Time for some more contour integrals. We have seen earlier examples of contour line integrals and now we look at a more detailed example:
We
want to integrate around the closed contour for which f(z) =
3x + 2iy:
We
check first to see if the function is analytic using the Cauchy –Riemann relations:
Recall
that we require: (see e.g. http://brane-space.blogspot.com/2013/10/looking-at-harmonic-conjugates-and.html )
i)
¶ u/ ¶x = ¶ v/ ¶y and ii) ¶ u/ ¶y
= - ¶ v/ ¶x
For
f(z) = 3x + 2iy = u(x,y) + iv(x,y)
We
have: ¶ u/ ¶x = 3 and
¶ v/ ¶y = 2,
so ¶ u/ ¶x ¹ ¶ v/ ¶y
And:
¶ u/ ¶y = 0 =
- ¶ v/ ¶x
Since
condition (i) is not fulfilled the function is not analytic, hence we cannot
evaluate using Cauchy’ theorem , i.e.
ò C f(z) dz = 0
ò C f(z) dz = 0
So
we must integrate line segment by line segment, viz.
I = ò C f(z) dz
= ò C (u + iv) (dx + idy) =
ò C (3x + 2iy) (dx + idy)
ò C (3x + 2iy) (dx + idy)
= ò C (3x dx
- 2y dy) + i ò C ( 2y dx + 3x dy)
I
= å3 n
= 1 [ò C (3x dx
- 2y dy) + iò C ( 2y dx + 3x dy)
On C1: 0 < x < 1, y = 3, dy = 0
ò C1 (3x dx
- 2y dy) + i ò C1 ( 2y dx + 3x dy)
= ∫ 0 1 3x dx
+ i∫ 0 1 6 dy = [3/2 x2] 0 1 +
i[6y] 0 1
= 3/2 + 6i
On
C2: 3 < y < 5, x = 1, and dx = 0
ò C2 (3x dx
- 2y dy) + i ò C2 ( 2y dx + 3x dy)
= ò3 5 (
-2y ) dy + i ò3 5 3 dy
= [- y2]
3 5 +
i[3y] 3 5
=
- 16 + 6i
On
C3: 1 < t < 0, x = t, and dx = dt, y = 2t +3,
dy = 2dt
Then:
ò C3 (3x dx
- 2y dy) + i ò C3 ( 2y dx + 3x dy)
= ∫ 1
0 [ 3t
dt - 2(2t + 3) 2 dt] + i ∫ 1
0 [ 2(2t + 3) dt + 3t (2dt)]
=
- ∫ 1 0
(5t + 12)
dt + i∫ 1 0
(10t + 6)
dt
= - [5t2
/ 2 + 12t] 1 0 + i[5t2 + 6t]
1 0 = 29/2 – 11i
Then
the contour integral value is:
I
= ( 3/2 + 6i) + (-16 + 6i) + (29/2 – 11i) = 0 + i
Problems
for Math Mavens:
1)
For the closed path shown in the diagram below, evaluate the contour integral. I.
Let
f(z) = z 2
f(z)
= 3x + i3y?
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