# anlytic function

• Nov 29th 2013, 12:52 PM
mami
anlytic 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.
• Nov 29th 2013, 12:58 PM
romsek
Re: 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.
• Nov 29th 2013, 01:15 PM
mami
Re: 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)

• Nov 29th 2013, 01:20 PM
mami
Re: anlytic function
to the best of my knowledge
if u(x,y)=(x^2)-(y^2) then

f=(x^2)-(y^2)+ i*2xy
• Nov 29th 2013, 01:28 PM
romsek
Re: anlytic function
Quote:

Originally Posted by mami
i taclked 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)

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
• Nov 29th 2013, 01:44 PM
mami
Re: 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
• Nov 29th 2013, 01:57 PM
romsek
Re: 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.
• Nov 29th 2013, 02:45 PM
mami
Re: anlytic function
would you mind, may i send e-mail to you if i couldn't tackle any problem?
• Nov 29th 2013, 02:50 PM
romsek
Re: anlytic function
Quote:

Originally Posted by romsek
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.

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).
• Nov 29th 2013, 03:01 PM
romsek
Re: anlytic function
posting them will be faster as I'm not always around.