Let z be a complex number such that \left| {{{z - 2i} \over {z + i}}} ight| = 2,z
e - i. Then z lies on the circle of radius 2 and centre :

Question

Let z be a complex number such that \left| {{{z - 2i} \over {z + i}}} ight| = 2,z 
e - i. Then z lies on the circle of radius 2 and centre :

Accepted Answer

Answer: A. (0, -2) \left|\frac{z-2 i}{z+i}ight|=2 \Rightarrow (z-2 i)(\bar{z}+2 i)=4(z+i)(\bar{z}-i) \Rightarrow z \bar{z}+2 i z-2 i \bar{z}+4=4(z \bar{z}-z i+\overline{z i}+1) \Rightarrow 3 z \bar{z}-6 i z+6 i \bar{z}=0 \Rightarrow z \bar{z}-2 i z+2 i \bar{z}=0 	herefore Centre (-2 i) or (0,-2) Other Method : \left|\frac{z-2 i}{z+i}ight|=2, z 
eq-i Put z=x+i y \begin{aligned} & \left|\frac{x+i y-2 i}{x+i y+i}ight|=2 \\ & \Rightarrow \left|\frac{x+i(y-2)}{x+i(y+1)}ight|^2=4 \\ & \Rightarrow x^2+(y-2)^2=4\left[x^2+(y+1)^2ight] \\ & \Rightarrow x^2+y^2+4-4 y=4\left[x^2+y^2+1+2 yight] \end{aligned} \Rightarrow x^2+y^2+4 y=0 or x^2+(y+2)^2=2^2 	herefore Centre is (0,-2).

Let $$z$$ be a complex number such that $$\left| {{{z - 2i} \over {z + i}}} \right| = 2,z \ne - i$$. Then $$z$$ lies on the circle of radius 2 and centre :

Answer

Explanation

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