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  1. #1
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    eulers

    The question is



    im not sure how to do this, i know that i=sqrt(-1) but im not sure how to compute this...

    thank you
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  2. #2
    MHF Contributor red_dog's Avatar
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    i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}

    z=\sqrt{i}=\cos\frac{\frac{\pi}{2}+2k\pi}{2}+i\sin  \frac{\frac{\pi}{2}+2k\pi}{2}, \ k=0,1

    k=0\Rightarrow z=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{\sqrt  {2}}{2}+\frac{\sqrt{2}}{2}i

    k=1\Rightarrow z=\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by acosta0809 View Post
    The question is



    im not sure how to do this, i know that i=sqrt(-1) but im not sure how to compute this...

    thank you
    You don't get a unique answer to this:

    a) \sqrt{\mathrm{i}}=\left(\mathrm{e}^{\mathrm{i}\lef  t(\frac{\pi}{2}+n\cdot 2\pi\right)}\right)^{1/2}= \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{4}+n\cdot\p  i\right)} where n\in \mathbb{Z}
    So, basically you have the two answers \sqrt{\mathrm{i}}= \mathrm{e}^{\mathrm{i}\frac{\pi}{4}} and \sqrt{\mathrm{i}}= \mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}
    Therefore, wirting an equality sign \sqrt{\mathrm{i}}=\ldots seems rather problematic. Rather, we would want to say that the equation z^2=\mathrm{i} has two solutions, z_1=\mathrm{e}^{\mathrm{i}\frac{\pi}{4}} and z_2=\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}.

    b) Just use r\cdot\mathrm{e}^{\mathrm{i\varphi}}=r\cdot\cos(\v  arphi)+\mathrm{i}r\cdot \sin(\varphi) for the above two solutions z_1, z_2.
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  4. #4
    Member alunw's Avatar
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    Complex square root without trig

    Quote Originally Posted by Failure View Post
    You don't get a unique answer to this:

    a) \sqrt{\mathrm{i}}=\left(\mathrm{e}^{\mathrm{i}\lef  t(\frac{\pi}{2}+n\cdot 2\pi\right)}\right)^{1/2}= \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{4}+n\cdot\p  i\right)} where n\in \mathbb{Z}
    So, basically you have the two answers \sqrt{\mathrm{i}}= \mathrm{e}^{\mathrm{i}\frac{\pi}{4}} and \sqrt{\mathrm{i}}= \mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}
    Therefore, writing an equality sign \sqrt{\mathrm{i}}=\ldots seems rather problematic. Rather, we would want to say that the equation z^2=\mathrm{i} has two solutions, z_1=\mathrm{e}^{\mathrm{i}\frac{\pi}{4}} and z_2=\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}.

    b) Just use r\cdot\mathrm{e}^{\mathrm{i\varphi}}=r\cdot\cos(\v  arphi)+\mathrm{i}r\cdot \sin(\varphi) for the above two solutions z_1, z_2.
    a)Almost every number has two distinct square roots. But \surd 2 conventionally denotes the positive solution to z^2=2. So the usual convention for square roots of complex numbers is to say that \surd x is the square root with a positive real part. If x is real and negative then \surd x is the root with positive imaginary part. This means there is a branch cut along the negative real axis.

    b) In fact you can find find square roots without resorting to trigonometry. The following pseudo-code, computes the square root (x,y) of the complex number (u,v)
    Code:
    if (u > 0) then
      x = sqrt((u+sqrt(u*u+v*v))/2)
      y = v/(2*x)
    else if (u < 0) then
      y = sign(v)*sqrt((-u+sqrt(u*u+v*v))/2)
      x = v/(2*y)
    else
      x = sqrt(abs(v)/2)
      if (v > 0)
        y = x
      else
        y = -x
    The point of testing the sign of u is to minimise round-off error.
    Last edited by alunw; August 7th 2009 at 11:47 PM. Reason: missing parenthesis in pseudo-code
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