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?

Answer 1

According to given in the question we have to find the value of $1 + z + {z^2} + ............. + {z^{11}}$ when Let $z = 1 + ai$be a complex number, $a > 0$, such that ${z^3}$ is a real number so, first of all we have to find the value of ${z^3}$ which can be obtained by find the square of the term z which we let $z = 1 + ai$.
Now, as we obtain the value of ${z^2}$ so, to obtain ${z^3}$ we have to multiply ${z^2}$ with z. After that we have to find the value of z as given in the question that ${z^3}$ is a real number where $a > 0$ and now, to solve the given expression we have to find the sum of given expression $1 + z + {z^2} + ............. + {z^{11}}$ by placing the value of z.

Formula used: $\Rightarrow {(a + b)^2} = ({a^2} + {b^2} + 2ab)...............(1)$
$\Rightarrow {i^2} = - 1.................(2)$
Sum$= \dfrac{{a({r^n} - 1)}}{{r - 1}}$……………….(3)

Complete step-by-step answer:
Step 1: First of all we have to find the real number ${z^3}$ for which first of all we have to find the value of ${z^2}$ as mentioned in the solution hint. Where $z = 1 + ai$
$\Rightarrow {z^2} = {(1 + ai)^2}$
Now, to solve the expression obtained just above we have to use the formula (1) as mentioned in the solution hint.
Hence,
$\Rightarrow {z^2} = {(1)^2} + {(ai)^2} + 2 \times 1 \times ai \\\ \Rightarrow {z^2} = 1 + {a^2}{i^2} + 2ai$
Now, to solve the above expression we have to use the formula (2) as mentioned in the solution hint.
$\Rightarrow {z^2} = 1 + {a^2}( - 1) + 2ai \\\ \Rightarrow {z^2} = 1 - {a^2} + 2ai$
Step 2: Now, to find the value of ${z^3}$ we have to multiply z with the ${z^2}$ as obtained in step 1.
Hence,
$\Rightarrow {z^3} = {z^2} \times z \\\ \Rightarrow {z^3} = (1 - {a^2} + 2ai)(1 + ai) \\\ \Rightarrow {z^3} = (1 - {a^2}) + 2ai + (1 - {a^2})ai - 2{a^2}$
${z^3}$ is a real number hence,
$\Rightarrow 2a + (1 - {a^2})a = 0$
On multiplying a with $1 - {a^2}$ in the expression obtained just above,

$$\Rightarrow 2a + a - {a^3} = 0 \\\ \Rightarrow 3a - {a^3} = 0 \\\ \Rightarrow a(3 - {a^2}) = 0$$

Now, as given $a > 0$ hence,
$\Rightarrow 3 - {a^2} = 0 \\\ \Rightarrow {a^2} = 3 \\\ \Rightarrow a = \sqrt 3$
Step 3: Now, with the help of the formula (3) as mentioned in the solution hint we can obtain the sum of the expression $1 + z + {z^2} + ............. + {z^{11}}$
$\Rightarrow 1 + z + {z^2} + ............. + {z^{11}} = \dfrac{{{z^{12}} - 1}}{{z - 1}}$
Where $(r > 1)$
On substituting the values obtained,
$= \dfrac{{{{(1 + \sqrt 3 i)}^{12}} - 1}}{{1 + \sqrt 3 i - 1}} \\\ = \dfrac{{{{(1 + \sqrt 3 i)}^{12}} - 1}}{{\sqrt 3 i}}$
On, multiplying and dividing with ${2^{12}}$ in the expression obtained just above,
$= {2^{12}}{\left( {\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i} \right)^{12}}$
Step 4: Now, as we know that:
$\cos \dfrac{\pi }{3} = \dfrac{1}{2}$and, $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
$= {2^{12}}{\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)^{12}} \\\ = {2^{12}}\left( {\cos \dfrac{{12\pi }}{3} + i\sin \dfrac{{12\pi }}{3}} \right) \\\ = {2^{12}}(\cos 4\pi + i\sin 4\pi )...........(4)$
Step 5: As we know that $\sin 4\pi = 0$hence, on substituting in the expression (4),
$= {2^{12}}$
Hence,
$= \dfrac{{{2^{12}} - 1}}{{\sqrt 3 i}} \\\ = \dfrac{{4095}}{{\sqrt 3 i}}$
On applying the rationalization in the expression just above,
$= \dfrac{{4095}}{{\sqrt 3 i}} \times \dfrac{{\sqrt 3 i}}{{\sqrt 3 i}} \\\ = - \dfrac{{4095\sqrt 3 i}}{3} \\\ = - 1365\sqrt 3 i$

Hence, with the help of formula (1), (2), and (3) we have obtained the sum of $1 + z + {z^2} + ............. + {z^{11}}$ $= - 1365\sqrt 3 i$

Note: In the expression $z = 1 + ai$ i is an imaginary term and on squaring it we can obtain it’s value which is ${i^2} = - 1$ and ${i^4} = 1$
To find the rationalization of a given fraction we have to multiply the term or it’s inverse as required given in denominator of a fraction with its numerator and denominator.