Let $k > 1$ be a real number. How may one find all complex numbers such that: $|\,1 - z\,| < k\ (1 - |\,z\,|\, )$?
................................................................................................................................................................ Will it form a circle, with radius $\frac{k}{\sqrt{1+k^2}}$ and with its center at $\left( \frac{1}{\sqrt{1+k^2}}, 0 \right)$?
Let $z=x+iy$. Then your inequality is $$ \sqrt{(1-x)^2+y^2}<k(1-\sqrt{x^2+y^2}). $$ It turns out, that it can be expressed in the following way: $$ \bigl\{(x,y)~|~(1-k)/(1+k)<x<1,\ -f(x,k)<y<f(x,k)\bigr\}, $$ with (this I got from the computer, I leave it to you to actually do the calculation) $$ f(x,k)=\frac{1}{1-k^2}\sqrt{k^4(1-x^2)-(1-x)^2+2k^2\Bigr(1-x+x^2-\sqrt{2k^2(1-x)+2x-1}\Bigr)}. $$
This is what the domain looks like for $k=1.1$, $k=2$, $k=3$ and $k=10$.
This never can be a circle. However :
$$ z= x +iy $$ $$ |1-z| =\sqrt{ (1-x)^2 +y^2 } $$ $$ 1- |z|=1-\sqrt{x^2 +y^2} $$ $$ (1-x)^2 +y^2 <(K-K\sqrt{x^2 +y^2})^2$$ $$ (1-x)^2 +y^2 -K^2 -K^2(x^2+y^2)<-2K^2\sqrt{x^2 +y^2}$$ $$ 1+x^2-2x+y^2-K^2-K^2x^2-K^2y^2<-2K^2\sqrt{x^2 +y^2}$$ $$1-K^2+x^2(1-K^2) + y^2(1-K^2)-2x<-2K^2\sqrt{x^2 +y^2}$$ $$1-\frac{2}{(1-K^2)}x+x^2 + y^2<\frac{-2K^2}{(1-K^2)}\sqrt{x^2 +y^2}$$ $$\frac{(1-\frac{2}{(1-K^2)}x+x^2 + y^2)^2}{x^2 +y^2}<\frac{4K^4}{(1-K^2)^2}$$ $$\frac{(1-\frac{2}{(1-K^2)}x+x^2 + y^2)^2}{x^2 +y^2}<\frac{4K^4}{(1-K^2)^2}$$
$$ x=rcos\theta $$ $$ r=rsin\theta$$
$$\frac{(1+r^2-\frac{2rcos\theta}{(1-K^2)})^2}{r^2}<\frac{4K^4}{(1-K^2)^2}$$
you can plot this expression to investigate what it really is. (The polar is optional)
Is not the circle.
The point $$\left(\frac{1+k}{\sqrt{1+k^2}},0\right)$$ is in that circle, but this point is outside the unit circle, since $1+k=\sqrt{1+2k+k^2}>\sqrt{1+k^2}$. And no point outside the unit circle can satisfy the inequality, becuase it makes the RHS negative.
Hint: you must consider the term $$f(a,b)=\frac{\sqrt{(1-a)^2+b^2}}{1-\sqrt{a^2+b^2}}$$ for such $f(a,b)$ must be $$f(a,b)<k$$
