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Math Help - Complex logarithm question

  1. #1
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    Complex logarithm question

    The question
    Find the cartesian form of the principle value of:

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

    My attempt
    e^{(1 - i)Log(e^{(\frac{\pi i}{3})}^{-3})
    e^{(1 - i)Log(e^{-\pi i}})
    e^{(1 - i)(ln|1| + i(-\pi))} //uh oh, this is on a branch cut!

    How do I proceed, if the principle Log is not defined on the negative real axis?

    Thank you.
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  2. #2
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    Re: Complex logarithm question

    Not sure there's a problem here. The only place Log isn't defined is at z=0.
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  3. #3
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    Re: Complex logarithm question

    Quote Originally Posted by Glitch View Post
    The question
    Find the cartesian form of the principle value of:

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

    My attempt
    e^{(1 - i)Log(e^{(\frac{\pi i}{3})}^{-3})
    e^{(1 - i)Log(e^{-\pi i}})
    e^{(1 - i)(ln|1| + i(-\pi))} //uh oh, this is on a branch cut!

    How do I proceed, if the principle Log is not defined on the negative real axis?

    Thank you.
    \displaystyle \left[\left(\frac{1 + i\sqrt{3}}{2}\right)^{-3}\right]^{1 - i} = \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i}


    \displaystyle \left|\frac{1 + i\sqrt{3}}{2}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1


    \displaystyle \arg{\left(\frac{1 + i\sqrt{3}}{2}\right)} = \arctan{\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2  }}\right)} = \arctan{\left(\sqrt{3}\right)} = \frac{\pi}{3}


    \displaystyle \begin{align*} \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i} &= \left(e^{\frac{\pi i}{3}}\right)^{-3 + 3i} \\ &= e^{(-1 + i)\pi i} \\ &= e^{-\pi - \pi i} \\ &= e^{-\pi}\cos{(-\pi)} + i\,e^{-\pi}\sin{(-\pi)} \\ &= -e^{-\pi}  \end{align*}
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    Re: Complex logarithm question

    Thanks, that's the answer I eventually got. However the book answer is -e^{\pi}. I think they just added 2\pi to the principle argument (not sure why).
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    Re: Complex logarithm question

    Quote Originally Posted by Glitch View Post
    Thanks, that's the answer I eventually got. However the book answer is -e^{\pi}. I think they just added 2\pi to the principle argument (not sure why).
    It's because the principal argument is in the domain \displaystyle (-\pi, \pi].
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    Re: Complex logarithm question

    Quote Originally Posted by Glitch View Post
    However the book answer is -e^{\pi}. I think they just added 2\pi to the principle argument (not sure why).
    I think the answer to that must be that they wanted you to evaluate the expression in the order indicated by the bracketing:

    \Bigl[\Bigl(\frac{1 + i\sqrt{3}}{2}\Bigr)^{-3}\Bigr]^{1 - i} = \left[(e^{i\pi/3})^{-3}\right]^{1-i} = (e^{-i\pi})^{1-i}.

    At that stage, they want you to replace e^{-i\pi} by its principal value e^{i\pi} before raising it to the power 1-i. Then you get (e^{i\pi})^{1-i} = e^{(i+1)\pi} = -e^\pi.
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    Re: Complex logarithm question

    Quote Originally Posted by Prove It View Post
    \displaystyle \left[\left(\frac{1 + i\sqrt{3}}{2}\right)^{-3}\right]^{1 - i} = \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i}


    \displaystyle \left|\frac{1 + i\sqrt{3}}{2}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1


    \displaystyle \arg{\left(\frac{1 + i\sqrt{3}}{2}\right)} = \arctan{\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2  }}\right)} = \arctan{\left(\sqrt{3}\right)} = \frac{\pi}{3}


    \displaystyle \begin{align*} \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i} &= \left(e^{\frac{\pi i}{3}}\right)^{-3 + 3i} \\ &= e^{(-1 + i)\pi i} \\ &= e^{-\pi - \pi i} \\ &= e^{-\pi}\cos{(-\pi)} + i\,e^{-\pi}\sin{(-\pi)} \\ &= -e^{-\pi} \end{align*}
    You got to be careful here; ({z^\alpha})^\beta \ne z^{\alpha \beta} in general.
    Last edited by ojones; June 15th 2011 at 07:07 PM.
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    Re: Complex logarithm question

    Quote Originally Posted by Glitch View Post
    Thanks, that's the answer I eventually got. However the book answer is -e^{\pi}. I think they just added 2\pi to the principle argument (not sure why).
    You can't just add stuff. Check what principal argument range the book is using.
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