Let z be a complex number such that |z-2iz+i|=2, z-i.

Question

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

Accepted Answer

$\left|\frac{z-2 i}{z+i}\right|=2$

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

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

$\Rightarrow 3 z \bar{z}-6 i z+6 i \bar{z}=0$

$\Rightarrow z \bar{z}-2 i z+2 i \bar{z}=0$

$\therefore$ Centre $(-2 i)$ or $(0,-2)$

Other Method :

$\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i$

Put $z=x+i y$

$$
\begin{aligned}
& \left|\frac{x+i y-2 i}{x+i y+i}\right|=2 \\
& \Rightarrow \left|\frac{x+i(y-2)}{x+i(y+1)}\right|^2=4 \\
& \Rightarrow x^2+(y-2)^2=4\left[x^2+(y+1)^2\right] \\
& \Rightarrow x^2+y^2+4-4 y=4\left[x^2+y^2+1+2 y\right]
\end{aligned}
$$

$\Rightarrow$ $x^2+y^2+4 y=0$

or $x^2+(y+2)^2=2^2$

$$ \therefore $$ Centre is $(0,-2)$.

Solutions · 1