Why is $|z_1| < 1$ and $|z_2| > 1$?

Question

Here are the values of $z_1$ and $z_2$

$$z_1 = \frac{-i + i \sqrt{1-r^2}}{r}$$

$$z_2 = \frac{-i - i \sqrt{1-r^2}}{r}$$

Where $0 < r < 1.$

Then, the writer said , then we have $|z_1| < 1$ and $|z_2|>1$ but I calculated $|z_1|$ and $|z_2|$ and I found them equals $$\frac{1 + 2 \sqrt{1 - r^2} + (1-r^2)}{r^2}$$ So, I do not know why the author said this actually. Here is the link for the question that contains this claim Several questions on calculating $\int_0^{2\pi} \frac{\mathrm{d} \theta}{1+r\sin \theta}$ via contours , could anyone clarify this to me please? and sorry if this was a very trivial question.

EDIT:

I do not see why $|z_1|$ looks like this. Mine is just $\sqrt{\frac{(-1)^2}{r^2} + (\frac{\sqrt{-1 + r^2}}{r})^2} = 1$

Can someone show me the detailed calculations please?

Answer 1

Actually, $|z_1|=\frac {1-\sqrt {1-r^{2}}} r$ and $|z_2|=\frac {1+\sqrt {1-r^{2}}} r$. Note that $1-r+\sqrt {1-r^{2}} >0$ (which gives $|z_2|>1$) and $\sqrt {1-r} <1<\sqrt {1+r} $ (so $(1-r)<\sqrt {(1-r)(1+r)}=\sqrt {1-r^{2}}$, which gives $|z_1| <1$).