Complex Numbers Spaced Around The Unit Circle

$\begin{aligned} z & = \frac{\sqrt{2+\sqrt{3}}}{2} + i \frac{\sqrt{2-\sqrt{3}}}{2} \quad \quad \small \color{#3D99F6}{\text{As } \left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)^2 + \left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)^2 = 1 \text{, we can assume}} \\ & = \cos \theta + i \sin \theta \end{aligned}$

$\begin{aligned} \Rightarrow \tan \theta & = \frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}} \\ & = 2-\sqrt{3} \\ \Rightarrow \theta & = \frac{\pi}{12} \\ \Rightarrow z & = e^{i\frac{\pi}{12}} \end{aligned}$

Therefore, $z$ is the $24^{\text{th}}$ root of unit, and that $z^{24n} = z^0 = 1$ , $\displaystyle \sum_{k=0}^{24n-1} z^k = 0$ and hence $\displaystyle \sum_{k=0}^{24n} z^j = 1$ , where $n$ is a natural number. Since $24|2016$ , we have $\displaystyle \sum_{k=0}^{2016} z^k = \boxed{1}$ .

$\begin{aligned}\\& z=\dfrac{\sqrt{4+2\sqrt 3}}{2\sqrt 2}+i\dfrac{\sqrt{4-2\sqrt 3}}{2\sqrt 2}\\&=\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})\\&=\large e^{\frac{i\pi}{12}}~~(\color{#3D99F6}{\small{De~ Moivre's ~Law}})\end{aligned}$ $\large \therefore \displaystyle \sum_{k=0}^{2016} z^k= \displaystyle \sum_{k=0}^{2016} e^{\frac{ki\pi}{12}}~~~~~~~~~~~~\\~~~~~~~~~~~~=\dfrac{e^{\frac{2017i\pi}{12}}-1}{e^{\frac{i\pi}{12}}-1}~~(\color{#3D99F6}{\small{Sum ~of~GP}}) \\~~~~=\huge \boxed 1$ $~~~~~~~~~(\because e^{\frac{2017i\pi}{12}}=e^{\frac{i\pi}{12}})$

Note : To show that $\cos \dfrac \pi{12} = \dfrac{\sqrt{4+2\sqrt3}}{2\sqrt2}$ , apply the double angle identity , $\cos(2x) = 2\cos^2(x) - 1$ for $x = \dfrac\pi{12}$ . And to calculate $\sin \dfrac\pi{12}$ , apply $\sin^2 x + \cos^2x = 1$ .

The given equation is the 12th root of unity. So z^12=1; z^11+z+^10+z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1=0; By using this two equations we can find that the answer is 1.