Question

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 :

• A. $1365\sqrt{3}i$
• B. $1250\sqrt{3}i$
• C. $-1250\sqrt{3}i$
• D. $-1365\sqrt{3}i$

Answer 1

The correct option is D $-1365\sqrt{3}i$

For such type of expressions , first try to check whether the given complex number $\left(z\right)$ is cube root of unity or not. In most cases it will be cube root of unity.
$z=1+ai{z}^3=1+3ai-3a^2-a^{3}i{z}^3=(1-3a^2)+a(3-a^2)i\text{Since,}{z}^3\text{is a real number.}\Rightarrow a(3-a^2)=0\Rightarrow a\ne 0,-\sqrt{3}(\because a>0)\therefore a=\sqrt{3}\Rightarrow z=1+\sqrt{3}i=-2\omega \Rightarrow |z|=2S=1+z+z^2+....+z^{11}=\frac{1(z^{12}-1)}{z-1}=\frac{2^{12}-1}{1+\sqrt{3}i-1}(\because z^3=|z{|}^3=2^3)=-1365\sqrt{3}i$