Let A=z C:|z-2-i|=3,B=z C:Re(z-iz)=2 and S=A B.

Question

Let $A={z\in C:|z-2-i|=3},B={z\in C:Re(z-iz)=2}$ and $S=A\cap B$. Then $\sum _{z\in S}|z{|}^{2}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 22

Accepted Answer

$\begin{gathered} z=x+iy \ |z-2-i|=3\Rightarrow (x-2{)}^{2}+(y-1{)}^{2}={3}^{2} \ Re(z-iz)=Re(x+iy-ix+y)=x+y\Rightarrow x+y=2 \ \Rightarrow A={(x,y):(x-2{)}^{2}+(y-1{)}^{2}={3}^{2},x,y\in R}, \ B={(x,y):x+y=2} \ \Rightarrow x-2=-y\Rightarrow {y}^{2}+(y-1{)}^{2}={3}^{2} \ \Rightarrow 2{y}^{2}-2y-8=0\Rightarrow {y}^{2}-y-4=0 \ {y}_{1}+{y}_{2}=1,{y}_{1}{y}_{2}=-4 \ \Rightarrow {y}_{2}^{1}+{y}_{2}^{2} \ ={({y}_{1}+{y}_{2})}^{2}-2{y}_{1}{y}_{2}=9 \ \Rightarrow {x}_{1}+{x}_{2}=4({y}_{1}+{y}_{2})=3, \ {x}_{1}{x}_{2}=(2-{y}_{1})(2-{y}_{2})=4-2({y}_{1}+{y}_{2})+{y}_{1}{y}_{2}=-2 \ \Rightarrow {x}_{2}^{1}+{x}_{2}^{2}={({x}_{1}+{x}_{2})}^{2}-2{x}_{1}{x}_{2}=13 \ \because S={({x}_{1},{y}_{1}),({x}_{2},{y}_{2})} \ \Rightarrow \sum _{z\in S}|z{|}^{2}=({x}_{2}^{1}+{y}_{2}^{1})+({x}_{2}^{2}+{y}_{2}^{2})=22 \end{gathered}$

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