### Complex Number Question 4

#### Question 4 - 25 January - Shift 2

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

(1) $(2,0)$

(2) $(0,0)$

(3) $(0,2)$

(4) $(0,-2)$

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Equation of a circle, Properties of modulus, Algebra of complex numbers

$(z-2 i)(\overline{z}+2 i)=4(z+i)(\overline{z}-i)$

$z \overline{z}+4+2 i(z-\overline{z})=4(z \overline{z}+1+i(\overline{z}-z))$

$3 z \overline{z}-6 i(z-\overline{z})=0$

$x^{2}+y^{2}-2 i(2 iy)=0$

$x^{2}+y^{2}+4 y=0$