Region in complex plane with $|1-z|\leq M(1-|z|)$ Let $M>0$. Describe the region in the complex plane such that $|1-z|\leq M(1-|z|)$.
To start, I take $M=1$. The inequality becomes $|1-z|\leq 1-|z|$. But by triangle inequality, we have $|1-z|+|z|\geq |(1-z)+z| = 1$. We must have equality, and it holds when $z\in [0,1]$.
For arbitrary $M$, the inequality becomes $|1-z|+M|z|\leq M$. I don't really know what to do with this, except that any $|z|>1$ is clearly ruled out because then $M|z|>M$.

Your region consists of all $z \in \mathbb C$, such that the ratio $$\frac{|1-z|}{1-|z|} $$ is bounded (by $M$). As you mentioned only points within the unit disk are admissible.
More precisely, the region is a subset of the unit disk, which is contained within a circular wedge of angle $\alpha=\alpha(M)$ (the Stolz angle). The higher $M$ gets, the wider $\alpha(M)$ is (as you mentioned $\alpha(0)=0)$.
I recommend this wolfram demonstration, for more insights.   
