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Complex Numbers, Hermitian matrix

Hi guys, Can you please help with the following problem:

Given: β„Žπ‘ (𝑛) 𝑒^(βˆ’π‘—(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0))+πœ”_𝑏 (𝑛)

Required Result: |𝑦(𝑛)|^2=|β„Ž|^2 |𝑠(𝑛)|^2+|πœ”_𝑏 (𝑛)|^2+2β„œ[β„Žπ‘ (𝑛) 𝑒^(βˆ’π‘—(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0))+πœ”_𝑏^𝐻 (𝑛)]

My Solution: I used the following rule of complex numbers: |𝑧_1+𝑧_2 |^2=|𝑧_1 |^2+|𝑧_2 |^2+2𝑅(𝑧_1 (𝑧_2 )Β Μ… ) I assumed that 𝑧_1=β„Žπ‘ (𝑛) 𝑒^(βˆ’π‘—(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0)), and 𝑧_2=πœ”_𝑏 (𝑛). So, |𝑧_1 |^2 γ€–=|β„Žπ‘ (𝑛) 𝑒^(βˆ’π‘—(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0)) |γ€—^2=|β„Ž|^2 |𝑠(𝑛)|^2 𝑒^(βˆ’2𝑗(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0)) |𝑧_2 |^2=|πœ”_𝑏 (𝑛)|^2 𝑧_1 (𝑧_2 )Β Μ…=|β„Ž|^2 |𝑠(𝑛)|^2 γ€–|πœ”_𝑏 (𝑛)|^2 𝑒〗^(βˆ’2𝑗(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0)) |𝑧_1+𝑧_2 |^2=|𝑧_1 |^2+|𝑧_2 |^2+2β„œ(𝑧_1 (𝑧_2 )Β Μ… )=|β„Ž|^2 |𝑠(𝑛)|^2 𝑒^(βˆ’2𝑗(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0))+|πœ”_𝑏 (𝑛)|^2+2β„œ(|β„Ž|^2 |𝑠(𝑛)|^2 γ€–|πœ”_𝑏 (𝑛)|^2 𝑒〗^(βˆ’2𝑗(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0)) ) but I couldn’t reach the same result in Green. (1) How was the first term of the real part β€œβ„Žπ‘ (𝑛) 𝑒^(βˆ’π‘—(2πœ‹βˆ†_𝑓 𝑛+πœƒ_0))β€œ obtained? (2) How was the second term of the real part β€œHermitian matrix πœ”_𝑏^𝐻 (𝑛")β€œ obtained"? I appreciate your help.