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:
$|\,1 - z\,| < k\ (1 - |\,z\,|\, )$?
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Will it form a circle, with radius $\frac{k}{\sqrt{1+k^2}}$ and with its center at $\left( \frac{1}{\sqrt{1+k^2}}, 0 \right)$?

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)
