Finding the magnitude of a complex exponential function

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SUMMARY

This discussion focuses on the process of finding the magnitude of a complex exponential function, specifically using the equation |a + jb| = sqrt(a^2 + b^2). The user initially attempted to simplify the expression by substituting z with e^(jw) and applying Euler's identity. After several algebraic manipulations, they sought confirmation on their approach and ultimately received guidance that clarified their method, leading to a successful resolution of the problem.

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  • Understanding of complex numbers and their representation
  • Familiarity with Euler's identity and its applications
  • Knowledge of algebraic manipulation of complex expressions
  • Experience with using computational tools like Wolfram Alpha
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Students studying electrical engineering, mathematicians working with complex analysis, and anyone interested in mastering the manipulation of complex exponential functions.

Malgrif
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Homework Statement


I want to know the steps involved in finding the magnitude of a complex exponential function. An example of the following is shown in this picture:

T2ASrR5.png

Homework Equations


|a+jb|=sqrt(a^2+b^2)

|x/y|=|x|/|y|

The Attempt at a Solution



For the denominator, I replaced z with e^jw and used euler identity to expand the terms.

1-2rcos(w)e^(-jw)+r^2e^(-2jw)

1-rcos(w)*(cos(w)-jsinw)+.5r^2(cos(2w)-jsin(2w))

After simplifying I get:

1-rcos^2(w)+.5r^2cos(2w) + j(rsin(w)cos(w)-.5r^2sin(2w)

from there I let a=1-r(cos^2(w)-.5r^2cos(2w)) and b=r(sin(w)cos(w)-.5rsin(2w))

and using wolfram alpha to solve for sqrt(a^2+b^2) I don't get the simplified expression shown in the picture above. Am I approaching the problem correctly?
 
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If z=e^(jw), you can use this directly in the first fraction, to get the factor of (1-r). The other factor is easy to handle then.
 
mfb said:
If z=e^(jw), you can use this directly in the first fraction, to get the factor of (1-r). The other factor is easy to handle then.

Ah, I see. Thank you very much. I got the answer =)
 

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