Let \(\left|z_{i}\right|=1\) for \(i=1,2,3\) satisfying \(\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=a+b \…
Let \(\left|z_{i}\right|=1\) for \(i=1,2,3\) satisfying \(\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=a+b \sqrt{2}\), where \(a, b\) an rational numbers such that \(\arg \left(z_1\right)=\frac{\pi}{4}, \arg \left(z_2\right)=0\) and \(\arg \left(z_3\right)=\frac{-\pi}{4}\), then find \((a, b)\)
\((5,-2)\)
\(\begin{aligned}&\text { Sol. }\\&& z_1=|1| e^{i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}+i \cdot \frac{1}{\sqrt{2}} \\& z_2=|1| e^{-(0)}= 1+0 i \\& z_3=|1| e^{-i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\& \bar{z}_1 z_2=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)(1) \\& \bar{z}_2 z_3=1\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\& \bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right) \\& \Rightarrow \bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\& +\left(\frac{1}{2}-\frac{1}{2}\right)+2 i\left(\frac{1}{2}\right) \\& =\sqrt{2}-\sqrt{2} i+i \\& \Rightarrow\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=|\sqrt{2}+i(-\sqrt{2}+1)|^2 \\& =\left(\sqrt{(\sqrt{2})^2+(1-\sqrt{2})^2}\right)^2 \\& =5-2 \sqrt{2} \\& (a, b)=(5,-2)\end{aligned}\)
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