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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
?

i had done similar problem, where i found dw/dx and then plugged in
x= z, y= 0
will this work here ?

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..

so my final answer should be
z^2 (z-i)
?

looks like for df/dy , the "i" component is minus

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

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

@hartnn yea yea that'll work too...

@hartnn It's essentially the same thing I just made it more complicated lol sorry about that :P

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.