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Thread: square root of i

  1. #1
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    square root of i

    Find the square roots of i

    ?

    would the modulus = i ? and not sure how to work out its argument?

    any help apprecited
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  2. #2
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    Re: square root of i

    Quote Originally Posted by Tweety View Post
    Find the square roots of i ?
    Well of course as with any complex number there are two square roots:
    $\displaystyle \sqrt {|z|} \exp \left( {\frac{{\arg (z)i}}{2} + \pi ki} \right),~k=0,~1$
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  3. #3
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    Re: square root of i

    $\displaystyle e^{i\theta} = \cos \theta + i \sin \theta$

    Plug in $\displaystyle \dfrac{\pi}{2} + 2\pi n$:

    $\displaystyle e^{i \left(\pi/2 + 2\pi n\right)} = i$

    Take the square root of both sides:

    $\displaystyle e^{i\left(\pi/4 + \pi n\right)} = \sqrt{i}$

    There are two solutions in the range $\displaystyle [0,2\pi)$:

    $\displaystyle e^{\pi i/4},e^{5\pi i/4}$
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  4. #4
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    Re: square root of i

    why plug in pi/2 ?
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  5. #5
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    Re: square root of i

    why cant I just work it out like this, the above looks quite complicated,;.....


    let z = 0 + 1*i

    than the modulus of z = $\displaystyle \sqrt{0^{2} + 1^{2}} = 1 $

    arg(z) = $\displaystyle tan \alpa = 1 = \frac{\pi}{4}} $

    but not sure how to go from here, the correct answers are

    $\displaystyle \frac{1}{\sqrt{2}}(1+i) $

    $\displaystyle -\frac{1}{\sqrt{2}}(1+i) $
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  6. #6
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    Re: square root of i

    First, what made you think, originally, that there were four square roots of i?

    Second, while the modulus of z= i is 1, its argument is NOT [itex]\pi/4[/itex]. I don't know where you got that "1" in "$\displaystyle arctan(1)$". Surely you don't think argument is the arctangent of the modulus? The argument of "a+ bi" is the arctangent of b/a, if a is not 0. Of course, if a= 0 then a+ bi= bi which has modulus $\displaystyle \frac{\pi}{2}$ if b> 0, $\displaystyle -\frac{\pi}{2}$ if b< 0. (0 has no "argument").

    So if z= i, $\displaystyle arg(z)= \frac{\pi}{2}$. HALF of that is $\displaystyle \frac{\pi}{4}$ and half of $\displaystyle \frac{\pi}{2}+ 2\pi= \frac{5\pi}{2}$ is $\displaystyle \frac{5\pi}{4}$. Those are the arguments of the two square roots of i which, of course, have modulus $\displaystyle 1^{1/2}= 1$, so the two square roots are $\displaystyle \frac{\sqrt{2}}{2}(1+ i)$ and $\displaystyle \frac{\sqrt{2}}{2}(1- i)$.
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