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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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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\(f'(z)\)
Milne Thompson Method ?
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 ?

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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 ?
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 ?
idek why i clicked on this. im am sooooo not smart enough for this lol.
Have you tried the partial derivatives ?
dw/dx or dw/dy are partial derivatives only....
what next after i find dw/dx (or dw/dy) ?
find partial derivative of f wrt x
right, we need to find both of those first
What would df/dx be ?
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)
df/dy = (-6xy +2x) + i(3x^2 +2y-3y^2)
replace x by z-iy
and all the y terms will cancel out... leaving only z terms (as it is analytic)
df/dx is good :)
do that in df/dx
i used milne thompson method for similar type of problem before...i would like to try same thing...
df/dy is good also
basically i plugged in x=z and y=0 and things sorted out....in this problem, it doesn't..
@Miracrown it is, but it'll make it a little more complicated. using f'(z)=df/dx is more straightforward,
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so my final answer should be z^2 (z-i) ?
looks like for df/dy , the "i" component is minus
@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?
The document I am looking at for this method, writes x^3 - 3xy^2 + 2xy as u(x,y)
"replace x by z-iy" what would df/dx convert to ??
3x^2 y - x^2 + y^2 - y^3 is written as v(x,y)
share the doc please ?
df/dx=3z^2-i2z
wot? you can't just plug in that on right side only ? can u ?
df/dz = ux(x,y) + ivx(x,y) , where
ux is the partial derivative of u with respect to x and v is the partial derivative of v with respect to x
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)
then, vx(x,y) = - uy(x,y)
I just checked it the answer is f'(z)=df/dx=3z^2-i2z
please wait
http://www.sciencedirect.com/science/article/pii/B9780080169392500214
i didn't know the answer, but after @klimenkov posted that screenshot, i tried plugging in x=z and y=0
@hartnn yea yea that'll work too...
so, basically, where we have df/dx , we input z for "x" and 0 for "y" , as you mentioned nd, that should work out .
@hartnn It's essentially the same thing I just made it more complicated lol sorry about that :P
sorry, my internet got disconnected.... did i do it correctly ?? or i integrated it one more time unnecessarily ???
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i should get f(z) as z^2(z-i) right ?
Did you try to divide \(f(z)\) by \(x+ iy\) ?
thats dw/dx is same as f'(z) and after i integrate it i directly get f(z) right ?
why would i do that division ?
@hartnn yes f(z)=z^3-i z^2.

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