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!)