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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?

  • 5
    Try to cancel $z-i$. – 2017-01-16
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    @A.G., How to do that? – 2017-01-16

4 Answers 4

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$\newcommand{\I}{\mathrm i}$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 $$ \lim_{z \to \I} \frac{z^{10}+1}{z^6 +1} = \lim_{z \to I} \frac{10\I}{6\I} = \frac{10\I}{6\I} $$

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I'm going to take a beating for this, but since it goes to $\frac{0}{0}$, you can use l'Hôpital's rule.

$$\lim_{z\to i}\frac{z^{10}+1}{z^6+1} = \lim_{z\to i}\frac{10z^{9}}{6z^5} = \lim_{z\to i}\frac{5z^4}{3} = \frac{5}{3}$$

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    How to cancel $z−i$? – 2017-01-16
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    I had started to work it out on paper, but Henry W. has done it. Probably ought to go with his answer; from what I gather, l'Hôpital's rule ain't welcome in these here parts. – 2017-01-16
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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

$${x^8-x^6+x^4-x^2+1\over x^4-x^2+1}\to {5\over 3}$$

  • 0
    This is probably a better idea than cancelling out $z - i$ since roots of a real polynomial comes in conjugate pairs. – 2017-01-16
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    If $i$ is a root so is $-i$ because the polynomial is real so cancelling $x^2+1$ comes obviously to mind – 2017-01-16
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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 $$ \frac{z^6-i^6}{z-i}=z^5+iz^4-z^3-iz^2+z+i $$