I am reading "Lectures on Complex Function Theory"(in Japanese) by Takaaki Nomura.
There is the following problem in this book.
Let $0 < \alpha < \frac{\pi}{2}$.
Let $\mathcal{D} := \{z \in \mathbb{C} \mid |z-1|<\cos \alpha, |\arg(1-z)| < \alpha \}$, where $-\pi < \arg(1-z) \leq \pi$.
Prove that if $z \in \mathcal{D}$, then $|z| < 1$ and $\frac{|1-z|}{1-|z|}<\frac{2}{\cos \alpha}$.
I can understand why $|z| < 1$ by the picture below.
But I want to prove $|z| < 1$ by logic.
Please give me a proof of $|z| < 1$.
Let $r := |z-1|$ and $\theta := \arg(1-z)$.
Then, $z \in \mathcal{D} \iff r < \cos \alpha$ and $|\theta| < \alpha$.
$1-z = r (\cos \theta + i \sin \theta)$.
$z = (1 - r \cos \theta) - i r \sin \theta$.
$|z|^2 = (1- r \cos \theta)^2 + r^2 \sin^2 \theta = 1 + r^2 - 2 r \cos \theta = 1 - r (2 \cos \theta - r)$.
Since $r < \cos \alpha < \cos \theta < 2 \cos \theta$, $|z|^2 < 1$.
So $|z| < 1$.