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$\gamma_r$ is the part of $|z-1|=r$ inside $|z|=1$, $\gamma_1$ is the part of |z|=1 outside $|z-1|<r$. prove that

1) If $|z|<1, |z-1|=r(0<r<1),$ then $$|\frac{\log (1-z)}{z}|\leq \dfrac{\log\frac1r}{1-r},$$ then show that $\lim\limits_{r\to0}\int_{\gamma_r}\frac{\log (1-z)}{z}dz=0$

2) definition $\int_{|z|=1}\frac{\log (1-z)}{z}dz=\lim\limits_{r\to0}\int_{\gamma_1}\frac{\log (1-z)}{z}dz $. Compute the integration

$$\int_{|z|=1}\frac{\log (1-z)}{z}dz$$

and then show that $\int_0^{2\pi}\log|1-e^{i\theta}|d\theta=0$

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For 2), $\frac{\log{(1-z)}}{z}$ has no poles in the region bounded by $\gamma_1$. Therefore, by Cauchy's integral theorem,

$$\oint_{\gamma_1} dz \frac{\log{(1-z)}}{z} = 0 $$

From part 1), this implies that

$$\oint_{|z|=1} dz \frac{\log{(1-z)}}{z} = 0 $$

Let $z=e^{i \theta}$, $dz/z = i d\theta$, then

$$i \int_0^{2 \pi} d\theta \log{(1-e^{i \theta})} = 0$$

Now

$$\log{(1-e^{i \theta})} = \log{|1-e^{i \theta}|} +i \arg{(1-e^{i \theta})}$$

It turns out that

$$\arg{(1-e^{i \theta})} = \arctan{\frac{\sin{\theta}}{1-\cos{\theta}}} = \arctan{(\cot{\frac{\theta}{2}})} = \frac{\pi}{2} - \frac{\theta}{2}$$

$$\int_0^{2 \pi} d\theta \: \arg{(1-e^{i \theta})}= \int_0^{2 \pi} d\theta \left ( \frac{\pi}{2} - \frac{\theta}{2} \right ) = \frac{\pi}{2} 2 \pi - \frac{1}{4} (2 \pi)^2 = 0 $$

$$\therefore \int_0^{2 \pi} d\theta \log{|1-e^{i \theta}|} = 0$$

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  • $\begingroup$ Edited and complete. $\endgroup$ – Ron Gordon Feb 21 '13 at 3:34
  • $\begingroup$ Thank you very much! I Edited the problem, now it is complete $\endgroup$ – ziang chen Feb 21 '13 at 3:40
  • $\begingroup$ Can you point out any faults with the solution? $\endgroup$ – Ron Gordon Feb 21 '13 at 4:15
  • $\begingroup$ oh! I am really sorry. we cannot use Cauchy's integral theorem on $|z|=1$ $\endgroup$ – ziang chen Feb 21 '13 at 4:40
  • $\begingroup$ I apologize, I misunderstood what you said. $\endgroup$ – Ron Gordon Feb 21 '13 at 4:42

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