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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

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

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

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

sorry $$-1/\sqrt2$$

9. anonymous

but how???

10. anonymous

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

11. anonymous

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

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