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    (Original post by thers)
    This has got to be the worst a level maths question I have seen lmfao. I scored 1/14 on it.

    http://filestore.aqa.org.uk/subjects...2-QP-JAN12.PDF

    Question 8. I can't make any progress past part (a). Can anyone explain what to do at all?
    The pdf looks little weird tbh.

    Does  z = e^{i \theta} ?
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    (Original post by thers)
    This has got to be the worst a level maths question I have seen lmfao. I scored 1/14 on it.

    http://filestore.aqa.org.uk/subjects...2-QP-JAN12.PDF

    Question 8. I can't make any progress past part (a). Can anyone explain what to do at all?
    I've not actually checked, but I imagine that the roots of \ z^5 - 1 =0 are the same as the roots of z^4+z^3+z^2+z+1=0
    given that it says hence
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    (Original post by joostan)
    I've not actually checked, but I imagine that the roots of \ z^5 - 1 =0 are the same as the roots of z^4+z^3+z^2+z+1=0
    given that it says hence
    They are apparently but why?
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    (Original post by thers)
    They are apparently but why?
    Because \dfrac{z^5-1}{z-1} = z^4 + z^3+z^2+z+1
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    (Original post by justinawe)
    No.

    \log (28x) - \log (9) = \log \left( \dfrac{28x}{9} \right), using basic log rules.

    So definitely not the same thing.
    Yeah silly me

    Thanks a lot
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    (Original post by joostan)
    Because \dfrac{z^5-1}{z-1} = z^4 + z^3+z^2+z+1
    Right, do you have any ideas for the next 2 parts?
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    (Original post by joostan)
    Because \dfrac{z^5-1}{z-1} = z^4 + z^3+z^2+z+1
    Actually I've figured out how to do (d) by using the given answer in (c). But can't do (c)
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    (Original post by thers)
    Right, do you have any ideas for the next 2 parts?
    You know (by de Moivre) that e^{i\theta} + e^{-i\theta} = 2cos(\theta) How does this help, do you think?
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    Lost 3 marks on Jan 2013 on this...

    I think I know why, but the mark scheme has no workings, just the values

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    Do I work out the gradient and make it negative to find a?
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    M5, June 2009. 99 UMS.

    This is what my life has become.
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    (Original post by joostan)
    You know (by de Moivre) that e^{i\theta} + e^{-i\theta} = 2cos(\theta) How does this help, do you think?
    So I've got z^2 + z^-2 + 2cos2pi/5 + 1 = 0
    What next?
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    (Original post by DJMayes)
    M5, June 2009. 99 UMS.

    This is what my life has become.
    Arithmetic errors?
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    (Original post by L'Evil Fish)
    Lost 3 marks on Jan 2013 on this...

    I think I know why, but the mark scheme has no workings, just the values

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Views: 58
Size:  28.6 KB

    Do I work out the gradient and make it negative to find a?
    Yep
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    (Original post by joostan)
    Yep
    Darn it!

    Lets hope I can smash Friday's exam now:cool: lets see what UMS I'd have had..

    2012: 100 UMS
    2013: 97 UMS

    I want 97 (birth year :cool:)
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    (Original post by thers)
    So I've got z^2 + z^-2 + 2cos2pi/5 + 1 = 0
    What next?
    What you're trying to do is factorise the quartic into 2 quadratics, then divide by z^2 to get the desired result
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    (Original post by L'Evil Fish)
    Darn it!

    Lets hope I can smash Friday's exam now:cool: lets see what UMS I'd have had..

    2012: 100 UMS
    2013: 97 UMS

    I want 97 (birth year :cool:)
    100 is a bit showy - same goes for you DJ
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    (Original post by joostan)
    100 is a bit showy - same goes for you DJ
    Showy? I want 97/98... So it looks nice... I'd love a 100 though, in one of C3/4 but not gonna happen:mmm:
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    (Original post by Boy_wonder_95)
    Arithmetic errors?
    Not even arithmetic, even worse. On a vectors question on systems of forces I forgot to multiply by -1 so forces would cancel out, -2 accuracy marks. Then, on a moments of inertia question with finding the force on a hinge, an absolute monster worth 9 marks, I did it all completely correctly, managed to avoid the 1001 different places I could have made an arithmetical error, but forgot to add an mg on the end of 1 of two things when expanding them out from brackets in the final line, so minus another mark. :facepalm:
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    (Original post by joostan)
    What you're trying to do is factorise the quartic into 2 quadratics, then divide by z^2 to get the desired result
    Right so I've got a quartic

    z^4 + (2cos(2pi/5) + 1 )z^2 + 1 =0

    How am I supposed to factorize this?
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    (Original post by thers)
    Right so I've got a quartic

    z^4 + (2cos(2pi/5) + 1 )z^2 + 1 =0

    How am I supposed to factorize this?
    Well you know the linear factors, you multiply two linear dactors together and you get a quadratic, you just need to pick which linear factors you want to mulitply to get the desired result.
 
 
 
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