Complex locus of circle

Question

In my book it is given

$\left\lvert \frac{z-z_1}{z-z_2}\right\rvert = k$ where k is not equal to 1 &0.

Here z represents the locus of circle .

But I could not understand here what is $z_1$ & $z_2$

Answer 1

The symbols $z_1$ and $z_2$ represent fixed numbers in the complex plane. The formula $$\left\lvert \frac{z-z_1}{z-z_2}\right\rvert = k$$ states that the distance between $z$ and $z_1$ is a constant multiple of the distance between $z$ and $z_2$ and your book asserts that the set of all $z$ satisfying this is a circle.

Here's one way to understand this. If $$\left\lvert \frac{z-z_1}{z-z_2}\right\rvert = k,$$ then $$\frac{z-z_1}{z-z_2} = k \, e^{it}$$ for some $t$. If we solve this for $z$, we obtain $$z = \frac{k \, e^{i t} z_2-z_1}{k \, e^{i t} - 1}.$$ Thus, $z$ is the image of a point on the unit circle under the Mobius transformation $$w \to \frac{k \, w \, z_2-z_1}{k \, w - 1}$$ and Mobius transformations are well known to map circles to circles or lines. In the case that $k=1$, we have that $1\to\infty$ and the image of the unit circle is a line. In fact, when $k=1$, we should get the line that bisects the line segment from $z_1$ to $z_2$.

Here's an illustration of the situation when $z_1=1+i$, $z_2=-2-i$, and $k=1/2$:

If we draw a line segment from $1+i$ to a point on the circle that segment should be half as long as a segment from the point $-2-i$ to the same point on the circle.