# How do you identify the locus in the complex plane given by: |z - 3i| = |z +2|?

Dec 19, 2015

$\left\{x + \left(- \frac{2}{3} x + \frac{5}{6}\right) i : x \in \mathbb{R}\right\}$

#### Explanation:

Let $z = x + i y$ where $x , y \in \mathbb{R}$

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

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

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

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

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

$\implies {x}^{2} + {y}^{2} - 6 y + 9 = {x}^{2} + 4 x + 4 + {y}^{2}$

$\implies y = - \frac{2}{3} x + \frac{5}{6}$

$\implies z = x + \left(- \frac{2}{3} x + \frac{5}{6}\right) i$

As we placed no restrictions on $x$ beyond $x \in \mathbb{R}$, this gives us the result

$\left\{z \in \mathbb{C} : | z - 3 i | = | z + 2 |\right\} = \left\{x + \left(- \frac{2}{3} x + \frac{5}{6}\right) i : x \in \mathbb{R}\right\}$