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:
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
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!
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