0
$\begingroup$

I need to prove that $\int_\gamma f'(z)/f(z)dz=0$ for any closed curve. It is given that f is holomorphic and satisfies $|f(z)-1|\lt1$ in the region. And we can assume $f'(z)$ is continuous.

I think that first I need to prove that $f'(z)/f(z)$ has a primitive, so that I can use the proposition to prove the integral is equal to 0.

But I don't know how to prove $f'/f$ has a primitive and how to use that inequality in my proof? Thanks.

$\endgroup$
  • $\begingroup$ The integrand is holomorphic; can you use the Cauchy integral theorem? $\endgroup$ – arkeet Sep 30 '16 at 20:33
  • $\begingroup$ So using the Cauchy integral theorem, we know that it's integral over the closed curve would be zero and that means $f'/f$ has a primitive? $\endgroup$ – J.doe Sep 30 '16 at 20:36
0
$\begingroup$

The quotient $\;\frac{f'}f\;$ is analytic wherever $\;f\neq0\;$, and since this happens in the integration region (why?) Cauchy's Theorem gives the answer at once (I'm assuming you meant closed, simple and rectifiable curve)

$\endgroup$
  • $\begingroup$ So using the Cauchy integral theorem, we know that it's integral over the closed curve would be zero and that means $f'/f$ has a primitive? $\endgroup$ – J.doe Sep 30 '16 at 20:41
  • $\begingroup$ @J.doe Well, yes: the primitive is easy: $\;\log f(z)\;$ , after choosing a suitable branch. Yet you don't even need to mention that when using Cauchy theorem... $\endgroup$ – DonAntonio Sep 30 '16 at 20:42
  • $\begingroup$ Wait, why do we need $logf(z)$? $\endgroup$ – J.doe Sep 30 '16 at 20:54
  • 1
    $\begingroup$ @J.doe Once again: if you use Cauchy's THeorem you don't even need to mention "primitive function", and of course we do use the inequality! How otherwise could we deduce that $\;f\neq0\;$ ? $\endgroup$ – DonAntonio Sep 30 '16 at 21:36
  • 1
    $\begingroup$ As long as $\gamma$ is contained in a simply connected open subset of the domain (on which $f \ne 0$), it doesn't matter whether it is simple or not. $\endgroup$ – arkeet Sep 30 '16 at 23:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.