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

Answer 1

The Correct Option is D

The correct answer is D:\(-1365\sqrt{3}i\)
Given that;
\(z=1+a i,\)
\(a > 0\)
\(z^{3}=1-3 a^{2}+\left(3 a-a^{3}\right) i\) is a real number
\(\Rightarrow 3 a-a^{3}=0\)
\(\Rightarrow a^{2}=3\)
\(\Rightarrow a=\sqrt{3},\)
\(a > 0\)
Then \(1+z+z^2+......z^4=\frac{z^{12}-1}{z-1}\)
\(=\frac{(1+\sqrt{3})^{12}-1}{(1+\sqrt{3}i)-1}\)
\(\Rightarrow z=1+\sqrt{3} i\)
\(=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\)
Now \(1+z+z^{2}+\ldots \ldots+z^{11}=\frac{1\left(1-z^{12}\right)}{1-z}=\frac{1-2^{12}(\cos 4 \pi+i \sin 4 \pi)}{1-(1+i \sqrt{3})}\)
\(=\frac{1-2^{12}}{-i \sqrt{3}}=\frac{4095}{i \sqrt{3}}=-1365 \,\sqrt{3} i\)