Let z be a complex number such that | (z - 2i)/(z + i) | = 2,z ≠ - i. Then z lies on the circle of radius 2 and centre :

Solution

Maths sir · 4 steps

Clear the fraction, put z = x + iy, and square both sides: the circle shows itself.

1. Remove the fraction. Since z ≠ -i, the condition becomes |z - 2i| = 2|z + i|. So the distance of z from (0, 2) is twice its distance from (0, -1).
2. Use x and y. Put z = x + iy. Then |z - 2i|² = x² + (y - 2)² and |z + i|² = x² + (y + 1)². Squaring both sides gives x² + (y-2)² = 4[x² + (y+1)²].
3. Expand and collect. The left side is x² + y² - 4y + 4 and the right side is 4x² + 4y² + 8y + 4. Moving everything to one side gives 3x² + 3y² + 12y = 0, that is x² + y² + 4y = 0.

   x² + (y + 2)² = 2²

4. So. Completing the square in y gives a circle of radius 2 with centre (0, -2), which is option (a).

The idea to keep. When |(z - a)/(z - b)| = k with k ≠ 1, the locus is a circle. Clear the fraction, square, collect terms, then complete the square to read off the centre.

Answer from the NTA final answer key. Solution by PrepWiser.