Let A = {z ∈ C : |z – 2 – i| = 3}, B = {z ∈ C : Re (z – iz) = 2} and S = A ∩ B.

Question

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

Answer 1

Answer is: 22 

Let z = x + iy

\( |z-2-i|=3 \Rightarrow(x-2)^2+(y-1)^2=3^2 \)

\( \operatorname{Re}(z-i z)=\operatorname{Re}(x+i y-i x+y)=x+y \Rightarrow x+y=2 \)

\( \Rightarrow A=\left\{(x, y):(x-2)^2+(y-1)^2=3^2, x, y \in R\right\}, \)

\( B=\{(x, y): x+y=2\} \)

\(\Rightarrow x-2=-y \Rightarrow y^2+(y-1)^2=3^2 \)

\(\Rightarrow 2 y^2-2 y-8=0 \Rightarrow y^2-y-4=0 \)

\( y_1+y_2=1, y_1 y_2=-4 \)

\( \Rightarrow y_1^2+y_2^2 \)

\(=\left(y_1+y_2\right)^2-2 y_1 y_2=9 \)

\( \Rightarrow x_1+x_2=4\left(y_1+y_2\right)=3, \)

\( x_1 x_2=\left(2-y_1\right)\left(2-y_2\right)=4-2\left(y_1+y_2\right)+y_1 y_2=-2 \)

\(\Rightarrow x_1^2+x_2^2=\left(x_1+x_2\right)^2-2 x_1 x_2=13 \)

\( \because S=\left\{\left(x_1, y_1\right),\left(x_2, y_2\right)\right\} \)

\( \Rightarrow \sum\limits_{z \in S}|z|^2=\left(x_1^2+y_1^2\right)+\left(x_2^2+y_2^2\right)=22\)