Find ##z## in the form ##a+bi## under Complex Numbers

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SUMMARY

The discussion focuses on finding the complex number ##z## in the form ##a+bi##, specifically through the expression ##z = \dfrac {3+i}{3-i} \cdot \dfrac {3+i}{3+i}##, which simplifies to ##z = \dfrac {4}{5} + \dfrac {3}{5}i##. The participants emphasize the importance of understanding the Argand plane for part (b) and clarify that the modulus of ##z## is 1. Additionally, they correct a misunderstanding regarding the modulus of ##z - z^*##, confirming it should be ##\dfrac {6}{5}##, not ##\frac{6}{5}i##.

PREREQUISITES
  • Understanding of complex numbers and their representation in the Argand plane
  • Familiarity with modulus and conjugate of complex numbers
  • Basic algebraic manipulation of complex fractions
  • Knowledge of complex number operations, including multiplication and simplification
NEXT STEPS
  • Study the properties of complex numbers in the Argand plane
  • Learn about the modulus and argument of complex numbers
  • Explore complex number operations, focusing on multiplication and division
  • Review examples of simplifying complex fractions
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chwala
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Homework Statement
see attached
Relevant Equations
Complex numbers
1646184520481.png


For part (a),
##z##=##\dfrac {3+i}{3-i}## ⋅##\dfrac {3+i}{3+i}##
##z##=##\dfrac {4}{5}##+##\dfrac {3}{5}i##

part (b) no problem as long as one understands the argand plane...

For part (c)
Modulus of ##z=1##
and Modulus of ##z-z^*##=##\frac{6}{5}i##
 
Last edited:
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The modulus is a real number, so it can't be ##\frac 6 5 i##.
 
PeroK said:
The modulus is a real number, so it can't be ##\frac 6 5 i##.
True, its supposed to be ##\dfrac {6}{5}##
 
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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