# Simple Complex Number Review Question

1. Sep 2, 2013

### bmb2009

1. The problem statement, all variables and given/known data
z=1 + e^(iθ) calculate z^2 and lzl^2

2. Relevant equations

3. The attempt at a solution

for z^2

(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

2. Sep 2, 2013

### Dick

It doesn't have just an e^(iθ) in it. It should also have an e^(-iθ). Check your expansion. Put them together and the result is not imaginary.

3. Sep 2, 2013

### bmb2009

Do you mind showing me the e^(-i@) + e^(-i@) cancellation?

4. Sep 2, 2013

### bmb2009

Do u mind showing me the cancellation written on?

5. Sep 2, 2013

### Dick

Express the sum e^(iθ)+e^(-iθ) (which is what you should have had in your expansion) in terms of sines and cosines.