# Determining whether functions could be the real part of a holomorphic function?

Which one of the following functions $u$ could be the real part of a holomorphic function $f(z)$ with $z = x + iy$? For those that could, find $f$.

1. $u(x, y) = \frac{x}{y}$
2. $u(x, y) = xy$

My attempt at a solution:

For (1), $u(x, y)$ cannot be defined as $\frac{x}{y}$ because it is not continuously differentiable (namely at $y = 0$).

For (2), we use the Cauchy-Riemann equations: $u_x = v_y; -v_x = u_y$

$u_x = y = v_y$

$u_y = x = -v_x$

Thus we can define $v(x, y) = -\frac{x^2}{2} + \frac{y^2}{2}$.

Thus, $f(z) = xy + i(\frac{y^2}{2} - \frac{x^2}{2})$. Is this correct?

• Your number 2 is pretty easy to check using the Cauchy-Riemann equations. I'm not going to say that you're wrong on number 1, but I don't follow your logic. You could say the same thing about $u(x,y)=x/(x^2+y^2)$, yet that's the real part of $1/z$, which is certainly holomorphic away from the origin. – Mark McClure Oct 13 '14 at 4:03
• @MarkMcClure I assume it's meant to be something holomorphic on the entire complex plane. – user98602 Oct 13 '14 at 4:03
• @MikeMiller There's a word for that - entire. Typically, a holomorphic function is a complex-valued function that is complex differentiable in a neighborhood of every point in its domain. – Mark McClure Oct 13 '14 at 4:05
• @MarkMcClure I'm aware, but when I first took complex analysis "holomorphic" without a specified domain usually meant entire. But it's a fair point that the first portion would be trivial with that interpretation. – user98602 Oct 13 '14 at 4:14
• @MikeMiller Well, that $u$ is not harmonic, so it is certainly not the real part of a holomorphic function on any open set. That seems more to the point. – Mark McClure Oct 13 '14 at 4:16

If by "real part of a holomorphic function" you mean "real part of a holomorphic function on the whole complex plane", then your answer to 1) is correct. If it means "real part of a holomorphic function defined on the subset of $\Bbb C$ for which $u$ is defined", you need justification (but your conclusion is still correct), as $u$ is continuously differentiable in that domain. Your answer to 2) is correct. Indeed, your $f(z) = \frac{z^2}{2i}$.