Question

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

Answer 1

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:
\[(1+i)^8=z^8=(\sqrt{2})^8(\cos(8{\pi \over 4})+i \sin(8{\pi \over 4}))=2^4(\cos(2\pi)+i \sin(2\pi))=16(1+i (0))=16\]

Answer 2

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 \(i^4=1\).

Answer 3

what if the question is (1-i)^11

Answer 4

Try to follow the same procedure and tell what you get.

Answer 5

what to do after this step
2(under root )^11 (cos-11/4+isin-11/4)

Answer 6

sorry
cos-11pie/4+isin-11pie/4

Answer 7

\(cos (-11\pi/4)=sin(-11 \pi/4)=1/\sqrt2\)

Answer 8

sorry \(-1/\sqrt2\)

Answer 9

but how???

Answer 10

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

Answer 11

There is a missing equality in the first line between the two cosines :)