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

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Answer 1 · 2016-08-27T19:27:04

If $z in RR$ then $z = 5$ and $z = 4$
If $z in CC$ then $(x-9/2)^2+y^2=(9/2)^2$

Considering $z in RR$ , $abs(2z-9)=1$ is solving with

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

but if $z in CC$ then

$abs(2(x+i y)-9) = abs(2x-9+i2y) = 1$ or

$((2x-9)^2+(2y)^2) = 1$ resulting in

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

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

or

$(x-9/2)^2+y^2=(9/2)^2$

a circle centered at $(9/2,0)$ and radius $9/2$

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