1no. not always.
it's correct only when imaginary part of z is 0.
14 Answers
1
1357It is not true
Left hand side is complex number
Right hand side is a real number
So this will be true only when z is a real number
24easy hai..........
lhs:
mod (z1+z2)2=mod(z1)2+mod(z2)2+2Re(z1.z2)
for LHS=RHS ,Re(z1.z2)=0
=>z1.z2 is purely imaginary
1357|z1+z2|2=(z1+z2)(z1+z2) as (zz=|z|2)
(z1+z2)(z1+z2)=z1z1+z2z2+z1z2+z2z1
=|z1|2+|z2|2+z1z2+z2z1
now z1z2+z2z1=0
let z1z2=z, then z2z1=z
z1z2+z2z1=0
implies z+z=0 implies 2Re(z)=0
so z=z1z2 is purely imaginary
1ohhh it shud b done in dis way......shit i was tryin in all diff methodz...n landed sum wer....thx a lot