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

Answer 1

Answer: (B)

$-1365\sqrt{3}i$

$z=1+ai$

$z^2=1-a^2+2ai$

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

$=(1-a^2)+2ai+(1-a^2)ai-2a^2$

$\because z^3$ is real $\Rightarrow 2a+(1-a^2)a=0$

$\Rightarrow a(3-a^2)=0\Rightarrow a=\sqrt{3}(a>0)$

$1+z+z^2.....z^{11}=\frac{z^{12}-1}{z-1}$ [Sum$=\frac{a(r^n-1)}{r-1}$ when $r>1$]

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

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

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