text: ## hartnn one year ago f(z)=(x^3-3xy^2+2xy)+ i(3x^2 y-x^2+y^2-y^3) is analytic, find \(f^' (z)\) & f(z) in terms of z
  
  1. hartnn
  
  \(f'(z)\)
  
  2. hartnn
  
  Milne Thompson Method ?
  
  3. hartnn
  
  w=(x^3-3xy^2+2xy)+ i(3x^2 y-x^2+y^2-y^3 ) what next ? would i find dw/dx or dw/dy ?
  
  4. hartnn
  
  i had done similar problem, where i found dw/dx and then plugged in x= z, y= 0 will this work here ?
  
  5. hartnn
  
  because with that i am getting the complex term too, with the 'i' ...... and if i try x=0, y=z (can i do this ?) then i get f'(z) as z^2 -z^3 correct or not ?
  
  6. RyGuy
  
  idek why i clicked on this. im am sooooo not smart enough for this lol.
  
  7. Miracrown
  
  Have you tried the partial derivatives ?
  
  8. hartnn
  
  dw/dx or dw/dy are partial derivatives only....
  
  9. hartnn
  
  what next after i find dw/dx (or dw/dy) ?
  
  10. nipunmalhotra93
  
  find partial derivative of f wrt x
  
  11. Miracrown
  
  right, we need to find both of those first
  
  12. Miracrown
  
  What would df/dx be ?
  
  13. hartnn
  
  f(z)=(x^3-3xy^2+2xy)+ i(3x^2 y-x^2+y^2-y^3) df/dx = (3x^2 -3y^2 +2y) + i (6xy -2x)
  
  14. hartnn
  
  df/dy = (-6xy +2x) + i(3x^2 +2y-3y^2)
  
  15. nipunmalhotra93
  
  replace x by z-iy
  
  16. nipunmalhotra93
  
  and all the y terms will cancel out... leaving only z terms (as it is analytic)
  
  17. Miracrown
  
  df/dx is good :)
  
  18. nipunmalhotra93
  
  do that in df/dx
  
  19. hartnn
  
  i used milne thompson method for similar type of problem before...i would like to try same thing...
  
  20. Miracrown
  
  df/dy is good also
  
  21. hartnn
  
  basically i plugged in x=z and y=0 and things sorted out....in this problem, it doesn't..
  
  22. nipunmalhotra93
  
  @Miracrown it is, but it'll make it a little more complicated. using f'(z)=df/dx is more straightforward,
  
  23. klimenkov
  
  24. hartnn
  
  so my final answer should be z^2 (z-i) ?
  
  25. Miracrown
  
  looks like for df/dy , the "i" component is minus
  
  26. nipunmalhotra93
  
  @hartnn I'm not really familiar with that method. But I think using the substitution x=z-iy in df/dx is fine too isn't it?
  
  27. Miracrown
  
  The document I am looking at for this method, writes x^3 - 3xy^2 + 2xy as u(x,y)
  
  28. hartnn
  
  "replace x by z-iy" what would df/dx convert to ??
  
  29. Miracrown
  
  3x^2 y - x^2 + y^2 - y^3 is written as v(x,y)
  
  30. hartnn
  
  31. nipunmalhotra93
  
  df/dx=3z^2-i2z
  
  32. hartnn
  
  wot? you can't just plug in that on right side only ? can u ?
  
  33. Miracrown
  
  df/dz = ux(x,y) + ivx(x,y) , where
  
  34. Miracrown
  
  ux is the partial derivative of u with respect to x and v is the partial derivative of v with respect to x
  
  35. nipunmalhotra93
  
  See we know that df/dz=df/dx right? (the one on the right is a partial one) So, once you've found the one on the right (using partial differentiation, just replace x by z-iy as x+iy=z is already known)
  
  36. Miracrown
  
  then, vx(x,y) = - uy(x,y)
  
  37. nipunmalhotra93
  
  I just checked it the answer is f'(z)=df/dx=3z^2-i2z
  
  38. hartnn
  
  39. Miracrown
  40. hartnn
  
  i didn't know the answer, but after @klimenkov posted that screenshot, i tried plugging in x=z and y=0
  
  41. nipunmalhotra93
  
  @hartnn yea yea that'll work too...
  
  42. Miracrown
  
  so, basically, where we have df/dx , we input z for "x" and 0 for "y" , as you mentioned nd, that should work out .
  
  43. nipunmalhotra93
  
  @hartnn It's essentially the same thing I just made it more complicated lol sorry about that :P
  
  44. hartnn
  
  sorry, my internet got disconnected.... did i do it correctly ?? or i integrated it one more time unnecessarily ???
  
  45. hartnn
  
  i should get f(z) as z^2(z-i) right ?
  
  46. klimenkov
  
  Did you try to divide \(f(z)\) by \(x+ iy\) ?
  
  47. hartnn
  
  thats dw/dx is same as f'(z) and after i integrate it i directly get f(z) right ?
  
  48. hartnn
  
  why would i do that division ?
  
  49. nipunmalhotra93
  
  @hartnn yes f(z)=z^3-i z^2.
  
  50. hartnn
  
  Thanks all :) @klimenkov @nipunmalhotra93 @Miracrown