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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let \(z=1+i\) 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: \[(1+i)^8=z^8=(\sqrt{2})^8(\cos(8{\pi \over 4})+i \sin(8{\pi \over 4}))=2^4(\cos(2\pi)+i \sin(2\pi))=16(1+i (0))=16\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By 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 \(i^4=1\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what if the question is (1i)^11

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try to follow the same procedure and tell what you get.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what to do after this step 2(under root )^11 (cos11/4+isin11/4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry cos11pie/4+isin11pie/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\(cos (11\pi/4)=sin(11 \pi/4)=1/\sqrt2\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\cos({11 \pi \over 4})=\cos(4\pi{11\pi \over 4})\cos({5\pi \over 4})\]. \(5\pi/4\) is the correspondent angle of \(pi/4\) in the third quadrant, Hence cos\((11\pi/4)=1/\sqrt2\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is a missing equality in the first line between the two cosines :)
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