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Show that $$\left| \oint_{|z| = R} \frac{\log z}{z^2}\,dz \right| \leq 2 \sqrt{2} \pi \frac{\log R}{R}, \quad R > e^\pi$$

I tried solving it, but I’m not sure if my methods are correct.

First, let $z =Re^{i \theta},\: 0 \leq \theta \leq 2 \pi$ and $R > e^\pi \Rightarrow dz = iRe^{i \theta} d \theta$.

\begin{align} \Biggl| \oint_{|z| = R} \frac{\log z}{z^2} dz\, \Biggr| &= \Biggl| \int_0^{2\pi} \frac{\log Re^{i \theta}}{(Re^{i\theta})^2} iRe^{i \theta} d\theta \,\Biggr| = \Biggl| \int_0^{2\pi} \frac{\log R + \log e^{i \theta}}{Re^{i\theta}} d\theta\, \Biggr| \\ &\leq \Biggl| \int_0^{2\pi} \frac{\log R + e^{i\theta} e^{i\theta}}{Re^{i \theta}} d\theta\,\Biggr| \leq\Biggl| \int_0^{2\pi} \frac{\log R \ e^{i\theta}}{R} d\theta \,\Biggr| \leq \Biggl|\int_0^{2\pi} \frac{\log R(1)}{R} d\theta \,\Biggr| \\ &= \frac{\log R}{R} \theta \biggm|_0^{2\pi} = \frac{\log R}{R}(2\pi) \leq 2 \sqrt{2} \pi \frac{\log R}{R} \end{align} Thanks in advance.

Boots
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user366669
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  • You have to specify which branch of logarithm you are using. It is not true that $\log(ab)=\log, a+\log, b$ so your way of handling $\log(Re^{i\theta})$ is not correct. – Kavi Rama Murthy Feb 26 '19 at 23:24
  • That’s true! Sorry, I’ll fix it accordingly after class. I goofed. – user366669 Feb 26 '19 at 23:59
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    @KaviRamaMurthy In fact, $\log(z_1z_2)=\log(z_1)+\log(z_2)$ where the equality is interpreted in terms of set equality. See THIS – Mark Viola Feb 27 '19 at 00:36
  • You also may want to remember that $\left| \oint_{|z| = R} \frac{\log z}{z^2},dz \right| \le \frac{\log(R)}{R^{2}}$ – Boots Feb 27 '19 at 02:13
  • Correction, I think it should be $\le \frac{| \log (z) |}{R^2}$ – Boots Feb 27 '19 at 02:20

1 Answers1

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The solution is not correct as $\log{e^{i\theta}}$ is just $i\theta$ so the first inequality doesn't make sense, plus you never use the hypothesis about $R > e^\pi$.

The way I would do it is like this; we parameterize the circle as $z =Re^{i \theta},\: -\pi \leq \theta \leq \pi$, so $\log{z} = \log{R} + i\theta$ and then:

$\Biggl| \oint_{|z| = R} \frac{\log z}{z^2} dz\, \Biggr| = \Biggl| \int_{-\pi}^{\pi} \frac{\log Re^{i \theta}}{(Re^{i\theta})^2} iRe^{i \theta} d\theta \,\Biggr| = \Biggl| \int_{-\pi}^{\pi} \frac{\log R + i\theta}{Re^{i\theta}} d\theta\, \Biggr| \\ \leq \ \int_{-\pi}^{\pi} \Biggl|\frac{\log R + i\theta }{Re^{i \theta}}\Biggr| d\theta\, = \frac{1}{R}\int_{-\pi}^{\pi}({\log^2{R} + \theta^2})^{\frac{1}{2}} d\theta \leq \frac{1}{R}\int_{-\pi}^{\pi}(2{\log^2{R})^{\frac{1}{2}}}d\theta = 2\sqrt{2} \pi \frac{\log R}{R}$

where we used $R > e^\pi$ to conclude $\log R > |\theta|$ for $-\pi \leq \theta \leq \pi$

Conrad
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