How do you solve abs(2z-9)=1?

Aug 27, 2016

If $z \in \mathbb{R}$ then $z = 5$ and $z = 4$
If $z \in \mathbb{C}$ then ${\left(x - \frac{9}{2}\right)}^{2} + {y}^{2} = {\left(\frac{9}{2}\right)}^{2}$

Explanation:

Considering $z \in \mathbb{R}$, $\left\mid 2 z - 9 \right\mid = 1$ is solving with

$2 z - 9 = 1$ giving $z = 5$
$- \left(2 z - 9\right) = 1$ giving $z = 4$

but if $z \in \mathbb{C}$ then

$\left\mid 2 \left(x + i y\right) - 9 \right\mid = \left\mid 2 x - 9 + i 2 y \right\mid = 1$ or

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

$4 {x}^{2} + 4 {y}^{2} - 36 x + 81 = 1$ or

${x}^{2} + {y}^{2} - 9 x + 20 = 0$

or

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

a circle centered at $\left(\frac{9}{2} , 0\right)$ and radius $\frac{9}{2}$