Let z=1+ai be a complex number, a0, such that z^3 is a real number.

Question

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

*(Online April 10, 2016)*

Accepted Answer

$z^2=(1-a^2)+2ai$, so $z^3=z^2\cdot z=\big[(1-a^2)-2a^2\big]+\big[2a+(1-a^2)a\big]i$. For $z^3$ real, the imaginary part vanishes:
$$2a+(1-a^2)a=0 \Rightarrow a(3-a^2)=0 \Rightarrow a=\sqrt3\ (a>0).$$

So $z=1+\sqrt3 i=2\left(\cos\tfrac\pi3+i\sin\tfrac\pi3\right)$.

$$1+z+\cdots+z^{11}=\frac{z^{12}-1}{z-1}=\frac{z^{12}-1}{\sqrt3 i}.$$

$z^{12}=2^{12}(\cos4\pi+i\sin4\pi)=2^{12}=4096$, so
$$\frac{4096-1}{\sqrt3 i}=\frac{4095}{\sqrt3 i}=-\frac{4095}{3}\sqrt3 i=-1365\sqrt3 i.$$

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