Question

DeMoivre's theorem

Answer 1

HI!!

Answer 2

haha hello! Give me a moment to post my question :P

Answer 3

express in form a+bi
[1- sq3]^5

Answer 4

do you have to write
\[a+bi\] as \[r\left(\cos(\theta)+i\sin(\theta)\right)\]?

Answer 5

yeah you do
that is
\[1-\sqrt3 i\] right?

Answer 6

Yep. So first I change it to polar form

Answer 7

right
you know how?

Answer 8

r= 2,
tan(theta) = -sq3/1 so, -sq3

Answer 9

so far so good

Answer 10

I can't find the exact theta though

Answer 11

oh ok lets go slow and careful

Answer 12

first off instead of writing
\[\tan(\theta)=-\sqrt3\] lets use
\[\cos(\theta)=\frac{a}{r},\sin(\theta)=\frac{b}{r}\]so you don't have to fret about what quadrant you are in

Answer 13

so
\[\cos(\theta)=\frac{1}{2},\sin(\theta)=-\frac{\sqrt3}{2}\]

Answer 14

a familiar point on the unit circle
do you know where that point is? i mean do you know what angle corresponds to that point?

Answer 15

pi/3?

Answer 16

ah no pi/6

Answer 17

if not, a unit circle cheat sheet would really help
no not \(\frac{\pi}{3}\) because sine is positive there

Answer 18

don't forget sine is negative

Answer 19

5pi/6

Answer 20

5pi/3 xD

Answer 21

look at the unit circle on the attached cheat sheet

Answer 22

yeah you got it

Answer 23

Sorry, I got confused because I mixed up and id (y,x) *facepalm.

Answer 24

so theta = 5pi/3?

Answer 25

lol don't forget x comes before y in the alphabet
you kinda have to sing that song all the way to the end though...
yeah \[\frac{5\pi}{3}\]

Answer 26

dude..ikr. xD

Answer 27

is that a fish?

Answer 28

...not..not quite. moving on :P back to the problem

Answer 29

in any case now it is really really easy to raise it to the power of 5
do you know how to do that?

Answer 30

if not i can show you
just asking

Answer 31

working it out on paper. just a sec.

Answer 32

ok

Answer 33

=32cis(25pi/3)

Answer 34

the last part can be simplified though

Answer 35

yeah that looks good

Answer 36

so
=32cis(pi/3)

Answer 37

yeah

Answer 38

looks like you got this right?

Answer 39

yep :) thanks so much!

Answer 40

ooh no wait, have to put it in a +bi form

Answer 41

btw it is a lot easier to use \[\cos(\theta)=\frac{a}{r},\sin(\theta)=\frac{b}{r}\] instead of tangent

Answer 42

yeah but you can do that, because you know what
\[\cos(\frac{\pi}{3})\] and
\[\sin(\frac{\pi}{3})\] are

Answer 43

err so it equals
16+ ..?

Answer 44

yeah 16 plus something ...

Answer 45

16 sq3?

Answer 46

16+16 sq3

Answer 47

yup

Answer 48

no..
it's 16+i(16 sq3)

Answer 49

lol yeah i forgot about the eye

Answer 50

kks

Answer 51

\[\color\magenta\heartsuit\]