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 $(a, b)$

Question

YESPYQ · Accepted Answer

Correct answer: c) $(5,-2)$. $\begin{aligned} &\text { Sol. }\ &\begin{aligned} & z_1=|1| e^{i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}+i \cdot \frac{1}{\sqrt{2}} \ & z_2=|1| e^{-(0)}= 1+0 i \ & z_3=|1| e^{-i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \ & \bar{z}_1 z_2=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)(1) \ & \bar{z}_2 z_3=1\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \ & \bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right) \ & \Rightarrow \bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \ & +\left(\frac{1}{2}-\frac{1}{2}\right)+2 i\left(\frac{1}{2}\right) \ & =\sqrt{2}-\sqrt{2} i+i \ & \Rightarrow\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=|\sqrt{2}+i(-\sqrt{2}+1)|^2 \ & =\left(\sqrt{(\sqrt{2})^2+(1-\sqrt{2})^2}\right)^2 \ & =5-2 \sqrt{2} \ & (a, b)=(5,-2) \end{aligned} \end{aligned}$

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 \…

Practice more JEE Maths PYQs