[UPLIFTJEE]

Let $\mathrm{z}_{1}, \mathrm{z}_{2}$ and $\mathrm{z}_{3}$ be three complex numbers on the circle $|\mathrm{z}|=1$ with $\arg \left(\mathrm{z}_{1}\right)=-\frac{\pi}{4}, \arg \left(\mathrm{z}_{2}\right)=0$ and $\arg \left(\mathrm{z}_{3}\right)=\frac{\pi}{4}$. If $\left|\mathrm{z}_{1} \overline{\mathrm{z}}_{2}+\mathrm{z}_{2} \overline{\mathrm{z}}_{3}+\mathrm{z}_{3} \overline{\mathrm{z}}_{1}\right|^{2}=\alpha+\beta \sqrt{2}, \alpha, \beta \in \mathrm{Z}$, then the value of $\alpha^{2}+\beta^{2}$ is [JEE (Main) 22 Jan 2025 S I]
(A) 24
(B) 41
(C) 29
(D) 31

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[UPLIFTJEE]

Solution

$\mathrm{z}_{1}=\mathrm{e}^{-\mathrm{i} \pi 4}, \mathrm{z}_{2}=1, \mathrm{z}_{3}=\mathrm{e}^{\mathrm{i} \pi / 4} \Rightarrow\left|\mathrm{z}_{1} \overline{\mathrm{z}}_{2}+\mathrm{z}_{2} \overline{\mathrm{z}}_{3}+\mathrm{z}_{3} \overline{\mathrm{z}}_{1}\right|^{2}=\left|\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}} \times 1+1 \times \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{e}^{\mathrm{i} \frac{\pi}{4}} \times \mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right|^{2}$

\[ \begin{aligned} & =\left|\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{e}^{\mathrm{i} \frac{\pi}{2}}\right|^{2}=\left|2 \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}+\mathrm{i}\right|^{2}=|\sqrt{2}-\sqrt{2} \mathrm{i}+\mathrm{i}|^{2}=(\sqrt{2})^{2}+(1-\sqrt{2})^{2}=2+1+2-2 \sqrt{2}=5-2 \sqrt{2} \\ & \alpha=5, \beta=-2 \Rightarrow \alpha^{2}+\beta^{2}=29 \end{aligned} \]

[UPLIFTJEE]

Answer

C

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