anonymous
  • anonymous
Find the magnitude of
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
http://assets.openstudy.com/updates/attachments/54b9abe2e4b04eed2e57186e-asaad123-1421454485085-img_20150117_041725.jpg
anonymous
  • anonymous
j=i imaginary part
ganeshie8
  • ganeshie8
magnitudes get divided when you divide two complex numbers : \[\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}\]

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ganeshie8
  • ganeshie8
so the magnitude of \(\large \eta^2 \) is given by : \[\large |\eta^2| = \dfrac{\omega\mu}{\sqrt{\sigma^2+\omega^2\epsilon^2}}\]
ganeshie8
  • ganeshie8
simply take the square root
anonymous
  • anonymous
Why did you take off the square root at the numerator only?
anonymous
  • anonymous
and how did the denominator turn to that form?
ganeshie8
  • ganeshie8
the magnitude of numerator, \(j\omega \mu\), is just \(|\omega \mu|\) right
ganeshie8
  • ganeshie8
its the same formula that you know magnitude of \(a+jb\) is given by \(\sqrt{a^2+b^2}\)
ganeshie8
  • ganeshie8
its just that mathematicians use \(i\) and physicists use \(j\) they both are same
anonymous
  • anonymous
I need time to digest it. Thanks for explanation :)
anonymous
  • anonymous
One more question: They ask for magnitude of \(\eta\) , not \(\eta^2\), why did you turn it to \(\eta^2\)?
ganeshie8
  • ganeshie8
\(|\eta| = \sqrt{|\eta^2|}\)
anonymous
  • anonymous
And after going around, you get back to \(|\eta|\), right? Thanks again. :)
ganeshie8
  • ganeshie8
Exactly, that wasn't supposed to be hard haha!
ganeshie8
  • ganeshie8
technically the answer is simply \( \sqrt{\dfrac{\omega\mu}{\sqrt{\sigma^2+\omega^2\epsilon^2}}}\)
anonymous
  • anonymous
:)

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