Q:

If $|f(z)-w|<|w|/2$ how do you conclude $|f(z)|>|w|/2$? $z$ and non-zero $w$ are complex numbers.

If $|f(z)-w|<|w|/2$ how do you conclude $|f(z)|>|w|/2$?

$z$ and non-zero $w$ are complex numbers. $f$ is a function of $z$. The inequality is defined on a domain $S$ in complex plane.

A:

You can prove it using reverse triangle inequality. You have
$$\frac{|w|}{2} > |w - f(z)| \ge ||w| - |f(z)|| \ge |w| - |f(z)|,$$
which, when rearranged, implies
$$|f(z)| > |w| - \frac{|w|}{2} = \frac{|w|}{2}.$$
This works wherever the triangle inequality works, i.e. in real numbers, complex numbers, normed linear spaces, etc.

