Find Analytic Functions with u(x,y)=(x^2)+(y^2)

[rasi]

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.

 

[lurflurf]

Require that v satisfy the Cauchy–Riemann equations
$$\dfrac{\partial v}{\partial x}=-\dfrac{\partial u}{\partial y}\\
\dfrac{\partial v}{\partial y}=\phantom{-}\dfrac{\partial u}{\partial x}$$

 

[rasi]

but it doesn't satisfy except (0,0)

 

[mathwonk]

note that the cauchy riemann equations imply that ∂^2(u)/∂x^2 + ∂^2(u)/∂y^2 = 0. but that is false for your example, so there are no such analytic functions. i.e. both u and v must be "harmonic" functions in order for u + iv to be analytic, and your u is not harmonic. try u = X^2 - Y^2.

 

[rasi]

is it anaytic at (0,0) ?

 

[HallsofIvy]

Is what analytic at (0,0)? You asked about a function u+ iv, with $u= x^2+ y^2$.

As lurflurf said, use the Cauchy-Riemann equations- if f(z)= u(x,y)+ iv(x,y), z= x+ iy is analytic then
$$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$$
$$\frac{\partial v}{\partial x}= -\frac{\partial u}{\partial y}$$

Here, $\partial u/\partial x= 2x$ and $\partial u/\partial y= 2y$ so we must have
$$\frac{\partial v}{\partial y}= 2x$$
$$\frac{\partial v}{\partial x}= -2y$$
From the second equation, $v= -2xy+ f(x)$ for some function, f, of x alone. Differentiating that with respect to x, $v_x= -2y+ f'(x)= 2x$ which is impossible. There cannot be an analytic function with real part $x^2+ y^2$.