3
$\begingroup$

proving this inequality $$\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \frac{4\pi}{3}$$

I tried with $$\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \int_{\left | z \right |=2}^{} \left | \frac{dz}{z^2+1} \right |$$

but I dont know what next

I think maybe we need to use triangular inequality

$\endgroup$
2
  • 5
    $\begingroup$ Find a lower bound for $\lvert z^2+1\rvert$ on the circle. $\endgroup$
    – Daniel Fischer
    Sep 4, 2013 at 15:38
  • $\begingroup$ Well, the integral can be evaluated quite easily. If you do it correctly, then you will obtain this very trivial inequality $$0\leq\frac{4\pi}{3}\,.$$ $\endgroup$
    – Batominovski
    Sep 16, 2018 at 17:13

2 Answers 2

Reset to default
1
$\begingroup$

$$\left|\;\;\oint\limits_{|z|=2}\frac{dz}{z^2+1}\;\;\right|\le\oint\limits_{|z|=2}\frac{dz}{|z^2+1|}\stackrel{\text{Est. Lem.}}\le\max_{|z|=2}\left(\frac1{|z|^2-1}\right)\cdot l\left(\{|z|=2\}\right)=$$

$$=\frac13\cdot4\pi=\frac{4\pi}3$$

Est.Lem. = Estimation Lemma of Cauchy

$\endgroup$
2
  • $\begingroup$ do you know another link help me with complexe integral inequality ? $\endgroup$
    – user79560
    Sep 4, 2013 at 15:50
  • $\begingroup$ @rama , simply google "Estimation Lemma" . $\endgroup$
    – DonAntonio
    Sep 4, 2013 at 15:52
0
$\begingroup$

Hint: Note that by the triangle inequality $$ |z^{2}+1|\geq||z^{2}|-|1||=|2^{2}-1|=3 $$

since $z$ satisfies $|z|=2$ on the path of the integration.

$\endgroup$

You must log in to answer this question.