Mathematics - Complex Number Question with Solution | TestHub

$z=1+ai$
Cubing both sides,
$z^3=\left(1-a^2\right)+2ai+\left(1-a^2\right)ai-2a^2$
$z^3=1-3a^2+(3a-a^3)i$
$\because$$z^3$is a real number.
$a\left(3-a^2\right)=0 \Rightarrow a= \sqrt{3}$$\left(a>0\right)$

$1+z+z^2\dots ..z^{11}=\frac{z^{12}-1}{z-1}=\frac{{\left(1+ \sqrt{3}i\right)}^{12}-1}{1+ \sqrt{3}i-1}=\frac{{\left(1+ \sqrt{3}i\right)}^{12}-1}{\sqrt{3}i}$

${\left(1+ \sqrt{3}i\right)}^{12}=2^{12}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^{12}=2^{12}{\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)}^{12}=2^{12} \left(\cos 4\pi +i\sin 4\pi \right)=2^{12}$

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