Finding the Cartesian equation of a line in the complex plane

Question

Find the Cartesian form in the complex plane for the following equations, given $z \in \textbb{C}$ ,
(a) $|z + 2| = |z - i|$
(b) $|z + 2| = |2z - i|$

Answer

Note that the Cartesian form of a complex number $z$ is $z = x + iy$ , where $i$ is called an $\textbf{imaginary number}$ , and $i^{2} = - 1$ , and the absolute $|z|$ , also called the modulus of $z$ , is defined as $|z| = \sqrt{x^{2} + y^{2}}$ . Thus solutions to the above problems are as follows

(a) $|z + 2| = |z - i|$ , $z \in \textbb{C}$ :

Let $z = x + iy$ , substituting $z$ in $|z + 2| = |z - i|$ , we get

$\renewcommand{\arraystretch}{1.5} \textsf{ \begin{tabular}{l r c l} & $|x + iy + 2)|$ &= &$|x + iy - i|$\\ & $|(x + 2) + iy|$ &= &$|x + i(y - 1)|$\\ & $\sqrt{(x + 2)^{2} + y^{2}}$ &= & $\sqrt{x^{2} + (y - 1)^{2}}$ \\ & $(x + 2)^{2} + y^{2}$ &= & $x^{2} + (y - 1)^{2}$ \\ &$x^2 + 4x + 4 + y^2$ &= & $x^2 + y^2 - 2y + 1$ \\ \therefore &$2y$ & = & $ - 4x - 3$ \\ \end{tabular} }$

$\textsf{ \implies \boxed{y = - 2x - \frac{3}{2}} }$

This is the Cartesian equation of a straight line with a gradient of $-2$ and a y-intercept of $-\frac{3}{2}$ (see Figure 1)

Cartesian form of the straight line |z+2|=|z-i|, where z is a complex number.

Figure 1

(b) $|z + 2| = |2z - i|$ , $z \in \textbb{C}$ :

Let $z = x + iy$ , substituting $z$ in $|z + 2| = |2z - i|$ , we get

$\renewcommand{\arraystretch}{1.5} \textsf{ \begin{tabular}{l r c l} & $|z + 2|$ &= &$|2z - i|$\\ & $|(x + 2) + iy|$ &= &$|2x + i2y - i|$\\ & $|(x + 2) + iy|$ &= &$|2x + i(2y - 1)|$\\ & $\sqrt{(x + 2)^{2} + y^{2}}$ &= & $\sqrt{(2x)^{2} + (2y - 1)^{2}}$ \\ & $(x + 2)^{2} + y^{2}$ &= & $4x^{2} + (2y - 1)^{2}$ \\ &$x^2 + 4x + 4 + y^2$ &= & $4x^2 + 4y^2 - 4y + 1$ \\ \therefore &$3y^{2} - 4y$ & = & $- 3x^{2} + 4x + 3$ \\ \therefore &$y^{2} - \frac{4}{3}y$ & = & $- x^{2} + \frac{4}{3}x + 1$ \\ \end{tabular} }$

Completing the square gives
$\renewcommand{\arraystretch}{1.5} \textsf{ \begin{tabular}{l r c l} &$y^{2} - 2\left(\frac{2}{3}\right)y + \frac{4}{9} - \frac{4}{9}$ & = & $- \left(x^{2} - 2\left(\frac{2}{3}\right)x + \frac{4}{9} - \frac{4}{9}\right) + 1$ \\ &$\left(y - \frac{2}{3}\right)^{2} - \frac{4}{9}$ & = & $- \left((x - \frac{2}{3})^{2} - \frac{4}{9}\right) + 1$ \\ \therefore &$\left(y - \frac{2}{3}\right)^{2} + (x - \frac{2}{3})^{2} & = & $+ \frac{4}{9} + \frac{4}{9}$ + 1$ \end{tabular} }$

$\textsf{ \implies \boxed{\left(y - \frac{2}{3}\right)^{2} + \left(x - \frac{2}{3}\right)^{2} & = & \frac{17}{9}} }$

This is the Cartesian equation of a circle with the center at $\left(\frac{2}{3},\frac{2}{3}\right)$ and a radius of $\frac{\sqrt{17}}{3}$ (see Figure 2)

Figure 2