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Thread: HSC 2018 MX2 Marathon

  1. #26
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by pikachu975 View Post
    So let z = x + 2i

    (x+2i)^2 = x^2 + 4xi - 4
    Since it is real then Im(z^2) = 0
    so 4x = 0
    x = 0

    Hence z = 2i
    thank you Pikachu!!

  2. #27
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by pikachu975 View Post
    So let z = x + 2i

    (x+2i)^2 = x^2 + 4xi - 4
    Since it is real then Im(z^2) = 0
    so 4x = 0
    x = 0

    Hence z = 2i
    how about if re(z) is 2Im(z) and z^2-4i is real

    I cant get the answer.. the answer is +- (2+i)

  3. #28
    Junior Member 1729's Avatar
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by sssona09 View Post
    how about if re(z) is 2Im(z) and z^2-4i is real

    I cant get the answer.. the answer is +- (2+i)
    Last edited by 1729; 8 Nov 2017 at 10:09 PM.
    sssona09 likes this.

  4. #29
    -insert title here- Paradoxica's Avatar
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    Re: HSC 2018 MX2 Marathon

    Prove the ellipses:



    where k is an arbitrary real number

    have the same area as the ellipse

    Last edited by Paradoxica; 20 Nov 2017 at 12:53 AM.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  5. #30
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Paradoxica View Post
    Prove the ellipses:



    have the same area as the ellipse

    Continuing on, prove that the area of the ellipse defined by



    has area


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