Let z be a complex number such that |z+2|=|z-2| and ( z+3z-i )= 4. Then |z|2 is equal to:
- 9
- 4
- 5
- 1
Answer
(1) 9
Solution
|z+2|=|z-2| z is equidistant from -2 and 2, so z lies on the perpendicular bisector, the imaginary axis.
z=iy, y R
z+3z-i= 3+iyi(y-1)= (3+iy)(-i)y-1= y-3iy-1= yy-1- 3y-1 i
= 4, the real and imaginary parts are equal and positive.
yy-1= -3y-1 y=-3
Check: yy-1= -3-4= 34>0
z=-3i |z|2=9
Answer: (1)