How find all complex numbers such that: $|\,1 - z\,| < k\ (1 - |\,z\,|\, )$? 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 $\left( \frac{1}{\sqrt{1+k^2}}, 0 \right)$?
 A: 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 $k=10$.

A: 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)
A: 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.
A: 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 $$f(a,b)<k$$
