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anonymous

  • 5 years ago

use the demoivre's theorem to find the answer (1-i)^11

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  1. dumbcow
    • 5 years ago
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    i think its -10(1+i) demoivre's theorem usually involves sin and cos though ??

  2. anonymous
    • 5 years ago
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    yes it involves sin theta and cos theta

  3. anonymous
    • 5 years ago
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    u r not sure about the answer???

  4. dumbcow
    • 5 years ago
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    no not positive but if you expand it out (1-i)^3 = -2(1+i) (1-i)^5 = -4(1+i) .... just continue the pattern

  5. anonymous
    • 5 years ago
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    can u tell me plz how this term can be open (/22)^11 underroot 2 with the power of 11

  6. dumbcow
    • 5 years ago
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    \[\sqrt{2}^{11}\] ???

  7. anonymous
    • 5 years ago
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    \[(1-i) = \sqrt{2} (\cos(-\frac{\pi}{4}) + isin(-\frac{\pi}{4}) )\]

  8. anonymous
    • 5 years ago
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    in polar form

  9. anonymous
    • 5 years ago
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    which can be expressed in exponential form as \[\sqrt{2} e^{ -\frac{i \pi}{4} }\]

  10. anonymous
    • 5 years ago
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    demoives theorm states that \[z^n = [r e^{i \theta}]^n = r^n e^{i n \theta } = r^n ( \cos(n \theta ) + i \sin(n \theta) )\]

  11. anonymous
    • 5 years ago
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    therefore \[(1-i)^{11} = (\sqrt{2})^{11} e^{ -11 \times \frac{\pi}{4}}\]

  12. anonymous
    • 5 years ago
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    now , get the angle back into the range -pi<theta<=pi

  13. anonymous
    • 5 years ago
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    if you add 2pi to the angle , then that will make the angle -3pi/4 , which is in the range that we want

  14. anonymous
    • 5 years ago
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    if you add 2pi to the angle , then that will make the angle -3pi/4 , which is in the range that we want

  15. anonymous
    • 5 years ago
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    so answer= \[(\sqrt{2})^{11} e^{-\frac{3\pi i }{4}} = (\sqrt{2})^{11} (\cos(-\frac{3 \pi}{4}) + isin( - \frac{3\pi}{4}) )\]

  16. anonymous
    • 5 years ago
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    \[= (\sqrt{2})^{11} ( -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} ) = - (\sqrt{2})^{10} (1+i) \]

  17. anonymous
    • 5 years ago
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    and\[(\sqrt{2})^{10} = 2^5=32 \]

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