Z=1+ai be a complex number, a>o such that z3 is a real number, Then the sum 1+z+z2+....+z11=

Z=1+ai,Z3=1−3a2+i3a−a3z3 Is real ⇒3a−a3=0⇒a=3Z=1+i3⇒|Z|=2⇒1+z+z2+….Z11=Z12−1Z−1=212−1i3=−13653i

Z=1+ai be a complex number, a>o such that z3 is a real number, Then the sum 1+z+z2+....+z11=

Z=1+ai,Z3=1−3a2+i3a−a3z3 Is real ⇒3a−a3=0⇒a=3Z=1+i3⇒|Z|=2⇒1+z+z2+….Z11=Z12−1Z−1=212−1i3=−13653i

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$Z = 1 + a i$ be a complex number, $a > o$ such that z³ is a real number, Then the sum $1 + z + z^{2} + .... + z^{11} =$

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$- 1250 \sqrt{3} i$

$1250 \sqrt{3} i$

$1365 \sqrt{3} i$

$- 1365 \sqrt{3} i$

answer is D.

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Detailed Solution

$\begin{array}{l} Z = 1 + a i, Z^{3} = 1 - 3 a^{2} + i (3 a - a^{3}) \\ z^{3} Is real \Rightarrow 3 a - a^{3} = 0 \Rightarrow a = \sqrt{3} \\ Z = 1 + i \sqrt{3} \Rightarrow | Z | = 2 \\ \Rightarrow 1 + z + z^{2} + \dots . Z^{11} = \frac{Z^{12} - 1}{Z - 1} \\ = \frac{2^{12} - 1}{i \sqrt{3}} = - 1365 \sqrt{3} i \end{array}$