If |z|=1 & z1, z2, z3 lies on this cirle. Arg (z1)=-π/4 Arg (z2)=0 & Arg(z3)=π/4

Question

If \(|z|=1\ \&\ z_1, z_2, z_3\) lies on this circle. \(\operatorname{Arg}\left(z_1\right)= -\frac{\pi}{4} \operatorname{Arg}\left(z_2\right)=0\) & \(\operatorname{Arg}\left(z_3\right)=\frac{\pi}{4} \&\) If \(\mid z_1 \bar{z}_2+ z_2 \bar{z}_3+\left.z_3 \bar{z}_1\right|^2=\alpha+\beta \sqrt{2}\) where \(\alpha, \beta\) are integers the find \(\alpha+\beta\) is

Answer 1

Answer is "3"    

\(Z_1=\frac{1}{\sqrt{2}}(1+i), Z_2=1, Z_3=\frac{1}{\sqrt{2}}(1-i) \)   

\(\left.\left|\overline{Z}_1 Z_2+\overline{Z}_2 Z_3+\overline{Z}_3 Z_1\right|=\frac{1}{\sqrt{2}}(1+i)+\frac{1}{\sqrt{2}}(1-i)+\frac{1}{2} \times 2 i \right\rvert\, \)   

\(=|\sqrt{2}+i(1-\sqrt{2})| \)   

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

\(=\sqrt{5-2 \sqrt{2}} \)   

\(\left|\overline{Z}_1 Z_2+\overline{Z}_2 Z_3+\overline{Z}_3 Z_1\right|^2=5-2 \sqrt{2}=\alpha+\beta \sqrt{2} \)

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