òC 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) such that (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 left side below:
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 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)) dq
= 14.44i ò p/2 -p/2 dq
= 14.44i ò p/2 -p/2 dq
= 14.44 (p/2 - (-p/2)) = 14.44i (p) = 14.44 pi
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 pi can also be
written: I = ò C dz/ z =
pi
Example(2):
The reference contour is shown in the graphic to the right in preceding diagram, 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
-i3x2 (z = x + iy)
The segment OA can be represented
parametrically as:
z = 0 + iy (0 < y < 1)
z = 0 + iy (0 < y < 1)
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
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 OB of the
line y = x then we have:
z = x + ix (0 < x < 1)
And
we can write:
ò C2 f(z) dz = ò 0 1 -i3x2( 1 +i) dx =
3(1-i) ò 0 1 x2dx = 1 – i
ò C2 f(z) dz = ò 0 1 -i3x2( 1 +i) dx =
3(1-i) ò 0 1 x2dx = 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
Problem For Math Mavens:
Redo Example (1) except change the
limits of the contour to have z = -2i to z = 2i.
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