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

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 $\sqrt{a^2 + b^2} \geq |a| \geq a$.
