Question

# If z1,z2,z3 and z4 are the roots of the equation z4+z3+z2+z+1=0, then Column I Column II (A)∣∣ ∣∣4∑i=1(zi)4∣∣ ∣∣is equal to (p)0(B)4∑i=1(zi)5is equal to (q)4(C)4∏i=1(zi+2)is equal to(r)1(D)Least value of [|z1+z2|]is(s)11 where [ ] represents greatest integer function Which of the following is the correct combination

A
(A)(p),(B)(r),(C)(s),(D)(q)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(A)(r),(B)(q),(C)(s),(D)(p)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(A)(q),(B)(r),(C)(s),(D)(p)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(A)(p),(B)(q),(C)(r),(D)(s)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B (A)→(r),(B)→(q),(C)→(s),(D)→(p)The given equation is z4+z3+z2+z+1=0 So, the equation is z5−1z−1 which means that z1,z2,z3,z4 are four out of five roots of unity except 1 (A) z41+z42+z43+z44+14=0 ⇒∣∣ ∣∣4∑i=1(zi)4∣∣ ∣∣=1 (B) z51+z52+z53+z54+1=5 ⇒4∑i=1(zi)5=4 (C) z4+z3+z2+z+1=(z−z1)(z−z2)(z−z3)(z−z4) Putting z=−2 on both the sides, we get ⇒4∏i=1(zi+2)=11 (D) |z1+z2|=√|z1|2+|z2|2+2|z1||z2|cos(θ1−θ2)⇒|z1+z2|=√2+2cos(θ1−θ2) ⇒|z1+z2|=2∣∣∣cos(θ1−θ22)∣∣∣⇒|z1+z2|=2∣∣ ∣∣cos(2π(k1−k2)2n)∣∣ ∣∣,k1,k2=1,2,3,4,[k1≠k2] Now n=5 and for least value k1−k2=3 ∴|z1+z2|=2∣∣∣cos3π5∣∣∣=2cos2π5=√5−12 whose greatest integer is 0

Suggest Corrections
5
Join BYJU'S Learning Program
Related Videos
De-Moivre's Theorem
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program