Let z1, z2 and z3 be three complex numbers on the circle |z|=1 with arg(z1)=-π/4, arg(z2)=0 and arg(z3)=π/4.

Question

Let \(\mathrm{z}_{\mathrm{l}}, \mathrm{z}_{2}\) and \(\mathrm{z}_{3}\) 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 \(\left|z_{1} \overline{\mathrm{z}}_{2}+\mathrm{z}_{2} \overline{\mathrm{z}}_{3}+\mathrm{z}_{3} \overline{\mathrm{z}}_{1}\right|^{2}=\alpha+\beta \sqrt{2}, \alpha, \beta \in \mathrm{Z},\) then the value of \(\alpha^{2}+\beta^{2}\) is:

(1) 24 

(2) 41 

(3) 31 

(4) 29

Answer 1

Correct option is (4) 29 

\(Z_{1}=e^{-i \pi / 4}, Z_{2}=1, Z_{3}=e^{i \pi / 4}\) 

\(\left|z_{1} \bar{z}_{2}+z_{2} \bar{z}_{3}+z_{3} \bar{z}_{1}\right|^{2}=\left|e^{-i \frac{\pi}{4}} \times 1+1 \times e^{-i \frac{\pi}{4}}+e^{i \frac{\pi}{4}} \times e^{i \frac{\pi}{4}}\right|^{2}\)

\(\left\lvert\, e^{-i \frac{\pi}{4}}+e^{-i \frac{\pi}{4}}+e^{\left.i \frac{\pi}{4}\right|^{2}}\right.\)

\(=\left|2 e^{-i \frac{\pi}{4}}+\mathrm{i}\right|^{2}=|\sqrt{2}-\sqrt{2} \mathrm{i}+\mathrm{i}|^{2}\)

\(=(\sqrt{2})^{2}+(1-\sqrt{2})^{2}=2+1+2-2 \sqrt{2}=5-2 \sqrt{2}\)

\(\alpha=5, \beta=-2\)

\(\Rightarrow \alpha^{2}+\beta^{2}=29\)