The sum of the square of the modulus of the elements in the set $\{z=\mathrm{a}+\mathrm{ib}: \mathrm{a}, \mathrm{b} \in \mathbf{Z}, z \in \mathbf{C},|z-1| \leq 1,|z-5| \leq|z-5 \mathrm{i}|\}$ is ________

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Aicharya · Accepted Answer

$\begin{aligned} & |\mathrm{z}-1| \leq 1 \\ & \Rightarrow|(\mathrm{x}-1)+\mathrm{iy}| \leq 1 \mid \\ & \Rightarrow \sqrt{(\mathrm{x}-1)^2+\mathrm{y}^2} \leq 1 \\ & \Rightarrow(\mathrm{x}-1)^2+\mathrm{y}^2 \leq 1 \ldots(1)\end{aligned}$ $\begin{aligned} & \text { Also }|z-5| \leq|z-5 i| \\ & (x-5)^2+y^2 \leq x^2+(y-5)^2 \\ & -10 x \leq-10 y \\ & \Rightarrow x \geq y.....(2)\end{aligned}$ $\begin{aligned} & \text { Solving (1) and (2) } \\ & \begin{array}{l}\Rightarrow(\mathrm{x}-1)^2+\mathrm{x}^2=1 \\ \Rightarrow 2 \mathrm{x}^2-2 \mathrm{x}=0 \\ \Rightarrow \mathrm{x}(\mathrm{x}-1)=0 \\ \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=1 \\ \quad \mathrm{y}=0 \text { or } \mathrm{y}=1\end{array}\end{aligned}$

Given $\mathrm{x}, \mathrm{y} \in \mathrm{I}$ Points $(0,0),(1,0),(2,0),(1,1),(1,-1)$ to find $\begin{aligned} & \left|z_1\right|^2+\left|z_2\right|^2+\left|z_3\right|^2+\left|z_4\right|^2+\left|z_5\right|^2 \\ & =0+1+4+1+1+1+1=9 \end{aligned}$

The sum of the square of the modulus of the elements in the set $\{z=\mathrm{a}+\mathrm{ib}: \mathrm{a},…

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