Let z1, z2 and z3 be three complex number on the circle |z| = 1 with arg(z1)=–π4, arg(z2)=0 and arg(z3)=π4. If |z1z2+z2z3+z3z1|2=α+β2, α,β∈Z, then the value of α2+β2 is : [2025]
(3)
Given, |z| = 1
Now, z1=eiπ4=12–i2=1–i2; z2=ei0=1;
z3=eiπ4=1+i2
Now, |z1z¯2+z2z¯3+z3z¯1|2=|1–i2·1+1·1–i2+(1+i2)2|2
=|2+(1–2)i|2=5–22=α+β2
⇒ α=5 and β=–2 ∴ α2+β2=29