0
$\begingroup$

Let $U$ be an open set and $f:U\to\mathbb{C}$ be a holomorphic function with real part $u(x,y)$ and imaginary part $v(x,y)$. Is it possible that $u(x,y)^2=1+v(x,y)^3$ for $x+iy\in U$

We have Cauchy-Riemann equations. But I don't know how to use them. Can anyone give any hint? thnx for your help.

$\endgroup$
  • $\begingroup$ Sure it's possible: Let $f(z) \equiv 1.$ $\endgroup$ – zhw. Dec 6 '15 at 19:05
1
$\begingroup$

Why not use constants for $u$ and $v$? Set $u=3$ and $v=2.$ There exist infinitely many solutions like that, parametrized by a real $t.$

On the other hand, taking the derivative of the given equation with respect to $x$ or $y$, and substituting the Cauchy-Riemann equations, yields a system of two homogeneous linear equations in $u_x$ and $u_y.$ That system can only have a nonzero solution (i.e., $u$ itself nonconstant) if the determinant $9v^4+4u^2$ vanishes. Substituting the given equation into that yields a fourth-degree polynomial equation with constant coefficients in $v$, so $v$ must be constant and therefore $u$ as well.

$\endgroup$
0
$\begingroup$

Hint: Assume $U$ is connected. The hypotheses imply $f(U)$ is a subset of the curve $y=(x^2-1)^{1/3}.$ Think about the open mapping theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.