I factored the denominator and got $\frac{z^2+i}{(z^2-1)(z^2+1)}$, now I can factor the denominator again and get $\frac{z^2+i}{(z-1)(z+1)(z-i)(z+i)}$, but I don't know how to factor the top. Any hints?
How to find the limit of $\lim \limits_{z \to i} \frac{z^2+i}{z^4-1}$
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complex-analysis
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2Hint: the only factor which vanishes at $z =i$ is in the denominator. – 2017-02-22
2 Answers
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The numerator can be factored some more (using the square roots of $i$), but that wouldn't be useful here. Instead consider what happens to the numerator and denominator as $z \to i$, and think about what the limit must then do.
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0I tried approaching $i$ from the imaginary axis, so I plugged in $2i$ and $3i$ and I got $\frac{4i^2+i}{15}$ and $\frac{9i^2+i}{80}$, but I don't see where the limit is approaching? – 2017-02-22
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Hint
You are trying to calculate $$\lim \limits_{z \to i} \frac{z^2+i}{z^4-1}$$ Perhaps the subsitution $x=z^4$ will help clear this up for you. Note that $x \to 1$ as $z \to i$ $$\lim_{x \to 1} \frac{i-x}{x-1}$$ $$=i\cdot\lim_{x \to 1} \frac{1}{x-1}-1$$ You now have a standard, real limit to evaluate