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Math Help - Another Complex Number Question

  1. #1
    Senior Member Stroodle's Avatar
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    Another Complex Number Question

    If z=x+yi determine the values of x and y such that z=\sqrt{3+4i}
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  2. #2
    MHF Contributor red_dog's Avatar
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    3+4i=4+4i-1=4+4i+i^2=(2+i)^2

    Then z_1=2+i, \ z_2=-2-i
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  3. #3
    Senior Member Stroodle's Avatar
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    Cool. Thanks for your help.
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  4. #4
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    mr fantastic's Avatar
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    Quote Originally Posted by Stroodle View Post
    If z=x+yi determine the values of x and y such that z=\sqrt{3+4i}
    Another approach:

    z^2 = 3 + 4i

    \Rightarrow (x + iy)^2 = 3 + 4i

    \Rightarrow (x^2 - y^2) + 2xy i = 3 + 4i.

    Therefore:

    x^2 - y^2 = 3 .... (1)

    2xy = 4 \Rightarrow xy = 2 \Rightarrow y = \frac{2}{x} .... (2)

    Substitute (2) into (1): x^2 - \frac{4}{x^2} = 3 \Rightarrow x^4 - 3x^2 - 4 = 0 \Rightarrow (x^2 - 4)(x^2 + 1) = 0.

    But since x is real, x = \pm 2.

    x = 2: y = 1.
    x = -2: y = -1.
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  5. #5
    Senior Member Stroodle's Avatar
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    Awesome. Thanks for that.
    This method will be much easier for me to reproduce for similar problems.
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