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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. anonymous

i think its -10(1+i) demoivre's theorem usually involves sin and cos though ??

2. anonymous

yes it involves sin theta and cos theta

3. anonymous

u r not sure about the answer???

4. anonymous

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

can u tell me plz how this term can be open (/22)^11 underroot 2 with the power of 11

6. anonymous

$\sqrt{2}^{11}$ ???

7. anonymous

$(1-i) = \sqrt{2} (\cos(-\frac{\pi}{4}) + isin(-\frac{\pi}{4}) )$

8. anonymous

in polar form

9. anonymous

which can be expressed in exponential form as $\sqrt{2} e^{ -\frac{i \pi}{4} }$

10. anonymous

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

therefore $(1-i)^{11} = (\sqrt{2})^{11} e^{ -11 \times \frac{\pi}{4}}$

12. anonymous

now , get the angle back into the range -pi<theta<=pi

13. anonymous

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

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

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

$= (\sqrt{2})^{11} ( -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} ) = - (\sqrt{2})^{10} (1+i)$

17. anonymous

and$(\sqrt{2})^{10} = 2^5=32$

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