Absolute value and complex number (2)

$|k^{2016i}|=|e^{2016(\ln k)i}|=1$

So the summation is simply $\displaystyle\sum_{k=2}^{2016}(1)=2015$ .

Now the product is:

$\left|\prod_{k=2}^{2016}k^{2016i}\right|=\left|\left(\displaystyle\prod_{k=1}^{2016}k\right)^{2016i}\right|=e^{\small{2016\ln\left(\displaystyle\prod_{k=1}^{2016}k\right)i}}=1$

Note:- Since $\large\color{#3D99F6}{\left|e^{i\theta}\right|=1}$ .

Hence the sum expression is simply $2015$ while product expression is $1$ . Hence answer is $2015+1=\boxed{2016}$ .

$\begin{aligned} S & = \sum_{k=2}^{2016} \left| k^{2016i} \right| + \left| \prod_{k=2}^{2016} k^{2016i} \right| \\ & = \sum_{k=2}^{2016} \left| e^{\color{#3D99F6}{2016(\ln k)}i} \right| + \left| \prod_{k=2}^{2016} e^{\color{#3D99F6}{2016(\ln k)}i} \right| & \small \color{#3D99F6}{\text{Let }\theta_k = 2016(\ln k)} \\ & = \sum_{k=2}^{2016} \left| \cos \color{#3D99F6}{\theta_k} + i \sin \color{#3D99F6}{\theta_k} \right| + \left| \exp \left( i\sum_{k=2}^{2016} \color{#3D99F6}{\theta_k} \right) \right| & \small \color{#3D99F6}{\text{Let }\sum_{k=2}^{2016} \theta_k = \Theta} \\ & = \sum_{k=2}^{2016} \color{#D61F06}{1} + \left| e^{ \color{#3D99F6}{\Theta}i} \right| & \small \color{#D61F06}{\text{Note that }\left| \cos \theta_k + i \sin \theta_k \right| = 1} \\ & = 2015 + \left| \cos \color{#3D99F6}{\Theta} + i \sin \color{#3D99F6}{\Theta} \right| \\ & = 2015 + 1 \\ & = \boxed{2016} \end{aligned}$