Imaginary number challenge

$(\frac{k-i}{i-k})^{k}$

$=(\frac{-(i-k)}{i-k})^{k}$

$=(-1)^{k}$

$i^{2016k}$

$=i^{4n}$ ,where $n$ is a positive integer.

$=1$

$\displaystyle\sum_{k=1}^{2016} ((\frac{k-i}{i-k})^{k} \times \frac{k}{i^k} \times i^{2016k} )$

$=\displaystyle\sum_{k=1}^{2016} ((-1)^{k} \times \frac{k}{i^k} )$

$=\displaystyle\sum_{k=1}^{2016} ((-1)^{k} \times \frac{k(i^k)((-i)^k)}{i^k} )$

$=\displaystyle\sum_{k=1}^{2016} (-1)^{k} \times k(-i)^{k}$

$=\displaystyle\sum_{k=1}^{2016} k((-i)(-1))^{k}$

$=\displaystyle\sum_{k=1}^{2016} k(i)^{k}$

$= i-2-3i+4+5i-6-7i+8 + \dots + 2013i-2014-2015i+2016$

$= (i-2-3i+4)+(5i-6-7i+8) + \dots + (2013i-2014-2015i+2016)$

$= (-2i+2)+(-2i+2)+\dots + (-2i+2)$

$= (\frac{2016}{4})(-2i+2)$

$= 1008-1008i$

$a+b = 1008+1008 = 2016$

$\begin{aligned} S & = \sum_{k=1}^{2016} \left(\color{#3D99F6}{\frac{k-i}{i-k}}\right)^k \frac k{i^k} \color{#D61F06}{i^{2016k}} \\ & = \sum_{k=1}^{2016} \left(\color{#3D99F6}{-\frac{k-i}{k-i}}\right)^k \frac k{i^k} \color{#D61F06}{i^{4\times 504k}} \quad \quad \small \color{#D61F06}{\text{We note that }i^4 = 1} \\ & = \sum_{k=1}^{2016} \left(\color{#3D99F6}{-1}\right)^k \frac k{i^k} \color{#D61F06}{(1)} \\ & = \sum_{k=1}^{2016} k \left(\frac{\color{#3D99F6}{-1}}i \right)^k \quad \quad \small \color{#3D99F6}{\text{We note that }-1 = i^2} \\ & = \sum_{k=1}^{2016} k \left(\frac{\color{#3D99F6}{i^2}}i \right)^k \\ & = \sum_{k=1}^{2016} k i^k \\ & = i - 2 - 3i + 4 + 5i - 6 - 7i + 8 + ...-2015i + 2016 \\ & = (\color{#3D99F6}{- 2 + 4} \color{#D61F06}{- 6 + 8} - 10+12-...+ 2016) + (\color{#3D99F6}{1-3}+\color{#D61F06}{5-7}+9-11+...-2015)i \\ & = \underbrace{(\color{#3D99F6}{2}+\color{#D61F06}{2}+2+...+2)}_{2\times 504} + \underbrace{(\color{#3D99F6}{-2} \color{#D61F06}{-2} -2-...-2)}_{-2\times 504}i \\ & = 1008 - 1008i \end{aligned}$

$\implies a + b = 1008+1008 = \boxed{2016}$