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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