How to find the limit $z\rightarrow i$? How to calculate the limit of $\frac{z^{10}+1}{z^6+1}$ as $z\rightarrow i$?
I tried to take the limit but the denominator becomes zero. Does this mean that the limit is infinite?
 A: I'm going to take a beating for this, but since it goes to $\frac{0}{0}$, you can use l'Hôpital's rule.
$$\lim_{z\to i}\frac{z^{10}+1}{z^6+1} = \lim_{z\to i}\frac{10z^{9}}{6z^5} = \lim_{z\to i}\frac{5z^4}{3} = \frac{5}{3}$$
A: To avoid L'Hopital rememeber that
$$\begin{align}x^{10}+1=&(x^2+1)(x^8-x^6+x^4-x^2+1)\\
x^6+1=&(x^2+1)(x^4-x^2+1)\end{align}$$
So the fraction simplifies to
$${x^8-x^6+x^4-x^2+1\over x^4-x^2+1}\to {5\over 3}$$
A: Hint:
$$
\frac{z^{10}-i^{10}}{z-i}=z^9+iz^8-z^7-iz^6+z^5+iz^4-z^3-iz^2+z+i
$$
and
$$
\frac{z^6-i^6}{z-i}=z^5+iz^4-z^3-iz^2+z+i
$$
A: $\newcommand{\I}{\mathrm i}$Thanks @A.G. for the idea of "cancel $z - \I$".
Using Polynomial long division,
$$ \left.\frac{z^{10}+1}{z-\I}\right|_\I = \left.\I + z - \I z^2 - z^3 + \I z ^4 + z^5 - \I z^6 - z^7 + \I z ^8 + z^9\right|_\I = 10\I
$$
while
$$ \left.\frac{z^6+1}{z-\I}\right|_\I = \left.\I + z - \I z^2 - z^3 + \I z ^4 + z^5\right|_\I = 6\I
$$
Hence
$$ \lim_{z \to \I} \frac{z^{10}+1}{z^6 +1} = \lim_{z \to I} \frac{10\I}{6\I} = \frac{10\I}{6\I}
$$
