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Math Help - 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:
    \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

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

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

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

    Take the square root of both sides:

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

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

    e^{\pi i/4},e^{5\pi i/4}
    Thanks from Tweety
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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 =  \sqrt{0^{2} + 1^{2}} = 1

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

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

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

      -\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 " 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 \frac{\pi}{2} if b> 0, -\frac{\pi}{2} if b< 0. (0 has no "argument").

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