7
$\begingroup$

My professor says given the real part $u$ of an analytic function $f$ defined on a domain $D\subset \mathbb{C}$, that we can't rule out the possibility that there could exist some other analytic function $g$, distinct from $f$ beyond just the addition of a constant, defined on a domain $E\subset \mathbb{C}$ either disjoint from, or not homeomorphic to, $D$, provided that $f$ is not analytic on $E$.

Since differentiating $u$ with respect to one variable and then integrating it with respect to the other completely determines the imaginary part, what this says to me is that $u$ would have to either produce different partial derivatives on $D$ and $E$ respectively, or $\frac{\partial u}{\partial x}$ different primitives.

The case of D and E being disjoint is trivial, but can anyone give me an example for D and E overlapping but non-homeomorphic?

  • 0
    The original domains we were discussing was an annulus centered at zero and the slit plain, thus overlapping but not-homeomorphic, and my professor said $log|z|$ is the real part of $log(z)$ on the slit plane but was not the real part of of some analytic function on the annulus, but that just because log(z) couldn't be extended analytically to the annulus didn't automatically allow us to conclude that there couldn't be some other analytic function defined on the annulus with $log|z|$ as its real part.2012-04-23

1 Answers 1

4

Let $f$ and $g$ be two analytic functions defined on the domains $D$ and $E$, respectively, and suppose $A=D\cap E$ is nonempty and connected. Suppose that $\Re f(z) = \Re g(z)$ for all $z\in A$. Then I claim that $f$ and $g$ differ by an additive (imaginary) constant, that is, $f(z) = g(z) + i\alpha$ for some $\alpha\in\mathbb R$. (Furthermore, by the uniqueness of analytic continuation, this relationship continues to hold wherever $f$ and $g$ are both defined.)

To prove the claim, we set $h=f-g$; then $\Re h(z)=0$ for all $z\in A$, and we need to show that $h$ is constant. But this follows immediately from the Cauchy-Riemann equations (the integral of the derivative of $0$ is constant), since the difference of two analytic functions is also analytic. (More precisely, the derivative being zero forces $h$ to be locally constant; since $A$ is connected, that implies that $h$ is constant.)

In particular, suppose that $g(z)$ were an analytic function defined on the annulus such that $\Re g(z) = \log |z|$ everywhere in the annulus. In particular, on the slit annulus, $\Re g(z) = \Re(\log z)$ (for your favorite branch of $\log$ defined on that slit annulus). Then on the slit annulus, $g(z) = \log z + i\alpha$ for some $\alpha\in\mathbb R$. However, the function $\log z + i\alpha$ is not continuous on the slit, contradicting the analyticity of $g$ there; hence there can be no such $g$.

  • 0
    Ok thanks Greg, this makes sense, it was certainly misleading of my professor to state things like he did.2012-04-23