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anonymous

  • 5 years ago

Use the demoivre's theorem to find the answer? (1+i)^8

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  1. anonymous
    • 5 years ago
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    Let \(z=1+i\) then: \[\left| z \right|=\sqrt{1^2+1^2}=\sqrt 2; \theta= \tan^{-1}({1 \over 1})=\pi/4\]. Writing z in polar coordinates: \[z=\sqrt2 (\cos({\pi \over 4})+i \sin({\pi \over 4}))\]. Now apply demoivre's theorem: \[(1+i)^8=z^8=(\sqrt{2})^8(\cos(8{\pi \over 4})+i \sin(8{\pi \over 4}))=2^4(\cos(2\pi)+i \sin(2\pi))=16(1+i (0))=16\]

  2. anonymous
    • 5 years ago
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    By the way, you can easily prove this by simple calculation: \[(1+i)^8=[(1+i)^2]^4=(1+2i+i^2)^4=(2i)^4=2^4(i^4)=16\] Notice here that \(i^2=-1\) and \(i^4=1\).

  3. anonymous
    • 5 years ago
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    what if the question is (1-i)^11

  4. anonymous
    • 5 years ago
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    Try to follow the same procedure and tell what you get.

  5. anonymous
    • 5 years ago
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    what to do after this step 2(under root )^11 (cos-11/4+isin-11/4)

  6. anonymous
    • 5 years ago
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    sorry cos-11pie/4+isin-11pie/4

  7. anonymous
    • 5 years ago
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    \(cos (-11\pi/4)=sin(-11 \pi/4)=1/\sqrt2\)

  8. anonymous
    • 5 years ago
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    sorry \(-1/\sqrt2\)

  9. anonymous
    • 5 years ago
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    but how???

  10. anonymous
    • 5 years ago
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    \[\cos({-11 \pi \over 4})=\cos(4\pi-{11\pi \over 4})\cos({5\pi \over 4})\]. \(5\pi/4\) is the correspondent angle of \(pi/4\) in the third quadrant, Hence cos\((-11\pi/4)=-1/\sqrt2\).

  11. anonymous
    • 5 years ago
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    There is a missing equality in the first line between the two cosines :)

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