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

Question

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.

Answer 1

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.

Answer 2

Because\begin{align}\bigl\lvert f(z)\bigr\rvert&=\bigl\lvert f(z)-w+w\bigr\rvert\\&\geqslant\left\lvert\bigl\lvert f(z)-w)\bigr\rvert-w\right\rvert\\&=\lvert w\rvert-\bigl\lvert f(z)-w\bigr\rvert\\&>\dfrac{\lvert w\rvert}2.\end{align}

Answer 3

Hint: Use the triangle inequality with $$\big||a|-|b|\big|\leq|a-b|.$$