How can I overestimate, $|\oint_{|z|=R} f(z) \mathrm{d}z |$ with $f(z)=\frac{z^a}{z^2+1}$, $0<|a|<1 \; \mathrm{with} \;a \in \mathbb R$ ?
I tried this: $|\oint_{|z|=R} f(z) \mathrm{d}z | <= \mathrm{length \;curve \times maximum\; modulus}$ =$2\pi R\times M $.
Unfortunately, I don't know how to find $M$.
Can someone please help me out?
If $\left|z\right|=R$, then, since $a$ is a real number :
$$\left|\dfrac{z^a}{z^2+1}\right|=\dfrac{\left|z^a\right|}{\left|z^2+1\right|}=\dfrac{R^a}{\left|z^2+1\right|}$$
Now, following Evan's indication, for all $z\in\mathbb C$ :
$$\left|z^2+1\right|\geqslant\left|\left|z^2\right|-1\right|\geqslant\left|z^2\right|-1=\left|z\right|^2-1$$
For $\left|z\right|=R$, you get :
$$\left|z^2+1\right|\geqslant R^2-1$$
If $R>1$ and $\left|z\right|=R$, you can invert both sides, reversing the inequality :
$$\dfrac 1{\left|z^2+1\right|}\leqslant \dfrac 1{R^2-1}$$
From there, you can easily find your $M$.