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#