Let S=z C:|(z-6i)/(z-2i)|=1 and |(z-8+2i)/(z+2i)|=(3)/(5).

Question

Let $S={z\in C:|\frac{z-6i}{z-2i}|=1$ and $|\frac{z-8+2i}{z+2i}|=\frac{3}{5}}$.

Then $\sum _{z\in s}|z{|}^{2}$ is equal to :

Accepted Answer

Given, $S={z\in C:|\frac{z-6i}{z-2i}|=1$ and $|\frac{z-8+2i}{z+2i}|=\frac{3}{5}}$

$|\frac{z-6i}{z-2i}|=1$

$\Rightarrow$ $|z-6i|=|z-2i|-(1)$

equation (1) represent $z$ is a point which is equidistant from point $(0,2)$ and $(0,6)$ so $z$ lies on the line segment which is perpendicular bisector of the line segment joining point $(0,2)$ \& $(0,6)$.

If $z=x+iy$ then solution of equation (1) is $y=\frac{6+2}{2}=4$ and $x\in R$.

So, solution is $z=x+4i$ \& $x\in R$.

$\begin{gathered} |\frac{z-8+2i}{z+2i}|=\frac{3}{5} \ z=x+4i \ \frac{|x+4i-8+2i|}{|x+4i+2i|}=\frac{3}{5} \end{gathered}$

$\Rightarrow$ $5|x+6i-8|=3|x+6i|$

$\Rightarrow$ $5|(x-8)+6i|=3|x+6i|$

$\Rightarrow$ $5\sqrt{(x-8{)}^{2}+{6}^{2}}=3\sqrt{{x}^{2}+{6}^{2}}$

Squaring

$\Rightarrow$ $25({x}^{2}+64-16x+36)=9{x}^{2}+324$

$\Rightarrow$ $25{x}^{2}-400x+2500=9{x}^{2}+324$

$\Rightarrow$ $16{x}^{2}-400x+2176=0$

$\Rightarrow$ ${x}^{2}-25x+136=0$

$\Rightarrow$ ${x}^{2}-17x-8x+136=0$

$\Rightarrow$ $x(x-17)-8(x-17)=0$

$\Rightarrow$ $(x-17)(x-8)=0$

$	herefore $ $x=17,x=8$

So set S contain two complex number ${z}_{1}=17+4i$ and ${z}_{2}=8+4i$

$\begin{gathered} \sum _{z\in S}|z{|}^{2}={|{z}_{1}|}^{2}+{|{z}_{2}|}^{2} \ =|17+4i{|}^{2}+|8+4i{|}^{2} \ ={17}^{2}+{4}^{2}+{8}^{2}+{4}^{2} \ =289+16+64+16=385 \end{gathered}$

Solutions · 1