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FP2 complex numbers question?

I'm having a bit of a mind blank, so if anyone could point me in the right direction with this question it would be greatly appreciated!

'Evaluate (cos pi/3 + i sin pi/3)^-4 in the form a+ib

Cheers!

I think I've just forgotten something basic as I've left complex numbers alone for a while :smile:
Original post by iMacJack
I'm having a bit of a mind blank, so if anyone could point me in the right direction with this question it would be greatly appreciated!

'Evaluate (cos pi/3 + i sin pi/3)^-4 in the form a+ib

Cheers!

I think I've just forgotten something basic as I've left complex numbers alone for a while :smile:


cos(x)+isin(x)=eix\cos(x)+i\sin(x)=e^{ix}

or just De Moivre's before evaluating the cosine/sine of the resulting angle.
(edited 7 years ago)
Original post by iMacJack
I'm having a bit of a mind blank, so if anyone could point me in the right direction with this question it would be greatly appreciated!

'Evaluate (cos pi/3 + i sin pi/3)^-4 in the form a+ib

Cheers!

I think I've just forgotten something basic as I've left complex numbers alone for a while :smile:


Use De Moivre's theorem. [r(cosθ+isinθ)]n=rn[cos(nθ)+isin(nθ)] [r(cos\theta +isin\theta)]^n = r^n[cos(n\theta) +isin(n\theta)], the argument of the trig functions can be evaluated in a calculator.
Reply 3
Original post by NotNotBatman
The argument of the trig functions can be evaluated in a calculator.


No need for a calculator. :tongue:
Original post by Zacken
No need for a calculator. :tongue:


I know, but at A level, I think most people don't remember trig values or bother with the triangles.
Reply 5
I got -1/2 + root3/2 i, thanks a lot! (I don't know if this is right because it's a zigzag sheet which I don't have the answers for)

Thanks :smile:
Original post by iMacJack
I got -1/2 + root3/2 i, thanks a lot! (I don't know if this is right because it's a zigzag sheet which I don't have the answers for)

Thanks :smile:


Correct.
Reply 7
Original post by RDKGames
cos(x)+isin(x)=eix\cos(x)+i\sin(x)=e^{ix}

or just De Moivre's before evaluating the cosine/sine of the resulting angle.


Original post by NotNotBatman
Use De Moivre's theorem. [r(cosθ+isinθ)]n=rn[cos(nθ)+isin(nθ)] [r(cos\theta +isin\theta)]^n = r^n[cos(n\theta) +isin(n\theta)], the argument of the trig functions can be evaluated in a calculator.


Original post by Zacken
No need for a calculator. :tongue:


Hey guys - thanks for the help earlier, I've a little question now however -

part A was show that z^n + z^-n = 2cosn theta and same for z^n - z^-n = 2isinN theta, that was no problems, part b is 'Hence deduce that cos^6 theta + sin^6 theta = 1/8(3cos4theta + 5)

I was going about my way of using the formulae for them but I, for whatever reason, don't seem to come up with what they want

Cheers guys!
Original post by iMacJack
Hey guys - thanks for the help earlier, I've a little question now however -

part A was show that z^n + z^-n = 2cosn theta and same for z^n - z^-n = 2isinN theta, that was no problems, part b is 'Hence deduce that cos^6 theta + sin^6 theta = 1/8(3cos4theta + 5)

I was going about my way of using the formulae for them but I, for whatever reason, don't seem to come up with what they want

Cheers guys!


Use the fact that z+z1=2cosθz+z^{-1}=2\cos\theta hence cos6θ=164(2cosθ)6=164(z+z1)6\cos^{6}\theta=\frac{1}{64}\left(2\cos \theta \right)^{6}=\frac{1}{64}(z+z^{-1})^{6}
A similar trick can be done with sine. You can then reduce it down as there will be some cancelling apart from the constant and (z3+z3)(z^{3}+z^{-3}) part of it.
(edited 7 years ago)
Reply 9
Original post by Cryptokyo
Use the fact that z+z1=2cosθz+z^{-1}=2\cos\theta hence cos6θ=164(2cosθ)6=164(z+z1)6\cos^{6}\theta=\frac{1}{64}\left(2\cos \theta \right)^{6}=\frac{1}{64}(z+z^{-1})^{6}
A similar trick can be done with sine. You can then reduce it down as there will be some cancelling apart from the constant and (z3+z3)(z^{3}+z^{-3}) part of it.


I'm just not seeming to get it right :frown:
Original post by iMacJack
I'm just not seeming to get it right :frown:


The expansion cos6θ=164(2cosθ)6=164(z+z1)6\cos^{6}\theta=\frac{1}{64}\left(2\cos \theta \right)^{6}=\frac{1}{64}\left(z+z^{-1}\right)^{6}
=164(z6+6z4+15z2+20+15z2+6z4+z6)=\frac{1}{64}\left(z^{6}+6z^{4}+15z^{2}+20+15z^{-2}+6z^{-4}+z^{-6}\right)
The expansion sin6θ=164(2isinθ)6=164(zz1)6\sin^{6}\theta=\frac{1}{64}\left(2i\sin \theta \right)^{6}=-\frac{1}{64}\left(z-z^{-1}\right)^{6}
=164(z66z4+15z220+15z26z4+z6)=\frac{1}{64}\left(z^{6}-6z^{4}+15z^{2}-20+15z^{-2}-6z^{-4}+z^{-6}\right)

By adding the two expansions you obtain:
164(12z4+12z4+40)\frac{1}{64}\left(12z^{4}+12z^{-4}+40\right)

You should be able to do it from here.

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