# Complex Numbers, Hermitian matrix

Given: βπ (π) π^(βπ(2πβ_π π+π_0))+π_π (π)

Required Result: |π¦(π)|^2=|β|^2 |π (π)|^2+|π_π (π)|^2+2β[βπ (π) π^(βπ(2πβ_π π+π_0))+π_π^π» (π)]

My Solution: I used the following rule of complex numbers: |π§_1+π§_2 |^2=|π§_1 |^2+|π§_2 |^2+2π(π§_1 (π§_2 )Β Μ ) I assumed that π§_1=βπ (π) π^(βπ(2πβ_π π+π_0)), and π§_2=π_π (π). So, |π§_1 |^2 γ=|βπ (π) π^(βπ(2πβ_π π+π_0)) |γ^2=|β|^2 |π (π)|^2 π^(β2π(2πβ_π π+π_0)) |π§_2 |^2=|π_π (π)|^2 π§_1 (π§_2 )Β Μ=|β|^2 |π (π)|^2 γ|π_π (π)|^2 πγ^(β2π(2πβ_π π+π_0)) |π§_1+π§_2 |^2=|π§_1 |^2+|π§_2 |^2+2β(π§_1 (π§_2 )Β Μ )=|β|^2 |π (π)|^2 π^(β2π(2πβ_π π+π_0))+|π_π (π)|^2+2β(|β|^2 |π (π)|^2 γ|π_π (π)|^2 πγ^(β2π(2πβ_π π+π_0)) ) but I couldnβt reach the same result in Green. (1) How was the first term of the real part ββπ (π) π^(βπ(2πβ_π π+π_0))β obtained? (2) How was the second term of the real part βHermitian matrix π_π^π» (π")β obtained"? I appreciate your help.

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