How do you graph #abs(z-i) = 2# in the complex plane?

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Answer 1 · 2016-02-03T23:30:38

This is a circle with radius #2# and centre #i#

To say #abs(z-i) = 2# is to say that the (Euclidean) distance between #z# and #i# is #2#.

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 #z = x+yi# where #x# and #y# are Real.

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 #2# with centre #(0, 1)#, i.e. #i#.