Show that $\displaystyle

\lim_{n \to \infty}(i)^\frac{1}{n}=1

$ where $\displaystyle (i)^\frac{1}{n}$ is the principal value of the nth root of $\displaystyle i$.

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- Nov 2nd 2006, 05:37 PMsplashlimits in the complex domain
Show that $\displaystyle

\lim_{n \to \infty}(i)^\frac{1}{n}=1

$ where $\displaystyle (i)^\frac{1}{n}$ is the principal value of the nth root of $\displaystyle i$. - Nov 2nd 2006, 05:58 PMQuick
- Nov 2nd 2006, 07:01 PMThePerfectHacker
I am not sure what you mean by $\displaystyle i^{1/n}$.

But given the equation,

$\displaystyle x^n=i=\cos \pi/2 +i\sin \pi/2$

Then, (by de Moiver's theorem)

$\displaystyle x=\cos \left( \frac{\pi}{2n}+\frac{2\pi k}{n} \right)+i\sin \left( \frac{\pi}{2n}+\frac{2\pi k}{n} \right)$ For $\displaystyle k=0,1,2..,n-1$

I assume by prinicpal you mean $\displaystyle k=0$

In that case,

$\displaystyle i^{1/n}=\cos \pi/2n +i\sin \pi/2n$

Thus, $\displaystyle n\to \infty$

$\displaystyle \cos 0+i\sin 0=1$