Estimation Theorem on a straight line in complex plane

Question

Apply the Estimat Theorem (ML inequality) to show that $$\bigg| \int_{[0, 1+i]}(z^2+1)^{-1}\bigg|\leq \sqrt 2$$

Note that $\gamma(t)=t(1+i), t\in [0, 1]$ and $\gamma'(t)=1+i$. Let $f(z)=(z^2+1)^{-1}$, so that \begin{align*} \bigg|\int_{\gamma}f(z)\ dz\bigg| \leq \int_{0}^{1}|f(\gamma(t))||\gamma'(t)|\ dt = \int_{0}^{1} \bigg|\frac{1+i}{2it^2+1}\bigg|\ dt = \int_{0}^{1}\frac{\sqrt{2}}{\sqrt{1+4t^4}}\ dt \end{align*} I am stuck at this point because I not sure I can just take the $\sqrt{2}$ and that is. I am also not sure it is integrable. Any help will be great. Thanks in advance.

Answer 1

You are making things unnecessarily complicated.

The length of $[0,1+i]$ is $\sqrt{2}$ and $(1+z^2)^{-1}$ has magnitude $\leq 1$ (achieved only at $z=0$). So ML-inequality gives the answer claimed.