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

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

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

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

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

Est.Lem. = Estimation Lemma of Cauchy

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

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.