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hartnn

  • 2 years 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

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  1. hartnn
    • 2 years ago
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    \(f'(z)\)

  2. hartnn
    • 2 years ago
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    Milne Thompson Method ?

  3. hartnn
    • 2 years ago
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    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
    • 2 years ago
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    i had done similar problem, where i found dw/dx and then plugged in x= z, y= 0 will this work here ?

  5. hartnn
    • 2 years ago
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    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
    • 2 years ago
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    idek why i clicked on this. im am sooooo not smart enough for this lol.

  7. Miracrown
    • 2 years ago
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    Have you tried the partial derivatives ?

  8. hartnn
    • 2 years ago
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    dw/dx or dw/dy are partial derivatives only....

  9. hartnn
    • 2 years ago
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    what next after i find dw/dx (or dw/dy) ?

  10. nipunmalhotra93
    • 2 years ago
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    find partial derivative of f wrt x

  11. Miracrown
    • 2 years ago
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    right, we need to find both of those first

  12. Miracrown
    • 2 years ago
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    What would df/dx be ?

  13. hartnn
    • 2 years ago
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    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
    • 2 years ago
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    df/dy = (-6xy +2x) + i(3x^2 +2y-3y^2)

  15. nipunmalhotra93
    • 2 years ago
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    replace x by z-iy

  16. nipunmalhotra93
    • 2 years ago
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    and all the y terms will cancel out... leaving only z terms (as it is analytic)

  17. Miracrown
    • 2 years ago
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    df/dx is good :)

  18. nipunmalhotra93
    • 2 years ago
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    do that in df/dx

  19. hartnn
    • 2 years ago
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    i used milne thompson method for similar type of problem before...i would like to try same thing...

  20. Miracrown
    • 2 years ago
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    df/dy is good also

  21. hartnn
    • 2 years ago
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    basically i plugged in x=z and y=0 and things sorted out....in this problem, it doesn't..

  22. nipunmalhotra93
    • 2 years ago
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    @Miracrown it is, but it'll make it a little more complicated. using f'(z)=df/dx is more straightforward,

  23. klimenkov
    • 2 years ago
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  24. hartnn
    • 2 years ago
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    so my final answer should be z^2 (z-i) ?

  25. Miracrown
    • 2 years ago
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    looks like for df/dy , the "i" component is minus

  26. nipunmalhotra93
    • 2 years ago
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    @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
    • 2 years ago
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    The document I am looking at for this method, writes x^3 - 3xy^2 + 2xy as u(x,y)

  28. hartnn
    • 2 years ago
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    "replace x by z-iy" what would df/dx convert to ??

  29. Miracrown
    • 2 years ago
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    3x^2 y - x^2 + y^2 - y^3 is written as v(x,y)

  30. hartnn
    • 2 years ago
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    share the doc please ?

  31. nipunmalhotra93
    • 2 years ago
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    df/dx=3z^2-i2z

  32. hartnn
    • 2 years ago
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    wot? you can't just plug in that on right side only ? can u ?

  33. Miracrown
    • 2 years ago
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    df/dz = ux(x,y) + ivx(x,y) , where

  34. Miracrown
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    then, vx(x,y) = - uy(x,y)

  37. nipunmalhotra93
    • 2 years ago
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    I just checked it the answer is f'(z)=df/dx=3z^2-i2z

  38. hartnn
    • 2 years ago
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    please wait

  39. Miracrown
    • 2 years ago
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    http://www.sciencedirect.com/science/article/pii/B9780080169392500214

  40. hartnn
    • 2 years ago
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    i didn't know the answer, but after @klimenkov posted that screenshot, i tried plugging in x=z and y=0

  41. nipunmalhotra93
    • 2 years ago
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    @hartnn yea yea that'll work too...

  42. Miracrown
    • 2 years ago
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    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
    • 2 years ago
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    @hartnn It's essentially the same thing I just made it more complicated lol sorry about that :P

  44. hartnn
    • 2 years ago
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    sorry, my internet got disconnected.... did i do it correctly ?? or i integrated it one more time unnecessarily ???

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  45. hartnn
    • 2 years ago
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    i should get f(z) as z^2(z-i) right ?

  46. klimenkov
    • 2 years ago
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    Did you try to divide \(f(z)\) by \(x+ iy\) ?

  47. hartnn
    • 2 years ago
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    thats dw/dx is same as f'(z) and after i integrate it i directly get f(z) right ?

  48. hartnn
    • 2 years ago
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    why would i do that division ?

  49. nipunmalhotra93
    • 2 years ago
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    @hartnn yes f(z)=z^3-i z^2.

  50. hartnn
    • 2 years ago
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    Thanks all :) @klimenkov @nipunmalhotra93 @Miracrown

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