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Math Help - complex plane

  1. #1
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    complex plane

    say if i have (p) |z+2|=4 and (q) Re(z)=0 which are two different lines in the hyperbolic plane and i want to find the point of intersection p interset q how would i do this?
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  2. #2
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    Hyperbolic plane! You mean complex plane. Note |z+2|=4 is the circle (x+2)^2 + y^2 = 4 while \Re (z) = 0 is the line x=0.
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  3. #3
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    sorry im rubbish i meant hyperbolic lines, ok so if i take what you said would i just substitute x=0 to get the intersection. i get confused as to how you get the circle from |z+2|=4

    z=x+iy
    | x+iy + 2|=4 group real and imaginary parts then mod

    sqrt( ((x+2)^2) + (iy)^2)=4
    and so on..
    to get
    ((x+2)^2) - y^2 =16

    where am i going wrong
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  4. #4
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    Quote Originally Posted by skystar View Post
    sorry im rubbish i meant hyperbolic lines, ok so if i take what you said would i just substitute x=0 to get the intersection. i get confused as to how you get the circle from |z+2|=4

    z=x+iy
    | x+iy + 2|=4 group real and imaginary parts then mod

    sqrt( ((x+2)^2) + (iy)^2)=4 Mr F says: And here's the mistake. {\color{red}|a + ib| = \sqrt{a^2 + b^2}}, NOT {\color{red}\sqrt{a^2 + (ib)^2}}.
    and so on..
    to get
    ((x+2)^2) - y^2 =16

    where am i going wrong
    |z + 2| = 4 <=> |z - (-2)| = 4. So the geometric interpretation is the set of all points in the complex plane that are a distance 4 from z = -2. This is a circle of radius 4 and centre at z = -2, that is, (-2, 0).
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