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    The complex number: a is given that:
    a = 2((cos(pi/8) +sin(pi/8))

    on your diagram, draw and label:
    (z - a) = 6

    the locus of points:
    arg(z - a) = (5pi)/8

    the first locus is just a circle, centre at z, with radius 6, right?

    Need help, espeically with the last bit. Is the last question a shaded regiion in and around the line from the origin to complex number a that stretches on forever? which is 5pi/6 distance away from the line than goes trough a and onwards, and since it's 5pi/8 distance away from every point on that infinitely long line it's really just a shaded region? if that made sense. help please lol.
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    Ermm... z is the variable so it should be a circle, centre a

    arg(z-a) is the angle between \overline{AZ} and the origin. You need to make that angle \frac{5 \pi}{8}. Therefore it will be a half line (I'll leave you to figure out where and at what angle)

    Also is a = 2 ( \cos \frac{\pi}{8} + i \sin \frac{\pi}{8} )?
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    yeah, it is
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    Still a bit confused. Is it, centre A, a circle that goes from arga to 5pi/8, so it's not a full circle, but one that goes to that angle then stops?
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    (Original post by Tallon)
    Still a bit confused. Is it, centre A, a circle that goes from arga to 5pi/8, so it's not a full circle, but one that goes to that angle then stops?
    |z-a|=6

    The locus of z is a circle centre a radius 6.

    This is because z-a is the line from a to z and the modulus signs indicate that the length of this line is 6.

    The second part is a completely separate question
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    Tallon,

    If you want to understand the circle centre, consider this:

    z=x+yi

    a=p+qi

    |z-a|=6

    |(x-p)+(y-q)i|=6

    \sqrt{(x-p)^2+(y-q)^2}=6

    (x-p)^2+(y-q)^2=6^2

    This is the circle centre a radius 6.
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    Mr M, a is not a real number. He made a mistake in the first post
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    (Original post by SimonM)
    Mr M, a is not a real number. He made a mistake in the first post
    I wondered why he was finding it difficult.

    My earlier post is now corrected in the light of this information.
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    I know the second part is a completely seperate question. that's the bit I'm stuck on lol
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    (Original post by Tallon)
    I know the second part is a completely seperate question. that's the bit I'm stuck on lol
    Private message me with an email address and I will send you a Powerpoint that will illuminate the matter for you.
 
 
 

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