Why is $\left| \int_{|z|=1} \frac{f(z)}{z}dz\right| \leq \int_{|z|=1} \frac{|f(z)|}{|z|}|dz|$?

Question

Why is

$$\left| \int_{|z|=1} \frac{f(z)}{z}dz\right| \leq \int_{|z|=1} \frac{|f(z)|}{|z|}|dz|$$

and not

$$\left| \int_{|z|=1} \frac{f(z)}{z}dz\right| = \int_{|z|=1} \frac{|f(z)|}{|z|}|dz|?$$

Answer 1

This is the triangle inequality in integral form, $$ \left\lvert\int_a^b f(x) \, dx \right\rvert \leqslant \int_a^b \lvert f(x) \rvert \, dx $$ Simple example of this: $$ \left\lvert \int_{-1}^1 x \, dx \right\rvert = \frac{1-1}{2} = 0, $$ whereas $$ \int_{-1}^1 \lvert x \rvert \, dx = 1, $$ which is easy to see, even by drawing the graph.

For an example like yours, take $f(z)=z^2$ Then $$ \int_{|z|=1} \frac{z^2}{z} \, dz = i \int_0^{2\pi} e^{i\theta} \, d\theta = 0, $$ but $$ \int_{|z|=1} \frac{|z^2|}{|z|} \, |dz| = \int_0^{2\pi} 1 \, d\theta = 2\pi. $$