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I just need to show that :

$$\int_0^{2\pi}\left|{\frac{i(Re^{i\theta})^\lambda}{1+Re^{i\theta}}}\right| d\theta \le \int_0^{2\pi} \frac{R^\lambda}{R-1}d\theta : 0 < \lambda <1 , R>1$$ Is there some trivial geometrical argument I don't see?

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@t.b. I would assume that this integral has come up in the context of some contour integration, and that probably the idea is to take $R \to \infty$. But maybe the OP should clarify this. –  Adrián Barquero Sep 12 '12 at 1:07
    
so technically we need R to tend to infinity .... because this is a step in computing a contour integral.... And yeah, you're right. –  Squirtle Sep 12 '12 at 1:08
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@AdriánBarquero: I certainly agree... I was just pointing out that some hypotheses were missing. –  t.b. Sep 12 '12 at 1:11

2 Answers 2

up vote 3 down vote accepted

You just have to use the following inequality, valid for complex numbers $a, b \in \mathbb{C}$

$$|a| - |b| \leq |a \pm b|$$

which in this case applied to $a = Re^{i\theta}$ and $b = 1$ becomes

$$ |Re^{i\theta}| - |1| \leq |Re^{i\theta} + 1| $$

so this implies that

$$ \frac{1}{|Re^{i\theta} + 1|} \leq \frac{1}{|Re^{i\theta}| - |1|} = \frac{1}{R - 1} $$

Then using this, proving the inequality with the integrals should be straightforward to you.

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he said $|1+Re^i\theta|$, not $|1+Re^{i\theta}|$ EDIT: its fixed now ^^ –  Integral Sep 12 '12 at 1:07

$|i(Re^{i\theta})^\lambda| = R^\lambda$

$|1+Re^{i\theta}| \geq |Re^{i\theta}| - |1| = R - 1\rightarrow \frac{1}{|1+Re^{i\theta}|}\leq\frac{1}{R-1}$

then

$$\frac{|i(Re^{i\theta})^\lambda|}{|1+Re^{i\theta}|}\leq\frac{R^\lambda}{R-1}$$

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