Question

use the demoivre's theorem to find the answer
(1-i)^11

Answer 1

i think its
-10(1+i)

demoivre's theorem usually involves sin and cos though ??

Answer 2

yes it involves sin theta and cos theta

Answer 3

u r not sure about the answer???

Answer 4

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

Answer 5

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

Answer 6

\[\sqrt{2}^{11}\] ???

Answer 7

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

Answer 8

in polar form

Answer 9

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

Answer 10

demoives theorm states that \[z^n = [r e^{i \theta}]^n = r^n e^{i n \theta } = r^n ( \cos(n \theta ) + i \sin(n \theta) )\]

Answer 11

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

Answer 12

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

Answer 13

if you add 2pi to the angle , then that will make the angle -3pi/4 , which is in the range that we want

Answer 14

if you add 2pi to the angle , then that will make the angle -3pi/4 , which is in the range that we want

Answer 15

so

answer=

\[(\sqrt{2})^{11} e^{-\frac{3\pi i }{4}} = (\sqrt{2})^{11} (\cos(-\frac{3 \pi}{4}) + isin( - \frac{3\pi}{4}) )\]

Answer 16

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

Answer 17

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