Without using Cauchy's theorems, how can we prove that $F'(z) = \frac{1}{z}$ implies that $F$ isn't holomorphic on a given annulus?

Question

I have been working on the following problem from Wong's Complex Analysis.

Prove that there exists no function $F(z)$ holomorphic on the annulus $D = \{ z \in \mathbb{C} : 1 < |z| < 2\}$ such that $$F'(z) = \frac{1}{z}, \ \ \ z \in D.$$

This question is given towards the beginning of this textbook, before any discussion about contour integrals, Cauchy's integral theorems, etc. This suggests that there may be an easy way of proving the above problem without the direct use of complex integration techniques.

Equivalent formulations of the above problem are given in the following Stack Exchange links:

Primitive of holomorphic Function $\frac{1}{z}$ on an Annulus.

Proving there is no holomorphic function on annulus

Holomorphic functions inside an Annulus

However, the solutions suggested in the above links are all based on complex integration techniques.

Is there an easier way of solving the above problem which does not directly rely on Cauchy's integral theorems, contour integrals, etc.? Can we solve this problem through the manipulation of a limit of the following form?
$$ \lim_{\Delta z \to 0} \frac{F(z + \Delta z) - F(z)}{\Delta z} = \frac{1}{z}. $$

Answer 1

I'm typing this up rather quickly, so hopefully there are no egregious mistakes. Regardless this should be the general idea. I believe the only thing I'm really using here are the Cauchy-Riemann equations.

Set $z=x+iy$. Suppose that such an $F(z)=u(z)+iv(z)$ exists. As a consequence of the Cauchy-Riemann equations, we must have $$ \frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}=1/z=F'(z)=u_x+iv_x $$ so that $$ u=\tfrac{1}{2}\log(x^2+y^2)+g(y)=\log(|z|)+g(\text{Im}(z)) $$ and (at least locally) $$ v=\arctan(y/x)+h(y)=\text{arg}(z)+h(\text{Im}(z)). $$ Therefore $$ F(z)=\log(|z|)+i\text{arg}(z)+g(\text{Im}(z))+ih(\text{Im}(z)). $$ But the problem is this function above (because of the $\text{arg}(z)$ term) cannot be defined in such a way that it is continuous on the whole annulus. If you pick a point, say $z=3/2$, this has $\text{arg}(3/2)=0$. As you travel around the annulus in a circle you will end up at $z=3/2$ again, except this time, $\text{arg}(3/2)=2\pi$. Does this help?

Answer 2

Assume that there is such a function $F$. Consider the $2\pi$-periodic function $$g(\phi):=-i\> F\bigl(\sqrt{2}e^{i\phi}\bigr)\qquad(\phi\in{\mathbb R}) $$ of the real variable $\phi$. One computes $$g'(\phi)=-i\>F'\bigl(\sqrt{2}e^{i\phi}\bigr)\>\sqrt{2}\>i\>e^{i\phi}={\sqrt{2}\>e^{i\phi}\over\sqrt{2}\>e^{i\phi}}=1\qquad(\phi\in{\mathbb R})\ .$$ It follows that $$0=g(2\pi)-g(0)=\int_0^{2\pi}g'(\phi)\>d\phi=2\pi \ .$$

Answer 3

Can be easily proved that for $D\subset\Bbb C$ open, $f:D\longrightarrow\Bbb C$ continuous:

$\exists F:D\longrightarrow\Bbb C$ s.t. $F' = f$ $\iff$ for all $\gamma:[a,b]\longrightarrow D$, $\displaystyle\int_\gamma f(z)\,dz$ depends only of $\gamma(a)$ and $\gamma(b)$.

For the $\impliedby$ part, define $$F(z) = \int_{z_0}^z f(w)\,dw = \int_\gamma f(w)\,dw$$ for any curve with $\gamma(a) = z_0$, $\gamma(b) = z$. The incremental quotient of $F$ can be easily calculated.