Assume that $f(z)$ is holomorphic and satisfies the inequality $|f(z)-1|<1$ in a region $|z|<R$. Show that $$\int_\gamma\frac{f'(z)}{f(z)}dz=0$$ for all closed curves $\gamma$ in $|z|<R$. (The continuity of $f'(z)$ is taken for granted.)

The anti-derivative of $\dfrac{f'(z)}{f(z)}$ is just $\ln(f(z))$, which is holomorphic as long as $\ln(f(z))$ avoids the log branch cut. The log branch cut is the negative real axis, $a\leq 0$.

I want to use the fact that: If $f$ is the derivative of a holomorphic function in the region $\omega$, then the integral $\int_\gamma fdz$ depends only on the endpoints of $\gamma$.

To use this, we only need the condition that $f(z)$ is not a non-positive real number. The condition $|f(z)-1|<1$ seems like an overkill. (Of course it does imply that $f(z)$ is not a non-positive real number.) But am I misunderstanding something?

  • $\begingroup$ Do you not know the Argument Principle? It's begging to be used... $\endgroup$ Sep 30, 2013 at 5:57
  • $\begingroup$ @AlexYoucis I'm using the fact which I just updated. My question is that the condition $|f(z)-1|<1$ in the problem seems unnecessarily strong. Or do I misunderstand something? $\endgroup$
    – PJ Miller
    Sep 30, 2013 at 6:01
  • $\begingroup$ I believe it's just so that $f(z)$ won't have any zeros in the region :) $\endgroup$ Sep 30, 2013 at 6:03
  • $\begingroup$ For $f$ to admit a logarithm on $U$ it suffices that $U$ is simply connected and $f$ is non-vanishing. I'm not sure what the non-negative real stuff is. That is a branch of the logarithm, but there are others. $\endgroup$ Sep 30, 2013 at 6:06
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    $\begingroup$ haha, no offense taken. If you're not happy with their answer, comment and let them know why--don't accept it. That'll just get you answers you don't want. But, if you feel as though AlexM helped you, and you are now understanding everything you can keep the answer accepted :) $\endgroup$ Sep 30, 2013 at 6:25

1 Answer 1


I will fill in Alex Youcis's suggestion. So, By the argument principle, the integral in question is equal to $2\pi i($number of zeros - number of poles). Since $f$ is holomorphic in the region, there are no poles. So there are only zeros. By Rouche, $f(z)$ has as many zeros as the function 1, i.e. zero.

  • $\begingroup$ Rouche is a bit of overkill. If $|f(z)-1|<1$, then certainly $f(z)\ne 0$, else $|0-1|<1$ :) +1 though $\endgroup$ Sep 30, 2013 at 6:07
  • $\begingroup$ @AlexYoucis Rouche just rolls off the tougne though $\endgroup$
    – AlexM
    Sep 30, 2013 at 6:08
  • $\begingroup$ Sort of like Alex, yeah? :) $\endgroup$ Sep 30, 2013 at 6:08
  • $\begingroup$ @AlexYoucis I know! We have the best name $\endgroup$
    – AlexM
    Sep 30, 2013 at 6:09
  • $\begingroup$ That we do, that we do. Don't forget it. $\endgroup$ Sep 30, 2013 at 6:10

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