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Here's one of the recommended exercises from our complex analysis class.

Prove or disprove: There is no analytic $f$ on $\mathbb{C} \setminus 0$ such that $exp(f(z)) = z^2$ for all nonzero $z \in \mathbb{C}$.

Via differentiating both sides of the equation, I get $f'(z) = \frac{2}{z}$. Integrating along the unit disk, $\int_{\partial B(0,1)} f'(z) = 0$.

Am I right in that my hunch is that there is no such $f$? To show this, I think I must show that this integral is non-zero. How do I do this?

Gerry Myerson
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user64464
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    When you integrate $2/z$ around a circle containing the origin, you don't get zero. – Gerry Myerson Mar 01 '13 at 03:31
  • That is what I want to show, that it can't be 0 around the unit circle, contradicting my earlier calculation that it is 0.

    Won't the integral of 2/z along any closed curve in the domain be zero since 2/z has an antiderivative in the domain?

     – user64464 Mar 01 '13 at 03:37
  • The antiderivative of $2/z$ is $2\log z$, which can't be defined in a neighborhood of zero. – Gerry Myerson Mar 01 '13 at 04:04

1 Answers1

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Just like there is no $\log$ on this domain.

As you said $f'(z)=2/z$.

Hence, integrating along the unit circle, $$ \int_\mathbb{T} f'(z)dz=2\int_\mathbb{T}\frac{dz}{z}. $$

Now the lhs yields $$ \int_0^{2\pi}f'(e^{i\theta})ie^{i\theta}d\theta=f(e^{2\pi i})-f(e^{\cdot 0i})=f(1)-f(1)=0 $$ while the rhs is $$ 2\int_0^{2\pi}\frac{ire^{i\theta}}{re^{i\theta}}d\theta=4i\pi. $$ Contradiction.

Julien
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