How do you show that if #abs(z) = abs(z-3i)# then #Im(z) = 3/2# ?

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Answer 1 · 2017-12-12T20:20:48

See explanation...

Note that if #a# is any complex number, then:

#abs(z-a)#

is the distance between #z# and #a# in the complex plane.

So if:

#abs(z-0) = abs(z-3i)#

then that means that the distance between #z# and #0# is the same as the distance between #z# and #3i#.

Hence #z# must lie on the perpendicular bisector of the line segment joining #0# and #3i#, namely the line #Im(z) = 3/2#

Answer 2 · 2017-12-12T22:38:19

See below.

#absz = sqrt(z * bar z)# with #z = x+i y# and #bar z = x-i y#

now

#u = z-3i = x+i(y-3)# and

#sqrt(z*bar z) = sqrt(u*bar u)#

Squaring both sides

#z*bar z = u*baru# but

#z * bar z = x^2+y^2#

#u*bar u = x^2+(y-3)^2#

then

#x^2+y^2 = x^2+(y-3)^2 rArr 9-6y = 0 rArr y = 3/2#