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

Hint: We will be using the condition i.e. ${z^3}$ is real, we will equate the imaginary part of ${z^3}$ equal to 0. Then we will get some condition, using that condition, we have to proceed further.

Complete step by step answer:
According to question,
$
  z = 1 + ai \\
  {z^2} = 1 - {a^2} + 2ai \\
  {z^2}.z = \left\{ {\left( {1 - {a^2}} \right) + 2ai} \right\}\left\{ {1 + ai} \right\} \\
   = \left( {1 - {a^2}} \right) + 2ai + \left( {1 - {a^2}} \right)ai - 2{a^2} \\
$
Given that ${z^3}$ is real
$\therefore 2a + \left( {1 - {a^2}} \right)a = 0$
$ \Rightarrow a(3 - {a^{^2}}) = 0 \Rightarrow a = \sqrt 3 (a > 0)$
We need to calculate the sum $1 + z + {z^2} + .... + {z^{11}}$
$1 + z + {z^2} + .... + {z^{11}} = \dfrac{{{z^{12}} - 1}}{{z - 1}}$ (Using sum of Geometric progression, Sum=$\dfrac{{a({r^n} - 1)}}{{r - 1}}$ when r>1)
$ \Rightarrow \dfrac{{{{(1 + \sqrt 3 i)}^{12}} - 1}}{{1 + \sqrt 3 i - 1}} \Rightarrow \dfrac{{{{(1 + \sqrt 3 i)}^{12}} - 1}}{{\sqrt 3 i}}$ …(1)
Solving ${(1 + \sqrt 3 i)^{12}} = {2^{12}}{(\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i)^{12}} = {2^{12}}{(\cos \dfrac{\Pi }{3} + i\sin \dfrac{\Pi }{3})^{12}} = {2^{12}}(\cos 4\Pi + i\sin 4\Pi ) = {2^{12}}$
Now keeping this value in above equation,
We obtain,
$\dfrac{{{2^{12}} - 1}}{{\sqrt 3 i}} = \dfrac{{4095}}{{\sqrt 3 i}} = - \dfrac{{4095}}{3}\sqrt 3 i = - 1365\sqrt 3 i$

So, the correct answer is “Option B”.

Note: For solving questions related to series and complex numbers, firstly we will find the relation, secondly, we will calculate the sum using sum of either AP or GP, and then using the relation of summation solve the expression of sum to get the result as shown.
Sum of an AP series $ \Rightarrow $${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$ where $a$ is the first term of AP and $d$ is the common difference of AP and $n$is the number of terms in AP.
Sum of a GP series $ \Rightarrow {S_n} = $$\dfrac{{a({r^n} - 1)}}{{r - 1}}$ (r>1) where $a$ is the first term of GP, $r$ is the common ratio and $n$ is the number of terms in GP.