How to find a complex number that satisfies $a|z| = |z-1|$?

Question

Let $a \in \mathbb{R}$ and $a \neq 1$, $a \neq 0$. Find $z \in \mathbb{C}$ where $a|z| = |z-1|$.

I tried this:

$ a\sqrt{x²+y²} = \sqrt{(x-1)² + y²} \implies a²(x²+y²) = (x-1)² + y² \implies (a²-1)x² + 2x + (a²-1)y² - 1 = 0$

Then I tried to use Bhaskara Formula, but got nowhere.

How can I solve this?

Answer 1

You can continue, divide by $a^2-1$ (you can, since $a\ne 1$) and you get:

$$\underbrace{x^2-{2\over a^2-1}x +\color{red}{1\over (a^2-1)^2}}-\color{red}{1\over (a^2-1)^2} +y^2 ={1\over a^2-1}$$

and then $$ \Big (x-{1\over a^2-1}\Big) ^2+y^2 = {a^2\over (a^2-1)^2}$$

So this is a circle with center at $({1\over a^2-1},0)$ and radius $r=\Big|{a\over a^2-1}\Big|$

Answer 2

There are no solutions if $a<0$. Otherwise an easier option is $a=|1-1/z|$ so $1-1/z=ae^{i\phi}$ with $\phi\in[0,\,2\pi)$. Hence $z=\dfrac{1}{1-ae^{i\phi}}$, which you can rewrite in Cartesian or polar form according to taste.

Answer 3

If $a<0$ you cannot find such a point $z$. If $a >0$ the $z =\frac 1 {1-a}$ is a solution.