Calculating $\int_0^{\infty}\frac1{1+z^4}dz$, how did we get this inequality?

Question

In a complex analysis lecture, which is related to Calculus of residues, I am shown how to calculate this improper integral using residues:

They compute $$\int_0^{\infty}\frac1{1+z^4}dz$$, and the argument goes like this:

$$\left|\int_{C_R}\frac1{1+z^4}dz\right|\le\int_{C_R}\left|\frac1{1+z^4}\right|\,d|z|\le\int_{C_R}\frac1{\left|z\right|^4}\,d\left|z\right|=\pi R\frac1{R^4}$$ where $C_R$ is positively oriented circle around the origin with radius $R$. Now, since as $R\to\infty$, this integral is zero, hence they use one of the residue theorems to compute the final result.

But, what I can't get in this argument is: How did they get $|z^4|\le|1+z^4|$ over $C_R$?

Answer 1

The inequality is wrong. The correct one is $|\frac 1 {1+z^{4}}| \leq \frac 1 {R^{4}-1}$ and this works fine in the proof.

[Recall that $|a+b| \geq |a|-|b|$. This gives $|1+z^{4}| \geq |z|^{4}-1=R^{4}-1$].