## 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