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I want to find the residue at $z=1$ of $\dfrac{1}{z^{100} -1}$

But I really have no idea how to go about

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    Hint: $z^{100}-1=(z-1)(z^{99}+z^{98}+\ldots+1)$ – 2017-01-30
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    Could you maybe tell us what you have tried so far? It's good practice to do that when asking a question on here. – 2017-01-30

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You need to evaluate the following limit:

$$\lim_{z\to1}\frac{z-1}{z^{100}-1}$$

and this is easily done with the geometric series:

$$\frac{z-1}{z^{100}-1}=\frac1{1+z+z^2+z^3+\dots+z^{99}}\to\frac1{100}$$

or L'Hospital's rule:

$$\lim_{z\to1}\frac{z-1}{z^{100}-1}=\lim_{z\to1}\frac1{100z^{99}}=\frac1{100}$$

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    But can't you only use the formula $Res(f,1) = \lim_{z\to1}\frac{z-1}{z^{100}-1}$ if there's a pole of order 1 at $z=1$. In our case we don't know if there's such a pole. – 2017-01-30
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    @SylvesterStallone Well, it's easy enough to see that it's a pole of order one...particularly out of practice for me, probably due to factoring of the denominator for you.... – 2017-01-30