Let S = \z C : | z-6iz-2i | = 1 and | z-8+2iz+2i | = 35 \. Then z S |z|2 is equal to :
- 413
- 398
- 385
- 423
Answer
(3) 385
Solution
| z-6iz-2i | = 1 |z-6i| = |z-2i|
z lies on the perpendicular bisector of the points 6i and 2i, i.e. on the line y = 4
Let z = x + 4i.
z-8+2i = (x-8) + 6i and z+2i = x + 6i
5|z-8+2i| = 3|z+2i| 25 [(x-8)2 + 36 ] = 9 (x2 + 36 )
16x2 - 400x + 2176 = 0
x2 - 25x + 136 = 0
(x-8)(x-17) = 0
z = 8 + 4i or z = 17 + 4i
z S |z|2 = (64+16) + (289+16)
= 80 + 305 = 385
Answer: (3)