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    Hi guys,
    I'm assuming you all are very good at maths, so if you could please help me with this FP2 question as a break from STEP practice, it would be really appreciated:
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    Part 9e) is what I can't seem to get.
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    (Original post by tanyapotter)
    Hi guys,
    I'm assuming you all are very good at maths, so if you could please help me with this FP2 question as a break from STEP practice, it would be really appreciated:
    Name:  image.jpeg
Views: 48
Size:  462.9 KB

    Part 9e) is what I can't seem to get.
    Haha I got stuck on that part and posted it on TSR as well a while ago XD.
    I believe you're supposed to try expanding the brackets to start.
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    (Original post by IrrationalRoot)
    Haha I got stuck on that part and posted it on TSR as well a while ago XD.
    I believe you're supposed to try expanding the brackets to start.
    Yes, I've expanded the brackets but I'm really not sure how to proceed. The complex number P has not been specified so my expansion is a huge mess of unknown and known complex numbers in the form re^ix
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    (Original post by tanyapotter)
    Yes, I've expanded the brackets but I'm really not sure how to proceed. The complex number P has not been specified so my expansion is a huge mess of unknown and known complex numbers in the form re^ix
    I'll have to take a look at the question properly to work it out, but I remember realising that I overlooked something minor which made me unable get the solution.
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    (Original post by tanyapotter)
    Hi guys,
    I'm assuming you all are very good at maths
    Part 9e) is what I can't seem to get.
    This is quite inappropriate; don't hijack threads. I'm moving your post to a thread in the maths forum.
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    (Original post by tanyapotter)
    Yes, I've expanded the brackets but I'm really not sure how to proceed. The complex number P has not been specified so my expansion is a huge mess of unknown and known complex numbers in the form re^ix
    Yeah I've got it now. You expand the brackets and notice that factorising the different bits will give you symmetric sums of alpha.beta,gamma. Note that these are the roots of z^3-(2+2i) and you should get the result from there.
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    (Original post by Zacken)
    This is quite inappropriate; don't hijack threads. I'm moving your post to a thread in the maths forum.
    I already posted it in the A-levels forum and didn't get any replies :/ Sorry
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    (Original post by IrrationalRoot)
    Yeah I've got it now. You expand the brackets and notice that factorising the different bits will give you symmetric sums of alpha.beta,gamma. Note that these are the roots of z^3-(2+2i) and you should get the result from there.
    Thank you so much!!!!!
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    (Original post by tanyapotter)
    I already posted it in the A-levels forum and didn't get any replies :/ Sorry
    That's because it needs to be in the maths forum, I've sorted it out for you - just remember that in the future!
 
 
 
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