The sum of the square of the modulus of the elements in the set z= a+ ib: a, b ∈ Z, z ∈ C,|z-1| ≤ 1,|z-5| ≤|z-5 i| is

$\begin{aligned} & |\mathrm{z}-1| \leq 1 \\ & \Rightarrow|(\mathrm{x}-1)+\mathrm{iy}| \leq 1 \mid \\ & \Rightarrow \sqrt{(\mathrm{x}-1)^2+\mathrm{y}^2} \leq 1 \\ & \Rightarrow(\mathrm{x}-1)^2+\mathrm{y}^2 \leq 1 \ldots(1)\end{aligned}$​∣z−1∣≤1⇒∣(x−1)+iy∣≤1∣⇒(x−1)2+y2​≤1⇒(x−1)2+y2≤1…(1)​ $\begin{aligned} & \text { Also }|z-5| \leq|z-5 i| \\ & (x-5)^2+y^2 \leq x^2+(y-5)^2 \\ & -10 x \leq-10 y \\ & \Rightarrow x \geq y.....(2)\end{aligned}$​ Also ∣z−5∣≤∣z−5i∣(x−5)2+y2≤x2+(y−5)2−10x≤−10y⇒x≥y.....(2)​ $\begin{aligned} & \text { Solving (1) and (2) } \\ & \begin{array}{l}\Rightarrow(\mathrm{x}-1)^2+\mathrm{x}^2=1 \\ \Rightarrow 2 \mathrm{x}^2-2 \mathrm{x}=0 \\ \Rightarrow \mathrm{x}(\mathrm{x}-1)=0 \\ \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=1 \\ \quad \mathrm{y}=0 \text { or } \mathrm{y}=1\end{array}\end{aligned}$​ Solving (1) and (2) ⇒(x−1)2+x2=1⇒2x2−2x=0⇒x(x−1)=0⇒x=0 or x=1y=0 or y=1​​

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