## anonymous 5 years ago use the demoivre's theorem to find the answer (1-i)^11

1. dumbcow

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

4. dumbcow

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. dumbcow

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

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