Download the app

Questions  

If z1=z3=z3=1 then value of z2z32+z3z12+z1z22 cannot exceed

a
3
b
6
c
9
d
12

detailed solution

Correct option is C

z1+z2+z3≥0⇒z12+z22+z32+2Re⁡z1z¯2+z2z¯3+z3z¯1≥0⇒Re⁡z1z¯2+z2z¯3+z3z¯1≥−32Now, z2−z32+z3−z12+z1−z22=6−2Re⁡z1z¯2+z2z¯3+z3z¯1≤6−(−3)=9

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

The complex number z, satisfies the condition z25z=24  The maximum distance from the  origin of coordinates to the point  z, is 


phone icon
whats app icon