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    Question 1 (part d)
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    So it says 'Given that beta is real...' but I didn't really see how that would actually help me? It isn't going to be the complex conjugate of alpha since the polynomial has complex coefficients.

    My answer to Question 1 (part d)
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    So it made sense to me to use the quadratic formula, I didn't know what else to do.
    But as you can see I end up getting two roots which appear to be imaginary. I tried to factorise and stuff to see if I could eliminate i somehow but it didn't work, so I entered them into my graphical calculator and it somehow gave me a real and imaginary root. This is where my confusion lies, as I can't see how i-sqrt(2i) is -1. Without my calculator I would have lost 2 marks lol.

    Question 1 mark scheme
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    I didn't even think of substituting in values for f(x) to find a root, surely I'd only bother doing that for cubic equations where there isn't a given equation to find the roots? But yeah anyway if I had done it that way I would have been able to figure out, finding something to multiply with (z+1) to obtain the given quadratic. Maybe they just expected me to find that 1 was a root purely by inspection?
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    Question 2 (part bii)
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    So my first thought was simply that it is equal to 1/2isinx, the result from part bi. I went from there and tried to somehow get to the result they give but I just couldn't at all, I don't think I even got close.

    I then gave up with that and tried to substitute the w's for e^ix and work from there, but I just got into a bit of a mess with that and didn't get a solution.

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    They did what I tried at first, but as you can see they just take a big leap from 1/2isinx to -i/2sinx. I just can't see what they did to get there, I'm probably just overlooking something really simple as it was only a 2 mark question.
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    They simply rationalized the denominator
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    (Original post by darkforest)
    Question 1 (part d)
    You can spot that z=-1 is a root, otherwise, the fact that \beta is real is important but not so much here. Alternative approach is the following:

    \beta^2 -2i \beta  - (1+2i) = 0 which is the same as (\beta^2-1) -2(\beta + 1)i = 0 which is only true if both real/im parts are 0, and obviously from the im part, \beta = -1 which can be checked with the real part as well.

    Question 2 (part bii)
    Use bi) and reciprocate both sides. Multiply top and bottom by i
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    (Original post by RDKGames)
    You can spot that z=-1 is a root, otherwise, the fact that \beta is real is important but not so much here. Alternative approach is the following:

    \beta^2 -2i \beta  - (1+2i) = 0 which is the same as (\beta^2-1) -2(\beta + 1)i = 0 which is only true if both real/im parts are 0, and obviously from the im part, \beta = -1 which can be checked with the real part as well.



    Use bi) and reciprocate both sides. Multiply top and bottom by i
    Yeah I just realised shortly after posting that I could use the fact that \beta + \alpha = 2i, with \beta represented by a+bi and \alpha by c. Doing the same with the product of the roots just leaves you to easily find a, b and c by comparing real and Im parts.

    For whatever reason I didn't think of that, despite messing around with the products and roots of the cubic to try and find them...

    ---

    Also, wow. Can't believe I didn't think of that to try and eliminate i from the denominator. Grrr.

    Thank you very much for the help, I'll be sure to look out for stuff like this next time.
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    (Original post by darkforest)
    Yeah I just realised shortly after posting that I could use the fact that \beta + \alpha = 2i, with \beta represented by a+bi and \alpha by c. Doing the same with the product of the roots just leaves you to easily find a, b and c by comparing real and Im parts.

    For whatever reason I didn't think of that, despite messing around with the products and roots of the cubic to try and find them...
    Yeah, though I think you meant to say \beta + \gamma = 2i then let \gamma = a+bi and \beta = c so (a+c)+bi = 2i and from product ac+bci = -(1+2i)

    There are a number of ways to do this question, I'd stick with my method as it's the nicest - no need for introducing new constants and solving for them tbh.

    As for the denominator with i, it is very common to get rid off it from there in many questions to a point where you should be able to recognise it and do it as naturally as rationalising the denominator due to some surd.
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    (Original post by RDKGames)
    Yeah, though I think you meant to say \beta + \gamma = 2i then let \gamma = a+bi and \beta = c so (a+c)+bi = 2i and from product ac+bci = -(1+2i)

    There are a number of ways to do this question, I'd stick with my method as it's the nicest - no need for introducing new constants and solving for them tbh.

    As for the denominator with i, it is very common to get rid off it from there in many questions to a point where you should be able to recognise it and do it as naturally as rationalising the denominator due to some surd.
    Ah yes, what you said.

    This was my second paper so I'm just familiarising myself with the intricacies and stuff so far. I'm aiming for an A* in FM so I can't be overlooking stuff like this in the actual exam.
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    (Original post by Radioactivedecay)
    They simply rationalized the denominator
    Hmm I didn't seem to get anywhere with that, but I instead tried to find the square root of 2i, which when squared and subbed into sqrt(2i) managed to do the trick.

    Edit: Oops I think you were talking about the second question.
 
 
 
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