wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If b+ic=(1+a)z and a2+b2+c2=1, where a,b,cR, then 1+iz1iz=a+ibk where k is equal to

A
1+a
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1+b
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1+c
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2+b
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1+c
Given
a2+b2+c2=1 (1)
and z=b+ic1+a
iz1=ibc1+a
Applying componendo and dividendo rule, we get
1+iz1iz=1+a+ibc1+aib+c=(1+ac)+ib(1+a+c)ib=(1+ac)+ib(1+a+c)ib×(1+a+c)+ib(1+a+c)+ib=(1+ac)(1+a+c)b2+ib(1+ac+1+a+c)(1+a+c)2+b2=1+a2+2ac2b2+2ib(1+a)1+a2+b2+c2+2a+2c+2ac=a2+2a+a2+2ib(1+a)2+2a+2c+2ac [a2+b2+c2=1]=2(1+a)(a+ib)2(1+a)(1+c)
=a+ib1+c
On comparing with a+ibk, we get
k=1+c

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon