Let z_1, z_2 and z_3 be three complex numbers on the circle |z|=1 with (z_1)=(-π)/(4), (z_2)=0 and (z_3)=(π)/(4) . If |z_1 z _2+z_2 z _3+z_3 z _1|^2=α+β 2 , α, β ∈ Z , then the value of α^2+β^2 is :

Correct answer: B. 29 aligned & |z|=1 & (z_1)=-(π)/(4), (z_2)=0, (z_3)=(π)/(4) & z_1=|1| e^ -(π)/(4) = 1 2 - i 2 & z_2=1+0 i & z_3= 1 2 + i 2 & z_1 z _2= 1-i 2 & z_2 z _3= 1-i 2 & z_3 z _1=((1+i)^2)/(2) & z_1 z _2+z_2 z _3+z_3 z _1= 2 +i(1- 2 ) & |z_1 z _2+z_2 z _3+z_3 z _1|^2=5-2 2 & α=5, β=-2 & α^2+β^2=29 aligned

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Complex Number

Let z_1, z_2 and z_3 be three complex numbers on the circle |z|=1...

Let

z_1, z_2 and z_3

be three complex numbers on the circle

|z|=1

with

\arg \left(z_1\right)=\frac{-\pi}{4}, \arg \left(z_2\right)=0 and \arg \left(z_3\right)=\frac{\pi}{4}

. If \left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2=\alpha+\beta \sqrt{2}, \alpha, \beta \in \mathbf{Z}

, then the value of

\alpha^2+\beta^2 is :

Options

A $24$
B $29$
C $41$
D $31$

Correct answer

B. $29$

Step-by-step solution

\begin{aligned} & |z|=1 \\ & \arg \left(z_1\right)=-\frac{\pi}{4}, \arg \left(z_2\right)=0, \arg \left(z_3\right)=\frac{\pi}{4} \\ & z_1=|1| e^{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\ & z_2=1+0 i \\ & z_3=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \\ & z_1 \bar{z}_2=\frac{1-i}{\sqrt{2}} \\ & z_2 \bar{z}_3=\frac{1-i}{\sqrt{2}} \\ & z_3 \bar{z}_1=\frac{(1+i)^2}{2} \\ & z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1=\sqrt{2}+i(1-\sqrt{2}) \\ & \left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2=5-2 \sqrt{2} \\ & \alpha=5, \beta=-2 \\ & \alpha^2+\beta^2=29\end{aligned}

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