Let A=\ z ∈ C:|z-2-i|=3\ , B=\ z ∈ C: Re(z-i z)=2\ and S=A B . Then ∑_ z ∈ S |z|^2 is equal to ________ .

Correct answer: 0 Let z=x+i y aligned & A:|z-2-i|=3 & |(x-2)+(y-1) i|=3 & (x-2)^2+(y-1)^2=9 ....(1) & B=Re(z-i z)=2 & Re((x+y)+i(y-x))=2 & x+y=2....(2) aligned . On solving (1) and (2) we get aligned & x= 3 ± 17 2 , y= 1 17 2 & ∑_ z ∈ ~s |z|^2=(1)/(4)[2 × 26+2 × 18] & (88)/(4)=22 aligned

JEE Main 20254 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Complex Number

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

Let

A=\{z \in C:|z-2-i|=3\}

B=\{z \in C: \operatorname{Re}(z-i z)=2\}

and

S=A \cap B

. Then

\sum_{z \in S}|z|^2

is equal to ________ .

Correct answer

Step-by-step solution

Let

z=x+i y

\begin{aligned}<br/>& A:|z-2-i|=3 \\ & |(x-2)+(y-1) i|=3 \\ & (x-2)^2+(y-1)^2=9 ....(1)\\ & B=\operatorname{Re}(z-i z)=2 \\ & \operatorname{Re}((x+y)+i(y-x))=2 \\ & x+y=2....(2)<br/>\end{aligned}

<br/>On solving (1) and (2) we get<br/>

\begin{aligned}<br/>& x=\frac{3 \pm \sqrt{17}}{2}, y=\frac{1 \mp \sqrt{17}}{2} \\ & \sum_{z \in \mathrm{~s}}|z|^2=\frac{1}{4}[2 \times 26+2 \times 18] \\ & \Rightarrow \frac{88}{4}=22<br/>\end{aligned}

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