Tuesday, February 18, 2014

Contour Integral Solutions (From Feb. 13)

1) Recall the contour defined:




We have f(z) = z 2


Then:   I  =   ò C  f(z) dz  =    

 
=  1 1+i    z2 dz  +   1+1 -1+i   z2 dz   +   -1+1 -1  z2 dz  +

   -1 1  z2 dz 

 
=  

[z 3  /3] 1 1+i       +   [z 3  /3] 1+i -1 +i   +   [z 3  /3] -1+i -1    +  [z 3  /3] -1 1       

 
=   (-1 + 2i/3)  + [(2/3 + 2i/3) – (-2/3 + 2i/3)] + [-1 – 2i/3] + [(1/3 – (-1/3)]


=   -2 + 4/3 + 2/3  =  -2 + 6/3 = -2 + 2 = 0

 

2)     Check Cauchy relations first.

 
i) u/ x = v/ y  and ii) u/ y   =  - v/ x

 

For f(z) = 3x + 3iy  =  u(x,y) + iv(x,y)

 
We have:  u/ x =  3   and v/ y    = 3,    so  u/ x =   v/ y  

 

And:

 
u/ y   = 0 =     - v/ x


So:   u/ y   =  - v/ x

 
Since the two Cauchy relations are satisfied then:

 
I  =   ò C  f(z) dz  =    0 

 
(Industrious readers can check this for themselves by doing the integration piece meal - as for the original example!)

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