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

- hartnn

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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- hartnn

\(f'(z)\)

- hartnn

Milne Thompson Method ?

- 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
?

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## More answers

- hartnn

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

- 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 ?

- RyGuy

idek why i clicked on this. im am sooooo not smart enough for this lol.

- Miracrown

Have you tried the partial derivatives ?

- hartnn

dw/dx
or dw/dy
are partial derivatives only....

- hartnn

what next after i find dw/dx (or dw/dy)
?

- nipunmalhotra93

find partial derivative of f wrt x

- Miracrown

right, we need to find both of those first

- Miracrown

What would df/dx be ?

- 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)

- hartnn

df/dy = (-6xy +2x) + i(3x^2 +2y-3y^2)

- nipunmalhotra93

replace x by z-iy

- nipunmalhotra93

and all the y terms will cancel out... leaving only z terms (as it is analytic)

- Miracrown

df/dx is good :)

- nipunmalhotra93

do that in df/dx

- hartnn

i used milne thompson method for similar type of problem before...i would like to try same thing...

- Miracrown

df/dy is good also

- hartnn

basically i plugged in x=z and y=0
and things sorted out....in this problem, it doesn't..

- nipunmalhotra93

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

- klimenkov

##### 1 Attachment

- hartnn

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

- Miracrown

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

- 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?

- Miracrown

The document I am looking at for this method, writes x^3 - 3xy^2 + 2xy as u(x,y)

- hartnn

"replace x by z-iy"
what would df/dx convert to ??

- Miracrown

3x^2 y - x^2 + y^2 - y^3 is written as v(x,y)

- hartnn

share the doc please ?

- nipunmalhotra93

df/dx=3z^2-i2z

- hartnn

wot?
you can't just plug in that on right side only ? can u ?

- Miracrown

df/dz = ux(x,y) + ivx(x,y) , where

- Miracrown

ux is the partial derivative of u with respect to x and v is the partial derivative of v with respect to x

- 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)

- Miracrown

then, vx(x,y) = - uy(x,y)

- nipunmalhotra93

I just checked it the answer is f'(z)=df/dx=3z^2-i2z

- hartnn

please wait

- Miracrown

http://www.sciencedirect.com/science/article/pii/B9780080169392500214

- hartnn

i didn't know the answer, but after @klimenkov posted that screenshot, i tried plugging in x=z and y=0

- nipunmalhotra93

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

- 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 .

- nipunmalhotra93

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

- hartnn

sorry, my internet got disconnected....
did i do it correctly ?? or i integrated it one more time unnecessarily ???

##### 1 Attachment

- hartnn

i should get f(z) as z^2(z-i)
right ?

- klimenkov

Did you try to divide \(f(z)\) by \(x+ iy\) ?

- hartnn

thats dw/dx is same as f'(z)
and after i integrate it i directly get f(z)
right ?

- hartnn

why would i do that division ?

- nipunmalhotra93

@hartnn yes f(z)=z^3-i z^2.

- hartnn

Thanks all :)
@klimenkov @nipunmalhotra93 @Miracrown

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