How can I find all analytic functions f=u+iv with u(x,y)=(x^2)+(y^2)

Thanks for the help. I appreciate it.

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- November 29th 2013, 01:52 PMmamianlytic function
How can I find all analytic functions f=u+iv with u(x,y)=(x^2)+(y^2)

Thanks for the help. I appreciate it. - November 29th 2013, 01:58 PMromsekRe: anlytic function
a) do you know what it means for a function to be analytic?

b) do you know what the Cauchy–Riemann equations are?

if not read up on those two and give it a shot. - November 29th 2013, 02:15 PMmamiRe: anlytic function
i tackled with the problem which is u(x,y)=(x^2)-(y^2)

but i need help for that one u(x,y)=(x^2)+(y^2)

i know your questions. thanks for your help. - November 29th 2013, 02:20 PMmamiRe: anlytic function
to the best of my knowledge

if u(x,y)=(x^2)-(y^2) then

f=(x^2)-(y^2)+ i*2xy - November 29th 2013, 02:28 PMromsekRe: anlytic function
ok, so you know that du/dx = dv/dy and du/dy = -dv/dx

you have u so compute it's partials

du/dx = 2x

du/dy = 2y

so dv/dy = 2x and thus v = 2xy + f(y)

dv/dx = -du/dy = -2y so v = -2xy + g(x)

2xy + f(y) = -2xy + g(x)

4xy = g(x) - f(y), and you can see that this has no solution - November 29th 2013, 02:44 PMmamiRe: anlytic function
thanks again. but I have a problem again. could you please check it. is it relevant our question? my mind confused...

Attachment 29824Attachment 29825 - November 29th 2013, 02:57 PMromsekRe: anlytic function
at (0,0) 4xy = 0 so the C-R equations happen to have a solution there so yes (x^2+y^2) + i*0 happens to satisfy the C-R equations at 0.

so yeah, you can say (x^2+y^2) + i 0 is analytic at 0 but nowhere else on the complex plane. - November 29th 2013, 03:45 PMmamiRe: anlytic function
would you mind, may i send e-mail to you if i couldn't tackle any problem?

- November 29th 2013, 03:50 PMromsekRe: anlytic function
I should be a bit more careful. You need more than satisfying the C-R equations to declare a function is analytic at a point. The first partials must exist and be continuous at that point. The first partials of (x^2+y^2) + i*0 do exist and are continuous everywhere and thus are at (0,0).

- November 29th 2013, 04:01 PMromsekRe: anlytic function
posting them will be faster as I'm not always around.