# limits in the complex domain

• November 2nd 2006, 05:37 PM
splash
limits in the complex domain
Show that $
\lim_{n \to \infty}(i)^\frac{1}{n}=1
$
where $(i)^\frac{1}{n}$ is the principal value of the nth root of $i$.
• November 2nd 2006, 05:58 PM
Quick
Quote:

Originally Posted by splash
Show that $
\lim_{n \to \infty}(i)^\frac{1}{n}=1
$
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 $\lim_{n \to \infty}i^{\frac{1}{n}}=i^0=1$
• November 2nd 2006, 07:01 PM
ThePerfectHacker
Quote:

Originally Posted by splash
Show that $
\lim_{n \to \infty}(i)^\frac{1}{n}=1
$
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, $n\to \infty$
$\cos 0+i\sin 0=1$