*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****)**) d**q****= 4i****ò**_{p}_{/2}^{-}^{p}^{/2}^{ }d**q**
= 4 i(

**p****/2 - (-****p****/2)) = 4i (****p****) = 4****p****i**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**

**p**

**i**can also be written:

I
=

**ò**_{C}^{ }dz/ z =**p****i**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 -i3x

^{2}(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 –i3x

^{2}) 1 dx =

**ò**

_{0}

^{1 }(1 – x)dx – 3i

**ò**

_{0}

^{1 }x

^{2}dx = 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 =

**ò**

_{0}

^{1 }(x –i3x

^{2}) 1 dx

Or:

**ò**

_{0}

^{1 }( x)dx – 3i

**ò**

_{0}

^{1 }x

^{2}dx = 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!

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