Question

Write the expression in a + bi form: (-√3 + i)^6

Answer 1

Have you been taught DeMoivre's Theorem or Binomial Theorem?

Answer 2

DeMoivre's Theorem:
If z = r(cos(theta) + isin(theta), then
z^n = r^n(cos(n*theta) + isin(n * theta))

Have you seen this formula before?

Answer 3

@AkashdeepDeb

Answer 4

:)

Answer 5

If you have been taught DeMoivre's Theorem then I can tell you the steps to solve this.

Answer 6

Okay i have

Answer 7

@ranga why do you keep saying you will help me then just leave...

Answer 8

Because I posted my first reply 21 hours ago and you wouldn't even bother to answer the question that I had to reask the question again now.

Answer 9

Oh im sorry i was waiting for a reply and then i went ofline

Answer 10

We need to put (-√3 + i) in the form r * ( cos(theta) + isin(theta) )
Find the magnitude of (-√3 + i). Factor it out.
Magnitude of a + ib = sqrt(a^2 + b^2).
I will wait for you to factor out the magnitude.

Answer 11

2

Answer 12

Right?

Answer 13

Correct. Factor the magnitude out.
(-√3 + i) = 2( -√3/2 + 1/2i )
Remember we need to put it in the form r * ( cos(theta) + isin(theta) )
Which angle theta will have cos(theta) = -√3/2 and sin(theta) = 1/2?
(look up the unit circle).

Answer 14

Okay hang on

Answer 15

1/2 is 90 right?

Answer 16

sin(90) = 1, cos(90) = 0. Neither one gives half.

Answer 17

oops sin(30)

Answer 18

sin(30) is 1/2 but cos(30) = +sqrt(3)/2.
We want cos(theta) = -√3/2 and sin(theta) = 1/2.
In a unit circle, the first coordinate is cos(theta) and the second coordinate is sin(theta). So look up unit circle and see which angle gives the coordinates (-sqrt(3)/2, 1/2).

Answer 19

The angle will be in the second quadrant because that is where cosine is negative and sine is positive.

Answer 20

150!! :)

Answer 21

Yes! That is in degrees. Here it will be easier if we use radians.
150 degrees = 5*pi/6.

Answer 22

\[(-\sqrt{3} + i) = 2(-\sqrt{3}/2 + 1/2i) = 2[ \cos(5*\pi/6) + isin(5*\pi/6) ]\]

Answer 23

\[(-\sqrt{3} + i)^6 = [2( \cos(5*\pi/6) + isin(5*\pi/6) )]^6 = \]

Answer 24

\[2^6 * ( \cos(5*\pi) + isin(5*\pi) ) = 64 * (\cos(\pi) + isin(\pi)) = 64 * -1 = -64\]

Answer 25

Okay still trying to understand all of it but why did the 5 just disappear?

Answer 26

In a unit circle, 2pi is one full rotation around a circle and so any additions or subtraction of 2*pi from the angle will land up in the same place.
5*pi = pi + 2pi + 2pi. We can subtract the two 2pi's because they are just two full rounds around the unit circle and you end up at the same angle.

Answer 27

Ohhh that makes sense :)

Answer 28

cos(5pi) is same as cos(3pi) is same as cos(pi)
is same as cos(7pi), cos(9pi), etc.

Answer 29

Hmm never knew that but i guess it makes sense because of what you said about 2pi :)

Answer 30

Can you help me with another similar one except it says to put it in a + bi form? ill open up a new question if you can help?

Answer 31

In degrees you can keep adding multiples of 360 degrees or in radians multiples of 2pi and ALL trig function will be unaltered.
sin(30) = sin(390) = sin(750) (where I keep adding 360 degrees).
tan(10) = tan(370) = tan(730), etc..
okay. post it as a separate question.

Answer 32

OKay thanks thats helpful

Answer 33

You are welcome. Close this question and post a new one.