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Contour integration can be loads of fun: constructing assorted loops, contours or connected segments in z-space (complex space - see the blogs on complex functions) then trying to graph them.

Contour integrals are integrals of complex functions f of
the complex variable z = x + iy. Such integrals are defined in terms of the
values of f(z) along a given contour C extending from a point z = z1 to a point
say, z = z2 in the complex plane. It is
therefore a line integral. This can be written as:

ò

_{C}^{ }f(z) dz
It may alternatively be written with limits, say from z1 to
z2, usually when the value of the integral is independent of the choice of
contour taken between the two end points.

Illustration: Suppose that the equation:

z = z(t) (a

__<__t__<__b)
represents
a contour C extending from some point z1 = z(a) to a point z2 = z(b). let the
function f(z) be piecewise continuous on C. That is, f(z(t)) is piecewise continuous on the interval a

__<__t__<__b. We define the line integral or contour integral of f along C as:**ò**

^{ }f(z) dz =

**ò**

_{a}

^{b }f[z(t)] z’(t) dz

Since C is a contour then z’(t), i.e. dz/dt is piecewise continuous on the interval a

__<__t

__<__b so the existence of the integral is assured.

__Example (1)__

This
is by reference to the contour shown on the graphic’s left side. Here C is the
right hand half of the circle êz ê = 3.8 from z = -3.8i to z = 3.8i. Hence we
want to find the integral of:

I
=

**ò**_{C}^{ }z* dz**For which z = 3.8 exp (i**

**q**

**) (-**

**p**

**/2**

__<__**q**

__<__**p**

**/2)**

Then:

I=

**ò**_{p}_{/2}^{-}^{p}^{/2}^{ }(3.8 exp(**i****q****))**(3.8 exp(-**i****q****)**) d**q****= 14.44i****ò**_{p}_{/2}^{-}^{p}^{/2}^{ }d**q**

= 14.44 (

**p****/2 - (-****p****/2)) = 14.44i (****p****) = 14.44****p****i**

Note that for such a point z on the circle êz ê = 3.8, it follows that zz* = 14.44 or z* =
14.44/z. So that the result

**14.44 p****i****can also be written:**
I
=

**ò**_{C}^{ }dz/ z =**p****i**
Example(2):

The
next contour is shown in the graphic to the right, along the path OAB, so the
integral can be solved as follows:

**ò**

_{C}

^{ }f(z) dz =

**ò**

_{OA}

^{ }f(z) dz +

**ò**

_{AB}

^{ }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

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 –i3x

^{2}) 1 dx =

**ò**

_{0}

^{1 }(1 – x)dx – 3i

**ò**

_{0}

^{1 }x

^{2}dx = ½ - i

Based on the
original contour definition (adding the
integrals for the segments OA and AB):

**ò**

_{C1}

^{ }f(z) dz = 1 – i/2

If
C2 denotes the segment 0B of the line y = x then we have:

z = x + ix (0

__<__x__<__1)**ò**

_{C2}

^{ }f(z) dz =

**ò**

_{0}

^{1 }-i3x

^{2}( 1 +i) dx =

3(1-i)

**ò**

_{0}

^{1 }x

^{2}dx = 1 – i

We can see from this that the integrals of f(z) have
different values though the two paths C1 and C2 have the same initial and
starting points. It follows from this that the integral of f(z) over the
simple closed contour OABO or C1 – C2 is:

**ò**

_{C1}

^{ }f(z) dz -

**ò**

_{C2}

^{ }f(z) dz = [½ - i] – (1 – i ) = -1 +i / 2

__Problems for the Math Maven:__

1) Redo Example (1) except change the limits of the contour to have z = -2i to z = 2i.

2) Redo Example (2) so that the contour segment C2 comes directly up from the line x = 0 instead of diagonally as shown in the graphic. Be sure to show all integrals and how you get the final contour.

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