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# what did i do wrong this time? FP2 'de moivre theorem' watch

1. Use de Moivre's theorem to evaluate (-1+i)^8

my working:

r = sqrt(2)
arg(-1+i) = -pi/4

therefore (-1+i)^8 = [sqrt(2)[cos(pi/4) - isin(pi/4)]]^8

using this dudes theorem: z^n = r^n (cos(nx) + isin(nx)

16(cos2-isin2) is what i got. expanded to get 16cos2 - 16isin2.

the answer is just 16 apparently. anyone know where i went wrong?

2. You forgot to include the pi value in the cosine and sine function
3. (Original post by Reml)
You forgot to include the pi value in the cosine and sine function
oh...

thx
4. There's an easier way to do this.
Any complex number can be written as Re^(i(theta)) where R is magnitude (in this case sqrt(2)) and theta is the argument (angle here is -pi/4) so that to the power 8 is 16e^(-2pi) and since 2 pi is one complete revolution its the same as an argument of 0 and e^0 = 1
5. (Original post by disco1000)
There's an easier way to do this.
Any complex number can be written as Re^(i(theta)) where R is magnitude (in this case sqrt(2)) and theta is the argument (angle here is -pi/4) so that to the power 8 is 16e^(-2pi) and since 2 pi is one complete revolution its the same as an argument of 0 and e^0 = 1
oh yeah, nice..

speaking of Re^(ix), can you help me out with this question?

(a) Express 4-4i in the form r(cosx + isinx) where r>0, -pi < x < pi, and r and x are exact values.

4sqrt(2) [rcos(pi/4) - isin(pi/4)]

(b) Hence, or otherwise solve the equation z^5 = 4 -4i leaving your answers in the form z = Re^ik(pi), where R is the modulus of z and k is a rational number such that -1 =< k =< 1.

mind giving me a tutorial on how to do this? would be much appreciated
6. (Original post by racist mathematici)
oh yeah, nice..

speaking of Re^(ix), can you help me out with this question?

(a) Express 4-4i in the form r(cosx + isinx) where r>0, -pi < x < pi, and r and x are exact values.

4sqrt(2) [rcos(pi/4) - isin(pi/4)]

(b) Hence, or otherwise solve the equation z^5 = 4 -4i leaving your answers in the form z = Re^ik(pi), where R is the modulus of z and k is a rational number such that -1 =< k =< 1.

mind giving me a tutorial on how to do this? would be much appreciated
It's basically finding the nth root of a complex number, using de Moivre's theorem.
7. (Original post by racist mathematici)
oh yeah, nice..

speaking of Re^(ix), can you help me out with this question?

(a) Express 4-4i in the form r(cosx + isinx) where r>0, -pi < x < pi, and r and x are exact values.

4sqrt(2) [rcos(pi/4) - isin(pi/4)]

(b) Hence, or otherwise solve the equation z^5 = 4 -4i leaving your answers in the form z = Re^ik(pi), where R is the modulus of z and k is a rational number such that -1 =< k =< 1.

mind giving me a tutorial on how to do this? would be much appreciated
Practically the same question, this time it becomes 4sqrt(2)*e^(-pi/4) which is 2^(5/2)*e^(-pi/4) now you take the 5th root which is just sqrt 2 * e (-pi/20) where k is - 1/20 + 2n(pi) for any value of n

hope that helps, its kind of hard to explain it all just written down, check the proof for why e^i.theta = cos theta + isin theta, it will help.

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