Sum!

$\Large z=e^{\small{\frac{i\pi}{3}}}$ Let $\mathcal{G}$ denote the given expression. $\Large\Rightarrow \mathcal{G} =\sum_{k=1}^{11}({z}^{k})= \sum_{k=1}^{11} e^{\small{\frac{k i\pi}{3}}}$ $\color{#456461}{\text{(GP)}}$ $\Large =e^{\small{\frac{ i\pi}{3}}}(\dfrac{e^{\small{\frac{11 i\pi}{3}}}-1}{e^{\small{\frac{ i\pi}{3}}}-1})$ $\Large =\dfrac{e^{\small{\frac{12 i\pi}{3}}}-e^{\small{\frac{ i\pi}{3}}}}{e^{\small{\frac{ i\pi}{3}}}-1}$ $\huge =-1$ $\huge\therefore \mathcal{G}=\color{#0C6AC7}{\boxed{\color{#EC7300}{{\boxed{\color{#007fff}{\textbf{-1}}}}}}}$

$\large z=e^{(\frac{i\pi}{3})}$ $\sum_{k=1}^{11}({z}^{k})=z^{1}+z^{2}+z^{3}+...+z^{11}~~(\color{#D61F06}{GP})$ $z^{1}+z^{2}+z^{3}+...+z^{11}=(\frac{z^{12}-z}{z-1})=$ $=(\frac{e^{(i4\pi)}-e^{(\frac{i\pi}{3})}}{e^{(\frac{i\pi}{3})}-1})=\frac{1-e^{(\frac{i\pi}{3})}}{e^{(\frac{i\pi}{3})}-1})=$ $\Large\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{-1}}}}$

$z = e^{\frac{i\pi}{3}}$ is a 6th primitive root of the unity. I mean $z^6 =1 \text{ and } z^5 + z^4 + z^3 + z^2 + z + 1 = 0 \Rightarrow$ $\sum_{k=1}^{11} z^k = z + z^2 + z^3 + z^4 + z^5 + z^6(1 + z + z^2 + z^3 + z^4 + z^5) =$ $= z + z^2 + z^3 + z^4 + z^5 = -1$