = x + iy)

The segment OA can be represented parametrically as:   z = 0 + iy (0 < y < 1)

And since x = 0 at all points on that segment the values of f there vary with the parameter y according to the equation:

f(z) = y(0 < y < 1)

Therefore:

ò OA  f(z) dz   =     ò 0 1   y idy = i ò 0 1   y dy = i/2

Meanwhile, on the segment AB, z = x + i(0 < x < 1) so that:

ò AB  f(z) dz   =    =     ò 0 1   (1 – x –i3x2) 1 dx =
ò 0 1   (1 – x)dx    3i ò 0 1    x2dx = ½ - i
Turning attention to contour C2: The segment OC can be represented parametrically as:
z = 0 + ix (0 < x < 1)
ò OC  f(z) dz     =     ò0 1   (x –i3x2) 1 dx
Or:
ò 0 1   ( x)dx    3i ò 0 1    x2dx = ½ - i
The vertical segment CB can be represented as:
z = 0 + iy (0 < y < 1)

And since x = 0 at all points on that segment the values of f there vary with the parameter y according to:

f(z) = y(0 < y < 1)
Therefore, we have:
ò CB  f(z) dz   =     ò 0 1   y idy = i ò 0 1   y dy = i/2

Then:   òOC  f(z) dz  +    òCB  f(z) dz   =  ½ - i + i/2  =  ½ - i/2

Finally:

òC1  f(z) dz   - òC2  f(z) dz    =   ½ - i/ 2  - [½ - i/2]  =  0
Is this result surprising? It shouldn't be if you examine the contours in the diagram!