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# De Moivres Question PleasE!!! Tweet

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IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
z^2 - z +2 - (1/z) + (1/(z^2)) = 4 cos ^2 theta -2cos theta

hence solve the quartic equation

z^4 - z^3 +2(z^2) - z + 1 = 0

2. Re: De Moivres Question PleasE!!!
can you solve the top equation in terms of cos theta? how does z relate to theta?

Notice how the two equations in z are related, can you make a nice factorisation?
3. Re: De Moivres Question PleasE!!!
(Original post by jj193)
can you solve the top equation in terms of cos theta? how does z relate to theta?

Notice how the two equations in z are related, can you make a nice factorisation?
I get stuck at I know that I can get thetha to be pi/2 or pi/3 but how do I proceed form there to get four values of z ?
4. Re: De Moivres Question PleasE!!!
you haven't solved for theta exhaustively if you want to consider this in polar co-ordinates with r>0

you should start slow
can zero be a solution? no
so you can express z as re^itheta r>0 zero in [0,2pi)
z=/=0 so has inverse
therefore

z^4 - z^3 +2(z^2) - z + 1 = 0 iff z^2 - z +2 - (1/z) + (1/(z^2)) =0

now in polar co-ordinates you know possible values of theta, can you figure out the values of r? (I honestly can't tell you how to do this atm) Maybe someone else will.
5. Re: De Moivres Question PleasE!!!
(Original post by member910132)
I get stuck at I know that I can get thetha to be pi/2 or pi/3 but how do I proceed form there to get four values of z ?

Alternatively,

Factorise a little more and you're done.

Using the original approach you need to use the fact that so that
6. Re: De Moivres Question PleasE!!!
(Original post by BabyMaths)
Alternatively,

Factorise a little more and you're done.

Using the original approach you need to use the fact that so that
Using the second approach after geting

what do we do ? Solve for theta and what ?

Edit: I think I got it now, after getting theta = pi/2 and pi/3 we just use to get the values of z right ?

My answers are z = \pm i , -2 and 1
Last edited by member910132; 28-05-2012 at 15:13.
7. Re: De Moivres Question PleasE!!!
(Original post by jj193)
you haven't solved for theta exhaustively if you want to consider this in polar co-ordinates with r>0

you should start slow
can zero be a solution? no
so you can express z as re^itheta r>0 zero in [0,2pi)
z=/=0 so has inverse
therefore

z^4 - z^3 +2(z^2) - z + 1 = 0 iff z^2 - z +2 - (1/z) + (1/(z^2)) =0

now in polar co-ordinates you know possible values of theta, can you figure out the values of r? (I honestly can't tell you how to do this atm) Maybe someone else will.
Sorry I don't understand this in the slightest way at all
8. Re: De Moivres Question PleasE!!!
(Original post by Taylor_R)
z^2 - z +2 - (1/z) + (1/(z^2)) = 4 cos ^2 theta -2cos theta

hence solve the quartic equation

z^4 - z^3 +2(z^2) - z + 1 = 0

What year is this from AQA right ?
9. Re: De Moivres Question PleasE!!!
(Original post by member910132)
My answers are z = \pm i , -2 and 1
You don't need to ask if you have the correct values. Sub in 1 for example and you get 1-1+2-1+1=2.