Let $z\in\mathbb{C}$ where $|z|<1$. Prove the inequality $$\left|-1+\sqrt{1-z^2}\right|<|z|.$$
I am having trouble showing this. All my attempts have failed. Here is one attempt that I believe is most promising, but I am stuck. $$ |z|=\left|i\sqrt{z^2}\right|=\left|1-1+\sqrt{-1+1-z^2}\right|$$ This must be greater than $\left|-1+\sqrt{1-z^2}\right|$ but I can't figure out how. Any insight would be greatly appreciated!
Hint: $\;\;\displaystyle\left|-1+\sqrt{1-z^2}\right| \cdot \left|-1-\sqrt{1-z^2}\right| = |z|^2\,$, therefore:
$$\left|-1+\sqrt{1-z^2}\right| \le |z| \;\;\iff\;\; \left|-1+\sqrt{1-z^2}\right| \le \left|-1-\sqrt{1-z^2}\right|$$
But the principal value of the square root $\,w = \sqrt{1-z^2}\,$ always lies in the right half-plane $\,\operatorname{Re}(w) \ge 0\,$, so it is geometrically obvious that the distance $\,w-1\,$ from $w$ to $+1$ is no larger than the distance $w+1$ from $w$ to $-1$.