CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
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)∣ ∣4i=1(zi)4∣ ∣is equal to (p)0(B)4i=1(zi)5is equal to (q)4(C)4i=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 z51z1 which means that z1,z2,z3,z4 are four out of five roots of unity except 1

(A) z41+z42+z43+z44+14=0

∣ ∣4i=1(zi)4∣ ∣=1

(B) z51+z52+z53+z54+1=5

4i=1(zi)5=4

(C) z4+z3+z2+z+1=(zz1)(zz2)(zz3)(zz4)

Putting z=2 on both the sides, we get

4i=1(zi+2)=11


(D) |z1+z2|=|z1|2+|z2|2+2|z1||z2|cos(θ1θ2)|z1+z2|=2+2cos(θ1θ2)
|z1+z2|=2cos(θ1θ22)|z1+z2|=2∣ ∣cos(2π(k1k2)2n)∣ ∣,k1,k2=1,2,3,4,[k1k2]
Now n=5 and for least value k1k2=3
|z1+z2|=2cos3π5=2cos2π5=512
whose greatest integer is 0

flag
Suggest Corrections
thumbs-up
5
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
De-Moivre's Theorem
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon