Find the magnitude of

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Find the magnitude of

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

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