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Math Help - limits in the complex domain

  1. #1
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    limits in the complex domain

    Show that <br />
\lim_{n \to \infty}(i)^\frac{1}{n}=1<br />
where (i)^\frac{1}{n} is the principal value of the nth root of i.
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  2. #2
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    Quote Originally Posted by splash View Post
    Show that <br />
\lim_{n \to \infty}(i)^\frac{1}{n}=1<br />
where (i)^\frac{1}{n} is the principal value of the nth root of i.
    I have no idea what to do, so if I'm wrong, than just ignore me. But I think this might help:

    \lim_{n \to \infty}\frac{1}{n}=0

    Thus \lim_{n \to \infty}i^{\frac{1}{n}}=i^0=1
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  3. #3
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    Quote Originally Posted by splash View Post
    Show that <br />
\lim_{n \to \infty}(i)^\frac{1}{n}=1<br />
where (i)^\frac{1}{n} is the principal value of the nth root of i.
    I am not sure what you mean by i^{1/n}.

    But given the equation,
    x^n=i=\cos \pi/2 +i\sin \pi/2
    Then, (by de Moiver's theorem)
    x=\cos \left( \frac{\pi}{2n}+\frac{2\pi k}{n} \right)+i\sin \left( \frac{\pi}{2n}+\frac{2\pi k}{n} \right) For k=0,1,2..,n-1
    I assume by prinicpal you mean k=0
    In that case,
    i^{1/n}=\cos \pi/2n +i\sin \pi/2n
    Thus, n\to \infty
    \cos 0+i\sin 0=1
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