Simple Complex Number Review Question

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SUMMARY

The discussion focuses on calculating the square of a complex number, z = 1 + e^(iθ), and its magnitude squared, |z|^2. The correct calculation for z^2 results in 1 + 2e^(iθ) + e^(i2θ), while the magnitude squared is computed as |z|^2 = 2(1 + e^(iθ)). A key point of confusion arises from the expectation that the final answer should not contain imaginary components, which is clarified by recognizing the presence of both e^(iθ) and e^(-iθ) in the calculations. The discussion emphasizes the importance of correctly expressing complex exponentials in terms of trigonometric functions.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with Euler's formula, e^(iθ) = cos(θ) + i sin(θ)
  • Knowledge of trigonometric identities
  • Basic algebraic manipulation of complex expressions
NEXT STEPS
  • Review the derivation of Euler's formula and its applications in complex analysis
  • Learn how to express complex exponentials in terms of sine and cosine functions
  • Study the properties of complex conjugates and their role in magnitude calculations
  • Explore the geometric interpretation of complex numbers on the Argand plane
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Students studying complex analysis, mathematics enthusiasts, and anyone seeking to deepen their understanding of complex number operations and their implications in various mathematical contexts.

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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