JEE 2025 — Mathematics PYQ

Given $z_1,z_2,z_3$ are three complex numbers and lie on circle $|z|=1$ with  

$\text{Arg}(z_1)=-\frac{\pi }{4}$  

$\text{Arg}(z_2)=0$  

$\text{Arg}(z_3)=\frac{\pi }{4}$

Using Euler’s formula,

$z=e^{i\theta }=\cos \theta +i\sin \theta$

$\therefore$

$z_1=\frac{1}{\sqrt{2}}(1-i),{\bar{z}}_1=\frac{1}{\sqrt{2}}(1+i)$

$z_2=1+0i,{\bar{z}}_2=1$

$z_3=\frac{1}{\sqrt{2}}(1+i),{\bar{z}}_3=\frac{1}{\sqrt{2}}(1-i)$

$\therefore |z_1{\bar{z}}_2+z_2{\bar{z}}_3+z_3{\bar{z}}_1{|}^2$

$={\left|\frac{1}{\sqrt{2}}(1-i)+\frac{1}{\sqrt{2}}(1-i)+\frac{1}{2}(1+i{)}^2\right|}^2$

$={\left|\sqrt{2}-\sqrt{2}i+\frac{1}{2}(2i)\right|}^2$

$={\left|\sqrt{2}+(1-\sqrt{2})i\right|}^2$

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

$=2+1+2-2\sqrt{2}$

$=5-2\sqrt{2}$

$\therefore \alpha =5,\beta =-2$

${\alpha }^2+{\beta }^2=25+4=29$