A complex number is such that the absolute value of z = the absolute value of (z-3i)
(a) Show that the imaginary part of z is 2/3.
I tried squaring both sides and then I find by equating both sides, that when z = a + bi, b satisfies the equation:
b^2 -7b +9 =0 however this does not give me b = 2/3.
Best Answer
You have $$a^2+b^2=a^2+(b-3)^2\iff6b=9\iff2b=3$$
or $$b^2-(b-3)^2=0\iff\{b-(b-3)\}\{b+(b-3)\}=0$$