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 ________

Correct answer: 0 aligned & |z-1| ≤ 1 & |(x-1)+iy| ≤ 1 & (x-1)^2+y^2 ≤ 1 & (x-1)^2+y^2 ≤ 1 (1) aligned aligned & Also |z-5| ≤|z-5 i| & (x-5)^2+y^2 ≤ x^2+(y-5)^2 & -10 x ≤-10 y & x ≥ y.....(2) aligned aligned & Solving (1) and (2) & array l (x-1)^2+x^2=1 2 x^2-2 x=0 x(x-1)=0 x=0 or x=1 y=0 or y=1 array aligned Given x , y ∈ I Points (0,0),(1,0),(2,0),(1,1),(1,-1) to find aligned & |z_1|^2+|z_2|^2+|z_3|^2+|z_4|^2+|z_5|^2 & =0+1+4+1+1+1+1=9 aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Complex Number

The sum of the square of the modulus of the elements in the set \...

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 ________

Correct answer

Step-by-step solution

\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{\text{ or } } \mathrm{x}=1 \\ \quad \mathrm{y}=0 \text{\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}

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