Prove that there exists no function $F(z)$ analytic is in the annulus $D: 1 < |z| < 2 $ such that $F'(z) = \frac{1}{z}$ for all $z$ in $D$.

Question

Prove that there exists no function $F(z)$ analytic is in the annulus $D: 1 < |z| < 2 $ such that $F'(z) = \frac{1}{z}$ for all $z$ in $D$.

Do I have to assume $F$ exists? Do I also have to show that for $z$ in $D$, $z$ is not a negative real number, $F(z) = \log z + c$, where $c$ is a constant?

Please help me with this proof?

Answer 1

You can do it by contradiction. If $1/z$ had a primitive, its integral over any closed curve would be zero. However, $$\oint_{\{ z\in \Bbb C\mid |z| = 3/2 \}} \frac{{\rm d}z}{z} \neq 0,$$as you can readily check.

Answer 2

You could also consider using general properties of Laurent expansions: if $F(z)$ is analytic on $1< |z|<2$ then there exists a unique Laurent expansion $$ F(z) = \sum_{n\in \mathbb Z} a_n z^n,\quad a_n\in \mathbb C$$ such that $$ \limsup_{n\to\infty} |a_n|^{1/n} \leq \frac12 \quad\text{and}\quad \limsup_{n\to\infty} |a_{-n}|^{1/n} \leq 1.$$ Then by uniform convergence of Laurent series on compact sets, we can differentiate termwise (iirc) so the derivative of $F$ satisfies $$F'(z) = \sum_{n\in \mathbb Z} na_n z^{n-1}.$$ Here the $z^{-1}$ term vanishes because $n=0$ at this term. Thus $F'(z) \neq \frac1{z}$ by uniqueness of Laurent expansions.

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To address your specific questions: in this proposed solution, you do assume $F$ exists but you do not have to say anything about how $F$ is related to the complex logarithm function. Hope this helps!