How do you graph #abs(z-i) = 2# in the complex plane?
1 Answer
Feb 3, 2016
This is a circle with radius
Explanation:
To say
graph{(x^2+(y-1)^2-4)(x^2+(y-1)^2-0.011) = 0 [-5.457, 5.643, -1.84, 3.71]}
Alternatively, use the definition:
#abs(z) = sqrt(z bar(z))#
Consider
Then
#2 = abs(z-i) = abs(x+yi-i) = abs(x+(y-1)i)#
#= sqrt((x+(y-1)i)(x-(y-1)i)) = sqrt(x^2+(y-1)^2)#
Square both ends and transpose to get:
#x^2+(y-1)^2 = 2^2#
which is the equation of a circle radius