Complex Modulus Is Less Than One Show that $|\frac{-a}{b}+\frac{\sqrt{a^2-b^2}}{b}|<1$ where $a>|b|>0$
This is just a minor part of a more full complex trigonometric integral proof that I'm working on. I'm pretty much set once I can show this inequality is true. Please Help!
What I was thinking was:
$|\frac{-a}{b}+\frac{\sqrt{a^2-b^2}}{b}|=|\frac{-a}{b}+\frac{a}{b}\sqrt{1-\frac{b^2}{a^2}}|$
Then $\sqrt{1-\frac{b^2}{a^2}}<1$, Which implies $\frac{a}{b}\sqrt{1-\frac{b^2}{a^2}}<\frac{a}{b}$
The problem is, this doesn't guarantee that first inequality is true, in less I'm missing something.

Hint:
$$
\sqrt{a^2-b^2}-a = \frac{a^2-b^2-a^2}{\sqrt{a^2-b^2}+a}$$
details:
$$
0\le a-\sqrt{a^2-b^2}
 = \frac{b^2}{\sqrt{a^2-b^2}+a}\le |b| \frac {|b|}a\le |b|$$
