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iop360

  • 2 years ago

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

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  1. ktnguyen1
    • 2 years ago
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    |dw:1353832796610:dw|

  2. ktnguyen1
    • 2 years ago
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    |dw:1353833084664:dw|

  3. iop360
    • 2 years ago
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    so is that just 2 solutions? shouldnt i have 4?

  4. cinar
    • 2 years ago
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    |dw:1353825968940:dw|

  5. yakeyglee
    • 2 years ago
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    It does have four solutions.

  6. cinar
    • 2 years ago
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    |dw:1353826101375:dw|

  7. iop360
    • 2 years ago
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    hmm it says the answers are |dw:1353818990720:dw|

  8. yakeyglee
    • 2 years ago
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    If 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.

  9. iop360
    • 2 years ago
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    how did you come up with the angles and value for r=4

  10. iop360
    • 2 years ago
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    i tried this method originally but im confused at that part

  11. iop360
    • 2 years ago
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    \[-16 = r^4e ^{i(4\theta)}\]

  12. yakeyglee
    • 2 years ago
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    Sorry...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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