## Thursday, February 13, 2014

### More Difficult Contour Integrals

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.)

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

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 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

2) For the example shown in the blog post (Fig. 1) , what would the value of I be if:

f(z) = 3x + i3y?