Denote by $Γ$ the circular arc {$z ∈ C : |z| = 3,\quad 0 ≤ Arg(z) ≤ \frac{π}{2}$} described in the anticlockwise direction. Consider the principal branch of $z^1/2$. Using the ML lemma, without evaluating the integral, show that $$\left|\int_{Γ}\frac{z^{\frac{1}{2}}}{z^2 + 1}dz\right|≤\frac{3√3π}{16}$$ I have an idea of what the ML lemma is and how to use it but I'm having some difficulty showing this one.
ML Lemma problem
1
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complex-analysis
complex-integration
integral-inequality
2 Answers
1
For $z\in\Gamma$: $$9=|z^2|=|z^2+1-1|\le|z^2+1|+1$$ So $\frac{1}{|z^2+1|}\le \frac{1}{8}$. Thus: $$\left|\frac{z^{1/2}}{z^2+1}\right|\le\frac{\sqrt 3}{8}$$
And by the M_L lemma: $$\left|\int_{\Gamma}\frac{z^{\frac{1}{2}}}{z^2 + 1}dz\right|\le \frac{\sqrt 3}{8}l(\Gamma)=\frac{\sqrt 3}{8}\frac{3\pi}{2}=\frac{3\pi\sqrt 3}{16}$$
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see https://en.wikipedia.org/wiki/Estimation_lemma for the ML-Lemma. Try now to show the inequalitiy.