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I have just started taking complex analysis course,The following problem is given in my class.Please help me solving it. Thnx in advance.

Suppose $f(z)$ is analytic and satisfies the relation $|f(z)-1| < 1 $ in a region $\Omega$ Show that $\displaystyle \int_\gamma \frac{f'(z)}{f(z)}=0$ for every closed curve $\gamma$ in $\Omega$

I am only taught upto Cauchy's Theorem which I think is applicable for disc. Now here nothing is mentioned about $\Omega$, only thing I know that it is open connected.So how can I apply Cauchy's Theorem here?

Being very new to complex analysis I appologise if I am missing something very simple or doing something very much wrong.

Please help me to solve this question. Thnx again.

Martin Sleziak
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usermath
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  • Hint: Can you think of an anti-derivative for $f'(z)/f(z)$ in $\Omega$? – Jyrki Lahtonen Aug 24 '14 at 11:49
  • But how can I ensure that antiderivative of f'(z)/f(z) exists in the given region? – usermath Aug 24 '14 at 11:51
  • Good! That doubt is at the core of this question. But if you have a candidate for the alternative, may be you can use what you know about the points where it might not be analytic? – Jyrki Lahtonen Aug 24 '14 at 11:53
  • Sorry.but perhaps I am missing something again.Will you please explain a little bit more? – usermath Aug 24 '14 at 11:55
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    Note that the inequality implies that $f(z)\not = 0$ on $\Omega$. – anomaly Aug 24 '14 at 11:57
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    You do remember what the indefinite integral $$\int\frac{f'(x)}{f(x)},dx$$ is, don't you? You have certainly seen it, when $f(x)=x$ and when $f(x)=x^2+1$. Most likely also, when $f(x)=\cos x$. – Jyrki Lahtonen Aug 24 '14 at 12:05
  • Arrggh! Should read: a candidate for the anti-derivative. – Jyrki Lahtonen Aug 24 '14 at 12:16
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    Is it lnf(z)?So existence of premitive implies the integral is zero?But can I say that lnf(z) exists in the given region?Sorry again.perhaps I am missing something very much simple. – usermath Aug 24 '14 at 12:18
  • Good. What do you know about regions, where $\ln z$ is analytic? – Jyrki Lahtonen Aug 24 '14 at 12:31
  • Is it whole complex plane except origin and one half(negative say) of the real line? – usermath Aug 24 '14 at 12:33
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    Correct. That's one possibility. So is it analytic in the open disk $$B(1,1)={z\in\Bbb{C}\mid |z-1|<1}?$$ If it is, then can you conclude that $\ln (f(z))$ is analytic in $\Omega$? – Jyrki Lahtonen Aug 24 '14 at 12:34
  • Yes,I think I got it.f(z) is in B(1,1) and hence lnf(z) is analytic and so integral is zero.Am I correct? – usermath Aug 24 '14 at 12:40
  • That's right! Well done. See, it wasn't too hard in the end (+1). – Jyrki Lahtonen Aug 24 '14 at 13:49
  • Yes it was not,I now understand.I was missing some facts which were easy indeed. Thank you for your help sir.It was really helpful for me.Thank you so much again. – usermath Aug 24 '14 at 13:54

3 Answers3

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This is a straightforward logarithmic integration, in your region $|f(z)-1|<1$, which means that the values of $f$ lie in a ball of center $1$ and radius $1$. There is a well defined branch of the log defined on this region, (in fact on any simply connected region not containing the origin). Therefore

$$\int \frac{f^{\prime}(z)}{f(z)}dz=\ln f(z)$$ a well defined and single valued function.

Now over a closed curve the initial and final points, $z_0$ are the same so by the fundamental theorem

$$\int_{\gamma} \frac{f^{\prime}(z)}{f(z)}dz=\ln f(z_0)-\ln f(z_0)=0$$

Rene Schipperus
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A hint:

The quotient ${f'\over f}$ rings a bell from real calculus. Maybe one can make the analogy work. By the way: There has to be a reason for the "technical assumption" in the problem $\ldots$

Christian Blatter
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Use principle of argument:

$$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=2\pi i (N_{zeros}-N_{poles}) .$$

As was said,

$$|f(z)-1|<1 \Leftrightarrow -1<f(z)-1<1\Leftrightarrow 0 <f(z)<2 $$ Since $f(z)>0$, it has no zeros in any $\operatorname{Int}\gamma$. As $f(z)<\infty$, it has no poles. Therefore, $$N_{zeros}=N_{poles}=0$$ $$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=0.$$

cool
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  • I don't know this formula yet.But it is really easy to solve using this formula.Thnx a lot for your answer. – usermath Aug 24 '14 at 12:42
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    To write $|f(z)-1| < 1$ as $-1 < f(z)-1 < 1$, aren't you implicitly assuming that $f$ is real-valued? – user193319 Dec 07 '21 at 03:01