How to show $\sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}} = 0$? I came across this sum while doing the homework my teacher gave us on series. It was originally $$a_k = \frac{k}{81}, S = \sum_{k=1}^{80} {a_k^2 \over 1+2a_k^2-2a_k}$$
I then decomposed it into partial fractions and ended up with
$40+ \sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}}$. I know the answer is 40 from WolframAlpha, however I don't know how to go about showing that the sum with the complex numbers is 0.
How should I proceed?

Notice that :
$$S = \sum_{k = 1}^{80} \dfrac{k^2}{81^2 + 2 k^2 - 2 \times 81 \times k} = \sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2}$$
and :
$$\sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2} = \sum_{k = 1}^{80} \dfrac{(81 - k)^2}{k^2 + (81 - k)^2}$$
which allows us to deduce that :
$$2 S = \sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2} + \sum_{k = 1}^{80} \dfrac{(81 - k)^2}{k^2 + (81 - k)^2} = \sum_{k = 1}^{80} \dfrac{k^2 + (81 - k)^2}{k^2 + (81 - k)^2} = \sum_{k = 1}^{80} 1 = 80$$
Hence :
$$S = 40$$
