If $G$ is an open connected subset of $\mathbb{C}$ that does not contain the origin, we call a continuous function $\alpha$ satisfying $\alpha(z) = \text{arg} z$ for all $z \in G$ a branch of arg $z$.

Prove there is no branch of arg z (and consequently no continuous complex logarithm) in the region $D^* =0 < |z| < 1$.

Attempt at a proof:

Assume there is a branch of arg $z$, i.e., there exists $\alpha : D^* \to [a,b] \subset \mathbb{R}$ such that $\alpha(z) = \text{arg} z$. Consider the path $\gamma : [0,1] \to D^*$ defined by $\gamma(t) = \frac{1}{2} e^{2 \pi i t}$.

Then, clearly $\gamma(0) = \gamma(1)$.

Moreover, the function $t \mapsto \alpha(\gamma(t)) = \text{arg}(\gamma(t))$ is a strictly increasing function. Hence, $\alpha(\gamma(0)) < \alpha(\gamma(1))$ contradicting the fact that $\gamma(0) = \gamma(1)$.

My concern with the outlined proof is that I don't know how to actually show that $t \mapsto \alpha(\gamma(t))$ is a strictly increasing function... other than the fact that intuitively this is so. Is it sufficient to say that arg($e^z$) = Im($z$)?

Thanks! Alternative proof styles are also welcome!

  • $\begingroup$ How did you defined $\arg z$ ? $\endgroup$
    – ts375_zk26
    Commented Jan 21, 2015 at 4:22
  • $\begingroup$ This is a problem from the text "Complex Function Theory" by Donald Sarason. If I am understanding it correctly, part of the point of the problem is that no matter how you choose to define arg $z$ it is not continuous on the given domain. Sarason does at one point in the preceding section define "arg $e^{b} =$ Im $b$." $\endgroup$
    – mlg4080
    Commented Jan 21, 2015 at 18:03
  • 2
    $\begingroup$ Then since $\gamma(t) = \frac{1}{2} e^{2 \pi i t}=e^{-\log 2+2\pi i t}$, we see that $\arg(\gamma(t))=2\pi t $ and that it is strictly increasing. $\endgroup$
    – ts375_zk26
    Commented Jan 22, 2015 at 10:09


You must log in to answer this question.

Browse other questions tagged .