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I am given some limits that exist, I'm supposed to find their values. Seems really simple however I am struggling.

Find the value of $\displaystyle \lim_{z\to\\i}\frac{z^4 - 1}{z-i} $.

My approach was to paramaterize $z$ to $it$ and transform the limit to something like:

$$\lim_{t\to\\1}\frac{(it)^4 - 1}{it-i} $$

Could anyone lend a hand as to how to solve something like this?

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    Hint: Can you factor the numerator $z^4 - 1$?2012-11-27
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    $i$ is a root of $z^4-1$ henve the polynomial $z-i$ divides the polynomial $z^4-1$. Find the polynomial which is their ratio and evaluate it at $z=i$.2012-11-27

1 Answers 1

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(1)L'Hospital directly

$$\lim_{z\to i}\frac{z^4-1}{z-i}=\lim_{z\to i}4z^3=-4i$$

(2) Factoring:

$$z^4-1=(z^2-1)(z-i)(z+i)\Longrightarrow \lim_{z\to i}\frac{z^4-1}{z-i}=\lim_{z\to i}(z^2-1)(z+i)=-2(2i)=-4i$$

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    thanks! I had forgotten L'Hospital, this serves to be a good reminder.2012-11-27
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    In the second method, how did you know that you know that you can just plug in $i$ in the end? How did you know the function is continous?2017-02-07
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    @Ovi Isn't it obvious? We have a complex polynomial there, which is an entire function...2017-02-07
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    Yes I strongly suspected that I just wanted to be 100% sure.2017-02-07