Question

using de Moivre's theorem to find (1-i)^6 in the form a+bi
(begin by expressing 1-i in polar forms)

Answer 1

sure

Answer 2

ok lets begin
first of all you need to write it as
\[r(\cos(\theta)+i\sin(\theta))\] so you need only two numbers, \(r\) and \(\theta\)

Answer 3

\(r\) is the easiest to find
it is \[r=\sqrt{a^2+b^2}\] which in our case is \[r=\sqrt{1^2+1^2}=\sqrt2\]

Answer 4

good so far?

Answer 5

yes

Answer 6

k now we need \(\theta\) which is actually obvious if you draw a picture, otherwise we have to do some annoying trig
your choice

Answer 7

picture?

Answer 8

good idea

Answer 9

what angle is that?

Answer 10

-45?

Answer 11

got it on the first try (if you are working in degree)

Answer 12

you could also say \(315\) it doesn't matter
lets stick with \(-45\)

Answer 13

ok

Answer 14

this gives you
\[1-i=\sqrt2(\cos(-45)+i\sin(-45))\]and you are done with the first part, writing it in "polar" form

Answer 15

unless polar form means in this case \[\large 1-i=\sqrt{2}e^{-\frac{\pi}{4}i}\]which i assume it does not

Answer 16

then from there is it is very very easy to raise it to the sixth power (that is why you write it in polar form)
all you need to do is find \((\sqrt2)^6\) and also multiply the angle by \(6\)

Answer 17

you good from there?

Answer 18

how did u get this graph?

Answer 19

\(1-i\) in the complex plane
right one, down one

Answer 20

no different from plotting the point \((1,-1)\)

Answer 21

btw the length of the line i drew is the absolute value of \(1-i\) which is pretty clearly \(\sqrt2\)

Answer 22

are you good for raising it to the power of 6?

Answer 23

yes, thanks for the help

Answer 24

yw
you want me to check your answer?

Answer 25

im just going through it again

Answer 26

kk

Answer 27

does it come to -8i

Answer 28

let me check

Answer 29

the \(8\) is right for sure

Answer 30

then \(-45\times 6=-270\) so it is
\[8(\cos(-270)+i\sin(-270))\]

Answer 31

but i think you are off by a minus sign

Answer 32

\[\cos(-270)=0,\sin(-270)=1\]

Answer 33

therefore it is \(8i\) not \(-8i\)

Answer 34

oh right! thanks alot!

Answer 35

i can show you a picture to make it clear that you get to \(8i\) if you like