Question

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

Answer 1

\(f'(z)\)

Answer 2

Milne Thompson Method ?

Answer 3

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
?

Answer 4

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

will this work here ?

Answer 5

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 ?

Answer 6

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

Answer 7

Have you tried the partial derivatives ?

Answer 8

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

Answer 9

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

Answer 10

find partial derivative of f wrt x

Answer 11

right, we need to find both of those first

Answer 12

What would df/dx be ?

Answer 13

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)

Answer 14

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

Answer 15

replace x by z-iy

Answer 16

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

Answer 17

df/dx is good :)

Answer 18

do that in df/dx

Answer 19

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

Answer 20

df/dy is good also

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

@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?

Answer 26

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

Answer 27

"replace x by z-iy"

what would df/dx convert to ??

Answer 28

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

Answer 29

share the doc please ?

Answer 30

df/dx=3z^2-i2z

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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)

Answer 35

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

Answer 36

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

Answer 37

please wait

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

so, basically, where we have df/dx , we input z for "x" and 0 for "y" , as you mentioned
nd, that should work out .

Answer 42

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

Answer 43

sorry, my internet got disconnected....

did i do it correctly ?? or i integrated it one more time unnecessarily ???

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

why would i do that division ?

Answer 48

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

Answer 49

Thanks all :)
@klimenkov @nipunmalhotra93 @Miracrown