Let $z=1+ai$ be a complex number, $a>0$, such that  $z^3$ is a real number. Then the sum $1+z+z^2........+z^{11}$is equal to

$z=1+ai,z^2=1-a^2+2ai$
$z^2\cdot z=\left\{\left(1-a^2\right)+2ai\right\}(1+ai)=\left(1-a^2\right)+2ai+\left(1-a^2\right)ai-2a^2$
$\therefore z^3$ is real $\Rightarrow 2a+\left(1-a^2\right)a=0$
$\Rightarrow a\left(3-a^2\right)=0\Rightarrow a=\sqrt{3}(\because a>0)$
Now, $1+z+z^2+\dots \dots +z^{11}=\frac{z^{12}-1}{z-1}$

$\begin{array}{l}=\frac{(1+\sqrt{3}i{)}^{12}-1}{1+\sqrt{3i-1}}=\frac{(1+\sqrt{3i}{)}^{12}-1}{\sqrt{3i}}\\ (1+\sqrt{3i}{)}^{12}=2^{12}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^{12}=2^{12}{\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)}^{12}\end{array}$
                              $-2^{12}(\cos 4\pi +i\sin 4\pi )=2^{12}$
  $\begin{array}{l}\therefore 1+z+z^2+\dots ..+z^{11}=\frac{2^{12}-1}{\sqrt{3}i}=\frac{4095}{\sqrt{3i}}\\ =-\frac{4095}{3}\sqrt{3}i=-1365\sqrt{3i}\end{array}$