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anonymous
 4 years ago
find all complex numbers z satisfying
z^4 = 16 in the form a+bi
anonymous
 4 years ago
find all complex numbers z satisfying z^4 = 16 in the form a+bi

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353832796610:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353833084664:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so is that just 2 solutions? shouldnt i have 4?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353825968940:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It does have four solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353826101375:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm it says the answers are dw:1353818990720:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you're familiar with complex exponentials, try this: \(\large z = re^{i \theta} \Rightarrow z^4 = r^4 e^{i(4\theta)} = 16 = 16 e^{i\pi}\) The values that satisfy this are \(\large r=4\) and \(\large \theta = \{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}\). Use \(\large e^{i \theta} = \cos \theta + i \sin \theta\) to convert to polar.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did you come up with the angles and value for r=4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i tried this method originally but im confused at that part

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[16 = r^4e ^{i(4\theta)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry...what I meant was \(r = 2\). We are assuming \(r\) to be positive and real, and we want it to correspond to a length of 16 since in the complex plane 16 has length (absolute value) 16=16. The angles are values \(0\le \theta<2\pi\) such that \(4\theta \text{ mod } 2\pi = \pi\).
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