Proof for $|1 - z| \geq 1 - |z|$ for $|z| < 1$, $z \in \mathbb{C}$

Question

I can prove it "by picture" by drawing a picture of a circle of radius $|z|$ centered at $(0, 1)$. Then $1 - |z|$ is the length from the origin to the intersection of the circle with the x-axis (to the left). $|1 - z|$ is length from the origin to any point on the circle.

However, I'm having trouble proving this algebraically / with words (aka not proof by picture). I tried using this picture again and the triangle equality, but that didn't work out.

Answer 1

It is well-known that the triangle inequality

$\vert a + b \vert \le \vert a \vert + \vert b \vert \tag{1}$

implies that

$\vert \vert a \vert - \vert b \vert \vert \le \vert a - b \vert. \tag{2}$

If we apply (2) to the case at hand, taking $a = 1$ and $b = z$, with $\vert z \vert < 1$, we find

$\vert 1 - \vert z \vert \vert \le \vert 1 - z \vert; \tag{3}$

but $\vert z \vert < 1$ implies

$\vert 1 - \vert z \vert \vert = 1 - \vert z \vert > 0, \tag{4}$

and using (4) in (3) yields

$1 - \vert z \vert \le \vert 1 - z \vert, \tag{5}$

which is the desired result.

Note: A proof that (2) follows from (1) may be found here. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 2

let $z = a + bi$ with $a^2 + b^2 < 1$, then $|1-z| = |1 - (a+bi)| = |1-a + (-b)i| = \sqrt{(1-a)^2 + b^2}$. So we need to prove: $\sqrt{(1-a)^2 + b^2} \geq 1 - \sqrt{a^2 + b^2}$. Squaring both sides of the above inequality we have: $(1-a)^2 + b^2 \geq 1 - 2\sqrt{a^2 + b^2} + a^2 + b^2 \iff -2a \geq -2\sqrt{a^2 + b^2} \iff \sqrt{a^2 + b^2} \geq a$ which is true since $\sqrt{a^2 + b^2} \geq |a| \geq a$.