1. ## complex number

the centre of circle represented by mod(z+1)=2mod(z-1) is

2. Hello, prasum!

$\text{The centre of circle represented by: }\;|z+1| \:=\:2|z-1|$

$|z+1|\:=\:|z - (\text{-}1)|$ is the distance from $\,z$ to (-1,0).

$|z-1|$ is the distance from $\,z$ to (1,0).

Code:
                |     P
|     o
| *    *
* |       *
A    *     |        * B
- - o - - - - + - - - - o - -
-1         |         1
|

We want points $P(x,y)$ so that: . $\overline{PA} \;=\;2\!\cdot\!\overline{PB}$
[This is known as the Circle of Apollonius.]

We have: . $\sqrt{(x+1)^2 + y^2} \;=\;2\sqrt{(x-1)^2+y^2}$

Square: . $(x+ 1)^2 + y^2 \;=\;4\left[(x-1)^2 + y^2\right]$

which simplifies to: . $3x^2 - 10x + 3y^2 \;=\;-3$

. . . . . . . . . . . . . . . . $x^2 - \frac{10}{3}x + y^2 \;=\;-1$

Complete the square: . $x^2 - \frac{10}{3}x + \left(\frac{5}{3}\right)^2 + y^2 \;=\;-1 + \left(\frac{5}{3}\right)^2$

. . . . . . . . . . . . . . . . . . . . . $\left(x - \frac{5}{3}\right)^2 + y^2 \;=\;\frac{16}{9}$

Therefore, the circle has center $(\frac{5}{3},\:0)$ radius $\frac{4}{3}$

3. thanks