geometric description of set of complex number

Question

A set of complex number: $$S=\{ z\in \Bbb C : |z|=\lambda |z-1|\}$$

what's the geometric description?

I try to draw it ... which seems like a circle but cannot find the equation to describe it..

Answer 1

Here is a conformal mappings perspective. Let $w=\dfrac{z}{z-1}$. Then the equation in the question implies that $|w|=\dfrac{|z|}{|z-1|}=\lambda$. This is equivalent to $w=\lambda e^{i\phi}$ for $\phi\in(-\pi,\pi)$; geometrically, these are circles of radius $\lambda$ in the complex $w$-plane. To see what this looks like in the complex $z$-plane, we solve for the inverse map $z=\dfrac{w}{w-1}$. The image of $w=\lambda e^{i\phi}$ is then $$z(\phi)=\dfrac{\lambda e^{i\phi}}{\lambda e^{i\phi}-1}\cdot\dfrac{\lambda e^{-i\phi}-1}{\lambda e^{-i\phi}-1}=\dfrac{\lambda^2-\lambda e^{i\phi}}{1-2\lambda \cos\phi +\lambda^2}=\frac{(\lambda^2-\lambda \cos\phi)-i\lambda \sin\phi}{1-2\lambda\cos\phi +\lambda^2}$$

Note that, since we can identify this to a curve $(x(\phi),y(\phi))\in\mathbb{R}^2$, this is no longer a problem of complex variables but rather of identifying a parametrized curve. (I'm having a spot of trouble of writing out said identification properly at the moment; I'll see if I can update soon.)

Answer 2

Hint: Suppose $\lambda >0$. Then, $|z|=\lambda|z-1|$ is same as $|z|^2=(\lambda)^2|z-1|^2$. Writing $z=x+iy$ we have,

$x^2+y^2=(\lambda)^2((x-1)^2+y^2)$. Brings the $x's$ and the $y's$ to the same side.

For $\lambda < 0$, there is no solution.

For $\lambda=0$, the solution is just one point.