Simple Complex Number Review Question

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Homework Help Overview

The problem involves complex numbers, specifically calculating \( z^2 \) and \( |z|^2 \) for the expression \( z = 1 + e^{i\theta} \). Participants are exploring the implications of their calculations and the expected forms of the results.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the expansion of \( z^2 \) and the calculation of \( |z|^2 \), questioning whether their results are correct given the presence of imaginary components. There are inquiries about the cancellation of terms in the expansion and the relationship between \( e^{i\theta} \) and \( e^{-i\theta} \).

Discussion Status

The discussion is ongoing, with participants seeking clarification on their calculations and the nature of the results. Some guidance has been offered regarding the need to consider both \( e^{i\theta} \) and \( e^{-i\theta} \) in the expansions, but no consensus has been reached on the correctness of the original poster's approach.

Contextual Notes

There is a concern regarding the presence of imaginary numbers in the final answers, as the original poster references a textbook that suggests the results should not include them. This raises questions about the assumptions made in the calculations.

bmb2009
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Homework Statement


z=1 + e^(iθ) calculate z^2 and lzl^2



Homework Equations





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
 
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bmb2009 said:

Homework Statement


z=1 + e^(iθ) calculate z^2 and lzl^2

Homework Equations


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

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.
 
Do you mind showing me the e^(-i@) + e^(-i@) cancellation?
 
Do u mind showing me the cancellation written on?
 
bmb2009 said:
Do you mind showing me the e^(-i@) + e^(-i@) cancellation?

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

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