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

Given: ℎ𝑠(𝑛) 𝑒^(−𝑗(2𝜋∆_𝑓 𝑛+𝜃_0))+𝜔_𝑏 (𝑛)

Required Result: |𝑦(𝑛)|^2=|ℎ|^2 |𝑠(𝑛)|^2+|𝜔_𝑏 (𝑛)|^2+2ℜ[ℎ𝑠(𝑛) 𝑒^(−𝑗(2𝜋∆_𝑓 𝑛+𝜃_0))+𝜔_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 ) ̅ ) I assumed that 𝑧_1=ℎ𝑠(𝑛) 𝑒^(−𝑗(2𝜋∆_𝑓 𝑛+𝜃_0)), and 𝑧_2=𝜔_𝑏 (𝑛). So, |𝑧_1 |^2 〖=|ℎ𝑠(𝑛) 𝑒^(−𝑗(2𝜋∆_𝑓 𝑛+𝜃_0)) |〗^2=|ℎ|^2 |𝑠(𝑛)|^2 𝑒^(−2𝑗(2𝜋∆_𝑓 𝑛+𝜃_0)) |𝑧_2 |^2=|𝜔_𝑏 (𝑛)|^2 𝑧_1 (𝑧_2 ) ̅=|ℎ|^2 |𝑠(𝑛)|^2 〖|𝜔_𝑏 (𝑛)|^2 𝑒〗^(−2𝑗(2𝜋∆_𝑓 𝑛+𝜃_0)) |𝑧_1+𝑧_2 |^2=|𝑧_1 |^2+|𝑧_2 |^2+2ℜ(𝑧_1 (𝑧_2 ) ̅ )=|ℎ|^2 |𝑠(𝑛)|^2 𝑒^(−2𝑗(2𝜋∆_𝑓 𝑛+𝜃_0))+|𝜔_𝑏 (𝑛)|^2+2ℜ(|ℎ|^2 |𝑠(𝑛)|^2 〖|𝜔_𝑏 (𝑛)|^2 𝑒〗^(−2𝑗(2𝜋∆_𝑓 𝑛+𝜃_0)) ) but I couldn’t reach the same result in Green. (1) How was the first term of the real part “ℎ𝑠(𝑛) 𝑒^(−𝑗(2𝜋∆_𝑓 𝑛+𝜃_0))“ obtained? (2) How was the second term of the real part “Hermitian matrix 𝜔_𝑏^𝐻 (𝑛")“ obtained"? I appreciate your help.