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Math Help - Simplifying an expression involving complex numbers.

  1. #1
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    Simplifying an expression involving complex numbers.

    The exercise is:
    write the expression in polar form: \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}
    Here is my work:
    \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}
    that is


    \frac{2(1+i)2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2  })(2\sqrt{3}-2i)}

    and that becomes:


    \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3  })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}


    and then


    \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3  })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}

    \frac{2\times 2 \times \sqrt{2}(cis(\frac{\pi}{4}))(cis(\frac{\pi}{3}))}{  3\times cis(\frac{\pi}{2})\times 2\times cis(\frac{11\pi}{12})}

    anyway the correct answer is -1 I get the absolute value wrong can anyone spot my error.?
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  2. #2
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    Quote Originally Posted by Henryt999 View Post
    The exercise is:
    write the expression in polar form: \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}
    Here is my work:
    \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}
    that is


    \frac{2(1+i)2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2  })(2\sqrt{3}-2i)}

    and that becomes:


    \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3  })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}


    and then


    \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3  })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}

    \frac{2\times 2 \times \sqrt{2}(cis(\frac{\pi}{4}))(cis(\frac{\pi}{3}))}{  3\times cis(\frac{\pi}{2})\times 2\times cis(\frac{11\pi}{12})}
    Here's your error. \sqrt{3}- i has argument -\pi/6 or 5\pi/6, not -\pi/12 or 11\pi/12.
    The right triangle formed by the lines from (0, 0) to (\sqrt{3}, -1), (0, 0) to \sqrt{3}, 0, and from (\sqrt{3}, 0) to (\sqrt{3}, -1) is half of an equilateral triangle. It vertex angle is 30 degrees or \pi/6 radians.

    anyway the correct answer is -1 I get the absolute value wrong can anyone spot my error.?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Here's your error. \sqrt{3}- i has argument -\pi/6 or 5\pi/6, not -\pi/12 or 11\pi/12.
    The right triangle formed by the lines from (0, 0) to (\sqrt{3}, -1), (0, 0) to \sqrt{3}, 0, and from (\sqrt{3}, 0) to (\sqrt{3}, -1) is half of an equilateral triangle. It vertex angle is 30 degrees or \pi/6 radians.
    And the modulus of \sqrt{3} - i has been forgotten in the denominator.
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