Finding V given U (complex variable function)

[mancini0]

Homework Statement

I can't seem to find my mistake when I work through the following problem:

f(z) = u(x,y) + iv(x,y) is holomorphic.
If u = x^2 + y^2, find v.

Homework Equations

I realize I must apply Cauchy - Riemann here.

The Attempt at a Solution

Since f(z) is holomorphic, it is differentiable. Since f is differentiable, its derivative must satisfy the Cauchy-Riemann Equations, that is du/dx = dv/dy and -du/dy = dv/dx.

Since they gave me u, finding the partials of u is easy enough:
du/dx = 2x, du/dy = 2y

So by Cauchy Riemann, dv/dy = 2x.
So if I integrate dv/dy with respect to y, I should get v + a constant, which may be a function of x:
Integral ( dv/dy) dy = Integral of 2x dy = 2xy + F(x)

So v = 2xy +F(x)
Then dv / dx = 2y + F'(x) which should equal -du/dy by Cauchy Riemann.
so 2y + F'(x) = -2y

Then F'(x) = -4y

Integrating F'(x) with respect to x gives F(x) = -4xy + c (my gut says I made my mistake here)

So v = 2xy -4xy

Check:

dx/dv = -2x
dy/dv = -2y

which do not satisfy the Cauchy Riemann Equations... so I did something wrong...

 

[Dick]

There's something very wrong here. x^2+y^2 isn't an harmonic function. It's laplacian isn't zero. It can't be the real part of an holomorphic function. This may explain your problem. Is there a typo? Did they mean x^2-y^2?

 

[mancini0]

No, I'm certain that u = x^2 + y^2.

So if I'm not mistaken, a harmonic function has 2nd order partial derivatives that sum to zero.
Uxx + Vyy = 0.

Why does this imply the function doesn't have a real part? We do not study harmonic functions until later.

Thanks,
Mike

 

[Dick]

If you take the x derivative of ux=vy you get uxx=vyx. Then take the y derivative of uy=(-vx). Get uyy=(-vxy). Now if you add the two and remember that vxy=vyx you get uxx+uyy=0. If that laplace equation isn't true then f(z) can't be holomorphic no matter what you pick for v(x,y).