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 = ò_C  z* dz

For which z = 2 exp (i q)  (- p/2 <   q  <  p/2)

Then:

I=  ò_p/2 ^-p/2  (2 exp(i q)) (2 exp(- i q))  dq  =  4i   ò_p/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 = ò_C  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:

ò_C1  f(z) dz  =   ò_OA  f(z) dz  +    ò_AB  f(z) dz  

 And for C2:

ò_C2  f(z) dz  =   ò_OC  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:   ò_OC  f(z) dz  +    ò_CB  f(z) dz   =  ½ - i + i/2  =  ½ - i/2

  Finally:

   ò_C1  f(z) dz   - ò_C2  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!