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Math Help - fp3 complex number

  1. #1
    Member ssadi's Avatar
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    fp3 complex number

    The number 8 question of the exercise says:
    Use the formula
    cos (x+iy)=cos x cos iy-sin x sin iy
    to find two imaginary numbers whose cosine is 3
    How do I use the formula here? I mean, why can't I just substitute (x+iy) in the place of z in the formula cos z =1/2 * (e^(iz)-e^-(iz))?

    I noticed when I convert "cos x cos iy-sin x sin iy" in terms of e and simplify it leads to the same term as the substitution mentioned above, hence the question.
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  2. #2
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    Quote Originally Posted by ssadi View Post
    The number 8 question of the exercise says:
    Use the formula
    cos (x+iy)=cos x cos iy-sin x sin iy
    to find two imaginary numbers whose cosine is 3
    How do I use the formula here? I mean, why can't I just substitute (x+iy) in the place of z in the formula cos z =1/2 * (e^(iz)-e^-(iz))?

    I noticed when I convert "cos x cos iy-sin x sin iy" in terms of e and simplify it leads to the same term as the substitution mentioned above, hence the question.
    You want to solve \cos z = 3 set z=x+iy so \cos (x+iy) = 3 .
    Expand LHS and compare real and imaginary parts.
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