complex numbers Watch

ihatePE
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I dont understand how they got the second line...how did they get (1+4n)/6 * pi?
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RDKGames
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(Original post by ihatePE)

I dont understand how they got the second line...how did they get (1+4n)/6 * pi?
What's the question?
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ihatePE
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(Original post by RDKGames)
What's the question?
find all possible values of z^(2/3) where z = 1 + j
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RDKGames
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(Original post by ihatePE)
find all possible values of z^(2/3) where z = 1 + j
Well if z = re^{i \theta} then z^2 = r^2 e^{i(2\theta)}.

Due to the cyclic behaviour of the complex numbers, we have that z^2 = r^2 e^{i(2\theta + 2\pi n)} for any n \in \mathbb{Z}.

The next step is to take cube root of both sides, so that exponent becomes \dfrac{2\theta + 2\pi n}{3} (omitting the i, but it's still there). Since \theta = \dfrac{\pi}{4}, the expression reduces to what you're uncertain about with the sufficient condition that n = 0, 1, 2 in order to capture all the roots.
Last edited by RDKGames; 4 weeks ago
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ihatePE
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(Original post by RDKGames)
Well if z = re^{i \theta} then z^2 = r^2 e^{i(2\theta)}.

Due to the cyclic behaviour of the complex numbers, we have that z^2 = r^2 e^{i(2\theta + 2\pi n)} for any n \in \mathbb{Z}.

The next step is to take cube root of both sides, so that exponent becomes \dfrac{2\theta + 2\pi n}{3} (omitting the i, but it's still there). Since \theta = \dfrac{\pi}{4}, the expression reduces to what you're uncertain about with the sufficient condition that n = 0, 1, 2 in order to capture all the roots.
thank you very much
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