Let $z_{1}, z_{2}$ and $z_{3}$ be three complex numbers on the circle $|z|=1$ with $\arg(z_{1})=\frac{-\pi}{4}, \arg(z_{2})=0$ and $\arg(z_{3})=\frac{\pi}{4}$. If $\left|z_{1} \overline{z}_{2}+z_{2} \overline{z}_{3}+z_{3} \overline{z}_{1}\right|^{2}=\alpha+\beta \sqrt{2}, \alpha, \beta \in \mathrm{Z}$, then the value of $\alpha^{2}+\beta^{2}$ is :

Complex Numbers — JEE Main · 2025 · 22 Jan 2025 · Shift 1 Question with Solution

Let

z_{1}, z_{2}

and

z_{3}

be three complex numbers on the circle

∣ z ∣ = 1

with

ar g (z_{1}) = \frac{- π}{4}, ar g (z_{2}) = 0

and

ar g (z_{3}) = \frac{π}{4}

. If

∣ z_{1} \overline{z}_{2} + z_{2} \overline{z}_{3} + z_{3} \overline{z}_{1} ∣^{2} = α + β 2, α, β \in Z

, then the value of

α^{2} + β^{2}

is :

Answer: (D)

With

∣ z ∣ = 1

and the given arguments,

z_{1} = e^{- iπ /4}

z_{2} = 1

z_{3} = e^{iπ /4}

, so

\overset{z}{ˉ}_{1} = e^{iπ /4}

\overset{z}{ˉ}_{2} = 1

\overset{z}{ˉ}_{3} = e^{- iπ /4}

z_{1} \overset{z}{ˉ}_{2} = e^{- iπ /4}, z_{2} \overset{z}{ˉ}_{3} = e^{- iπ /4}, z_{3} \overset{z}{ˉ}_{1} = e^{iπ /4} \cdot e^{iπ /4} = e^{iπ /2} = i .

Adding:

z_{1} \overset{z}{ˉ}_{2} + z_{2} \overset{z}{ˉ}_{3} + z_{3} \overset{z}{ˉ}_{1} = 2 e^{- iπ /4} + i = 2 (\frac{2}{2} - i \frac{2}{2}) + i = 2 + i (1 - 2) .

Taking the modulus squared,

∣ z_{1} \overset{z}{ˉ}_{2} + z_{2} \overset{z}{ˉ}_{3} + z_{3} \overset{z}{ˉ}_{1} ∣^{2} = (2)^{2} + (1 - 2)^{2} = 2 + (1 - 22 + 2) = 5 - 22 .

α = 5, β = - 2

, and

α^{2} + β^{2} = 25 + 4 = 29.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option