# If  z in C, then what does the equation 2|z+3i|-|z-i|=0 represent?

Mar 18, 2018

This is the equation of a circle, center $\left(0 , - \frac{13}{3}\right)$ and radius $= \frac{8}{3}$

#### Explanation:

The equation is

$2 | z + 3 i | = | z - i |$

The modulus of $\left(z - i\right)$ is twice the modulus of $\left(z + 3 i\right)$

Let $z = x + i y$

Then,

$2 | x + i y + 3 i | = | x + i y - i |$

$2 | x + i \left(y + 3\right) | = | x + i \left(y - 1\right) |$

Then,

$2 \sqrt{{x}^{2} + {\left(y + 3\right)}^{2}} = \sqrt{{x}^{2} + {\left(y - 1\right)}^{2}}$

Squaring both sides

$4 \left({x}^{2} + {\left(y + 3\right)}^{2}\right) = \left({x}^{2} + {\left(y - 1\right)}^{2}\right)$

$4 \left({x}^{2} + {y}^{2} + 6 y + 9\right) = \left({x}^{2} + {y}^{2} - 2 y + 1\right)$

$4 {x}^{2} - {x}^{2} + 4 {y}^{2} - {y}^{2} + 24 y + 2 y + 36 - 1$

$3 {x}^{2} + 3 {y}^{2} + 26 y + 35 = 0$

$= 3 {x}^{2} + 3 \left({y}^{2} + \frac{26}{3} y + \frac{169}{9}\right) = - 35 + 3 \cdot \frac{169}{9}$

$= {x}^{2} + {\left(y + \frac{13}{3}\right)}^{2} = \frac{64}{9} = {\left(\frac{8}{3}\right)}^{2}$

This is the equation of a circle, center $\left(0 , - \frac{13}{3}\right)$ and radius $= \frac{8}{3}$

graph{x^2+(y+13/3)^2-64/9=0 [-7.6, 10.18, -7.55, 1.34]}