z=1 + e^(iθ) calculate z^2 and lzl^2
The Attempt at a Solution
(1+e^(iθ))(1+e^(iθ)) = 1 + 2e^(iθ) + e^(i2θ).. is that the final answer? i expanded it into cosines and sines as well but that doesn't simplify anymore i don't believe.
for lzl^2 = (1+e^(iθ))*(1+e^(-iθ))= 1 + e^(iθ) + e^(iθ) + e^0 = 2 + 2e^(iθ) = 2(1+e^(iθ)) = 2z
but did I do something wrong because the book says the final answer should not have an imaginary number in it? but z has e^(iθ) in it? thanks for the help