# Determining an Analytic Function from its Real Part

I'm trying to prove that $log|z|$ is not the real part of an analytic function defined on an annulus centered at zero. Due to the Cauchy-Riemann Equations, I've been under the impression that given a harmonic function, such as $log|z|$, its role as the real part of an analytic function is unique, and thus an analytic function is completely determined, up to the addition of a constant, by its real (or imaginary) part.

Thus I feel like since $log(z)$ is an analytic function which cannot be defined on an annulus centered at zero whose real part is $log|z|$, then I can conclude that $log|z|$ is not the real part of an analytic function defined on said annulus.

Can someone help clarify my understanding? Thanks.

• Have you tried to see what happens if you use Cauchy-Riemann to find the putative analytic function of which $\log|z|$ is the real part? – Gerry Myerson Apr 23 '12 at 1:42
• Well yes and I get $-tan^{-1}(x/y)$, which if I remember correctly is the definition of $Arg(z)$. Thus giving me $log(z)$. – Thoth Apr 23 '12 at 1:43

Here is one way to make the connection you are hinting at more rigorous. Throughout this answer, $\log$ denotes only the real logarithm (defined for positive real numbers only, so there is no ambiguity in how it is interpreted).

Recall that $$|e^w| = e^{\operatorname{Re} w}, \qquad w \in \mathbb{C}.$$ If there were an annulus $A$ centered at $0$ and an function $g$ on $A$ with the property that $$\operatorname{Re} g(z) = \log |z|, \qquad z \in A,$$ then the function $h$ on $A$ given by $h(z) = e^{g(z)}/z$ for all $z \in A$ would be analytic on $A$ and would satisfy $$|h(z)| = \frac{|e^{g(z)}|}{|z|} = \frac{e^{\operatorname{Re} g(z)}}{|z|} = \frac{e^{\log |z|}}{|z|} = 1, \qquad z \in A.$$ It would follow from this that $$\overline{h(z)} = \frac{|h(z)|^2}{h(z)} = \frac{1}{h(z)},$$ being a quotient of analytic functions, is also analytic on $A$.

It easily follows from the Cauchy-Riemann equations that $h$ must be constant on $A$ (generally, whenever both a function and its complex conjugate are analytic on a connected open set, the function must be constant on that set).

So there is $\omega \in \mathbb{C}$ with $|\omega| = 1$ and $$e^{g(z)} = \omega z, \qquad z \in A.$$ Choosing a real number $s$ with $\omega = e^{is}$ it follows that the function analytic $L$ on $A$ given by $L(z) = e^{g(z) - is}$ satisfies $$e^{L(z)} = z, \qquad z \in A.$$ To summarize: beginning with an analytic function having $\log |z|$ as its real part on $A$, we have constructed an analytic branch of the logarithm on $A$. So if we happen to know that no such thing exists, we have our contradiction, and are done.

It is, of course, generally true that if $g$ and $k$ are analytic functions on a connected open set $G$ and $\operatorname{Re} g(z) = \operatorname{Re} k(z)$ holds for all $z \in G$, then there must be a real constant $C$ with $g(z) = k(z) + iC$. (Proof: consider the Cauchy-Riemann equations for the difference $g - k$ to deduce that $g - k$ must be constant, and clearly the constant must have zero real part.) So the intuition that an analytic function having $\log |z|$ as its real part must be "essentially the same thing" as a branch of the logarithm is correct. But in translating this intuition into a nonexistence proof, one must be careful, as you probably sensed when you were asking this question. (Working hastily and without careful thought, one might, for example, try to prove what you want by applying the result just mentioned with $G$ equal to the annulus, and one of the functions $g$ or $k$ equal to a branch of the logarithm--- conveniently forgetting for a moment that the result being appealed to concerns analytic functions on $G$, and there isn't an analytic logarithm on the annulus.)

To get an actual proof along these lines, one must proceed more carefully: let $A$ be the annulus, and let $k(z)$ denote e.g. the principal branch of the logarithm. Then $k$ is analytic on the connected open set $G = A \setminus (-\infty,0]$. If $g$ is an analytic function on $A$ satisfying $\operatorname{Re} g(z) = \log |z|$ for all $z \in A$, the result of the previous paragraph implies that there is a real constant $C$ with the property that $g(z) = k(z) + iC$ holds for all $z \in G$ (not all $z \in A$). Hence $k(z) = g(z) - iC$ holds for all $z \in G$, and as $g$ is continuous on all of $A$, we deduce from this equation that for any $c$ in the nonempty set $A \cap (-\infty,0)$, the limit $\lim_{z \to c, z \in G} k(z)$ exists (and is $g(c) - iC$). This contradicts the fact, obvious from the explicit formulas for $k$, that $k$ has a jump discontinuity at every point on the negative real axis.

You're right on man! Namely, (i) if $f,g: \Omega \rightarrow \mathbb{C}$ are holomorphic, and $Re \; f = Re \; g$, then $f = g$, to see this, taking the difference, it suffices to check that if a holomorphic function $f-g$ has real part $\equiv 0$, it's constant.

So suppose we have $f$ on an annulus such that $Re f = \log |z|$, by uniqueness of the real part on any branch cut (i), $f = \log z$ up to a constant, so $f$ is a holomorphic extension of a branch of the logarithm to a whole annulus, which is messed up o_O.

There's a bunch of ways to see (i): the open mapping theorem says holomorphic functions send open sets to open sets of $\mathbb{C}$, and if the image has constant real part, it's confined to super-skinny line $Re z = \lambda$, which can't contain an open ball!

More directly, you can use Cauchy-Riemann, we have $u \equiv 0$, and $0 = \frac{du}{dx} = \frac{dv}{dy}$, $0 = \frac{du}{dy} = - \frac{dv}{dx}$, so $v: \mathbb{R}^2 \rightarrow \mathbb{R}$ is smooth, with 0 derivative everywhere, hence constant too!

• In the first sentence of your posting you mean '...then $f = g + ai$ for some real $a$', don't you? – user20266 Apr 23 '12 at 7:13
• yes I do lol. good catch man – uncookedfalcon Apr 24 '12 at 10:09