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    Let's try making a community MS again.
    Please PM me to be given editor permissions.

    https://onedrive.live.com/redir?resi...nt=file%2cdocx
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    Thankfully I don't think that paper was too bad. (Nothing like the horrendous M2 from Wednesday). I'll have dropped a mark for sure on the intersection points of the bisector and the circle on the argand diagram question because I couldn't remember how to do it, but everything else was fine.

    (Original post by Gander01)
    The roots of the z after the substitution were 9+40i and 9-40i I believe, so you needed to use both other parts of the question to get a solution. Mine did involve some nasty fractions though, something over 41 I think, so I doubt I got it fully correct.
    I can't remember my exact answer, but got a similar fraction, so that sounds right to me.
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    Posting the questions I remember - Starting to forget them already, might leave it at this for now, hopefully it helps at least.

    Question 1: Find The sum from r=1 to n of (3r+1)(r-1) in terms of n

    Question 2 i: Find z given arg(z) = - (1/3) pi and mod z = 2 sqrt(3) (unsure about this one, not sure I remembered the modulus correctly)
    ii: 1/(z*-5i)^2

    Question 3: This was some simple matrix arithmetic, I can't remember the values in the matrices so if anyone can that would be appreciated. There were 3 matrices, A and B being 1 by 3 and C being 3 by 1. The first part was something like 2A+3B

    Question 4: Find the square roots of 9+40i, and show that it is a root of x^2+18x+1681=0, hence find the roots of 1+18u+1681u^2
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    It was awful

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    I think I made so many stupid mistakes
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    Question 6 was the most confusing for me.
    I think I got |z - 3 - 3i| = 5 for the equation, and 1+4i and 5+2i for the point of intersections.
    Did anyone else get those answers?
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    (Original post by SonicTails19)
    Question 6 was the most confusing for me.
    I think I got |z - 3 - 3i| = 5 for the equation, and 1+4i and 5+2i for the point of intersections.
    Did anyone else get those answers?
    I got root 5 as radius
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    Did everyone get q is reflection in the line y=-x
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    (Original post by Bobby21231)
    I got root 5 as radius
    Correct it was root5 radius because the diameter was 2root5
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    (Original post by Bobby21231)
    Did everyone get q is reflection in the line y=-x
    Yes that's what I got.
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    (Original post by Bobby21231)
    Did everyone get q is reflection in the line y=-x
    Pretty sure that's what I put
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    (Original post by Mathematicus65)
    Yes that's what I got.
    (Original post by Stylesw)
    Pretty sure that's what I put
    Yay I might get some marks
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    One of the easiest papers i think for 90 UMS it will be about 110%
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    (Original post by SonicTails19)
    Question 6 was the most confusing for me.
    I think I got |z - 3 - 3i| = 5 for the equation, and 1+4i and 5+2i for the point of intersections.
    Did anyone else get those answers?
    First part I got |z-(3+3i)|=sqrt(5) for. That's the question which I struggled on though, so don't take my word for it.

    IIRC the question was that A= 1+2i and B= 5+4i so that AB is the diameter of a circle.

    I got the center as (5+1)/2, (4+2)/2 = 3,3 and the radius as sqrt((3-1)^2+(3-2)^2))
    Hence |z-(3+3i)|=sqrt(5)

    I couldn't do the intersection of the bisector though.
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    I think for 6 it was actually 1+18u^2+1681u^4, I also can't remember if the 18 was a plus or minus

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    (Original post by sdhand)
    First part I got |z-(3+3i)|=sqrt(5) for. That's the question which I struggled on though, so don't take my word for it.

    IIRC the question was that A= 1+2i and B= 5+4i so that AB is the diameter of a circle.

    I got the center as (5+1)/2, (4+2)/2 = 3,3 and the radius as sqrt((3-1)^2+(3-2)^2))
    Hence |z-(3+3i)|=sqrt(5)

    I couldn't do the intersection of the bisector though.
    Got the same for that, remember my two points of intersection were 2root 5 apart from one another and so I think they were correct, 4+i was one of them I think


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    Is it just me or will those boundaries be disgustingly high?
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    It was plus, just checked on wolfram alpha and this is what I put so I'm fairly certain the answers were +/-((4/41)+(5/41)i) and +/-((4/41)-(5/41)i)

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    Tbh the paper was relatively harder so I reckon grade boundaries will be about 57/72. Also, How many marks do I lose for not rationalizing my answers to the very last question?
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    (Original post by airborne454)
    Tbh the paper was relatively harder so I reckon grade boundaries will be about 57/72. Also, How many marks do I lose for not rationalizing my answers to the very last question?
    Hopefully they will be like this because I probably made so many stupid mistakes
 
 
 
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