Let z = 1 + ai be a complex number,a 0,such | vedclass

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(B) $z = 1 + ai$$z^3 = (1 + ai)^3 = 1^3 + 3(1)^2(ai) + 3(1)(ai)^2 + (ai)^3$$z^3 = 1 + 3ai - 3a^2 - a^3i = (1 - 3a^2) + i(3a - a^3)$Since $z^3$ is a re

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$-1365\sqrt{3}i$

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(B) $z = 1 + ai$$z^3 = (1 + ai)^3 = 1^3 + 3(1)^2(ai) + 3(1)(ai)^2 + (ai)^3$$z^3 = 1 + 3ai - 3a^2 - a^3i = (1 - 3a^2) + i(3a - a^3)$Since $z^3$ is a real number,the imaginary part must be zero:$3a - a^3 = 0 \Rightarrow a(3 - a^2) = 0$Given $a > 0$,we have $a^2 = 3$,so $a = \sqrt{3}$.Thus,$z = 1 + \sqrt{3}i = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$.The sum is a geometric series: $S = \frac{z^{12} - 1}{z - 1}$.$z^{12} = [2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})]^{12} = 2^{12}(\cos 4\pi + i \sin 4\pi) = 2^{12} = 4096$.$S = \frac{4096 - 1}{1 + \sqrt{3}i - 1} = \frac{4095}{\sqrt{3}i} = \frac{4095}{\sqrt{3}i} 	imes \frac{i}{i} = \frac{4095i}{-\sqrt{3}} = -\frac{4095}{3} \sqrt{3}i = -1365\sqrt{3}i$.

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

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