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Let $$z$$ be a complex number such that $$\left|\frac{z-2i}{z+i}\right| = 2$$, $$z \neq -i$$. Then $$z$$ lies on the circle of radius 2 and centre
Given: $$\left|\frac{z - 2i}{z + i}\right| = 2$$
Let $$z = x + iy$$.
$$|z - 2i|^2 = 4|z + i|^2$$
$$x^2 + (y-2)^2 = 4[x^2 + (y+1)^2]$$
$$x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4$$
$$0 = 3x^2 + 3y^2 + 12y$$
$$x^2 + y^2 + 4y = 0$$
$$x^2 + (y+2)^2 = 4$$
This is a circle with centre $$(0, -2)$$ and radius 2.
The correct answer is Option 4: (0, -2).
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