$z=1+a i$

$z^{2}=1-a^{2}+2 a i$

$z^{2} \cdot z=\left\{\left(1-a^{2}\right)+2 a i\right\}\{1+a i\}$

$=\left(1-a^{2}\right)+2 a i+\left(1-a^{2}\right) \quad a i-2 a^{2}$

$\because \quad z^{3}$ is real $\Rightarrow 2 a+\left(1-a^{2}\right) a=0$

${a\left( {3 – {a^2}} \right) = 0 \Rightarrow a = \sqrt 3 (a > 0)}$

${1 + z + {z^2} \ldots  \ldots  \ldots {z^{11}} = \frac{{{z^{12}}}}{{z – 1}}}$

${ = \frac{{{{(1 + \sqrt 3 i)}^{12}} – 1}}{{1 + \sqrt {3i}  – 1}}}$

${ = \frac{{{{(1 + \sqrt 3 i)}^{12}} – 1}}{{\sqrt {3i} }}}$

${{{(1 + \sqrt 3 i)}^{12}} – {2^{12}}{{\left( {\frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)}^{12}}}$

${ = {2^{12}}{{\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right)}^{12}}}$ ${ = {2^{12}}(\cos 4\pi  + i\sin )}$ ${ = {2^{12}}}$

$\Rightarrow \quad \frac{2^{12}-1}{\sqrt{3} i}=\frac{4095}{\sqrt{3} i}=-\frac{4095}{3} \sqrt{3} i=-1365 \sqrt{3} i$