Show that the set $z$ satisfying $|z-z_0|=\rho|z-z_1|$ ($\rho \neq 1$) is a circle. Prove that for $\rho \ge 0$, $\rho \neq 1$ and fix $z_0,z_1 \in \Bbb C$. Show that the set $z \in \Bbb C$ satisfying $|z-z_0|=\rho|z-z_1|$ is a circle.
From the given condition I have reached that $(1-\rho^2)|z|^2 -2\mathfrak{Re} (\bar z (z_0-\rho^2 z_1)) + |z_0|^2 -\rho^2|z_1|^2=0$
But I do not get my desired result.
 A: From where you left off
$$(1-\rho^{2})|z|^{2}-\lbrace z^{*}(z_{0}-\rho^{2}z_{1})+z(z_{0}-\rho^{2}z_{1})^{*}\rbrace=\rho^{2}|z_{1}|^{2}-|z_{0}|^{2}$$
Dividing by $(1-\rho^{2})$ and completing the square
$$|z-\frac{1}{(1-\rho^{2})}(z_{0}-\rho^{2}z_{1})|^{2}=\frac{\rho^{2}|z_{1}|^{2}-|z_{0}|^{2}}{1-\rho^{2}}+|\frac{1}{(1-\rho^{2})}(z_{0}-\rho^{2}z_{1})|^{2} $$
A: WOLOG, we may assume $z_1 =0$. Then we are looking at $|z-z_0|=\rho |z|$. Next, by rotation, we may assume that $a=z_0$ is real and positive.
Now, let $z=x+iy$ and square the equation to get
$(x-a)^2 +y^2=\rho^2 (x^2 +y^2)$. When this is expanded, a conic section is obtained with equal coefficients for $x$ and $y$. This is a circle. 
Alternatively, and using what you wrote, let $z=x+iy$ and find that the coefficients of $x^2$ and $y^2$ are the same, and there is $xy$ term. This gives a circle.
A: Hint..you can write $z=x+iy$, $z_0=a+ib$, $z_1=c+id$ for real $x,y,a,b,c,d$ and apply your condition so that you get$$(x-a)^2+(y-b)^2=\rho^2[(x-c)^2+(y-d)^2]$$ which will lead to the equation of a circle
A: Begin by noting that $\langle x+iy,a+ib \rangle = xa+yb = \text{Re}\left((x+iy)(a-ib)\right)$ so if we think of $z$ and $w$ as two-dimensional vectors then the complex algebra permits a natural formulation of the Euclidean inner-product in the plane:
$$ \langle v,w \rangle  = \text{Re} (v \bar{w}) $$
On the other hand, we have $|z|^2 = z \bar{z}$ and the many wonderful properties of the complex conjugate. Suppose $|z-z_o| = \rho|z-z_1|$ for some $\rho>0$ (if I say $\rho>0$ then it must be real !). Square the given condition and use properties of complex conjugation:
$$ (z-z_o)(\bar{z}-\bar{z_o}) = \rho^2(z-z_1)(\bar{z}-\bar{z_1})$$
or
$$ z\bar{z}-z\bar{z_o}-z_o\bar{z}+z_o\bar{z_o} = \rho^2\left( z\bar{z}-z\bar{z_1}-z_1\bar{z}+z_1\bar{z_1}\right)$$
which gives
$$ |z|^2-z\bar{z_o}-z_o\bar{z}+|z_o|^2 = \rho^2\left( |z|^2-z\bar{z_1}-z_1\bar{z}+|z_1|^2\right)$$
which we can rearrange to
$$ \rho^2z\bar{z_1}+\rho^2z_1\bar{z}-z\bar{z_o}-z_o\bar{z} = 
(\rho^2-1)|z|^2 +\rho^2|z_1|^2- |z_o|^2$$
Ok, I'm tired of $\rho$, I'm setting $\rho=1$ in which case:
$$ z\bar{z_1}+z_1\bar{z}-z\bar{z_o}-z_o\bar{z} = 
|z_1|^2- |z_o|^2$$
which gives, using $\text{Re}(z) = \frac{1}{2}(z + \bar{z})$,
$$ z(\overline{z_1-z_o})+\bar{z}(z_1-z_o) = 2\text{Re}\left(z(\overline{z_1-z_o})\right) = |z_1|^2- |z_o|^2$$
or,
$$ \langle z, z_1-z_o \rangle = \frac{1}{2}\left(|z_1|^2- |z_o|^2\right) $$
Now, if we have $|z_o| = |z_1|$ then it follows:
$$ \langle z, z_1-z_o \rangle = 0 $$
which says $z$ points in a direction perpendicular to the line-segment from $z_0$ to $z_1$. In any event, you can use the things I've done here to find the answer.
