Question

Let , $z={{\left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \right)}^{5}}$ . If $R\left( z \right)$ and $I\left( z \right)$ respectively denote the real and imaginary parts of $z$ , then
(a) $R\left( z \right)>0$ and $I\left( z \right)>0$
(b) $R\left( z \right)<0$ and $I\left( z \right)>0$
(c) $R\left( z \right)=3$
(d) $I\left( z \right)=0$

Answer 1

Consider , the equation $z={{\left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \right)}^{5}}$ and expand the powers . After separating the powers , you will get a complex number . Convert the complex number to the general form of writing a complex number i.e. $z=a+ib$ , $a=R\left( z \right),b=I\left( z \right)$ ; . At this stage you have the values of $R\left( z \right)$ and $I\left( z \right)$ , observe these values and hence choose the option wisely.

Complete step-by-step answer:
We are given a complex number of the form
$z={{\left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \right)}^{5}}$
And we have to observe its real and imaginary parts so first we have to simplify it .
Consider this complex number and expand the terms involving powers to simplify the complex number , and get
$\begin{aligned} & z=\dfrac{1}{{{2}^{5}}}{{\left( \sqrt{3}+i \right)}^{5}}+\dfrac{1}{{{2}^{5}}}{{\left( \sqrt{3}-i \right)}^{5}} \\\ & =\dfrac{1}{{{2}^{5}}}\left( {{\left( \sqrt{3}+i \right)}^{2}}{{\left( \sqrt{3}+i \right)}^{2}}\left( \sqrt{3}+i \right) \right)+\dfrac{1}{{{2}^{5}}}\left( {{\left( \sqrt{3}-i \right)}^{2}}{{\left( \sqrt{3}-i \right)}^{2}}\left( \sqrt{3}-i \right) \right) \\\ & =\dfrac{1}{{{2}^{5}}}\left( {{\left( 2+2\sqrt{3}i \right)}^{2}}\left( \sqrt{3}+i \right) \right)+\dfrac{1}{{{2}^{5}}}\left( {{\left( 2-2\sqrt{3}i \right)}^{2}}\left( \sqrt{3}-i \right) \right) \\\ & =\dfrac{1}{{{2}^{3}}}\left( {{\left( 1+\sqrt{3}i \right)}^{2}}\left( \sqrt{3}+i \right) \right)+\dfrac{1}{{{2}^{3}}}\left( {{\left( 1-\sqrt{3}i \right)}^{2}}\left( \sqrt{3}-i \right) \right) \\\ & =\dfrac{1}{{{2}^{2}}}\left( \left( -1+\sqrt{3}i \right)\left( \sqrt{3}+i \right) \right)+\dfrac{1}{{{2}^{2}}}\left( \left( -1-\sqrt{3}i \right)\left( \sqrt{3}-i \right) \right) \\\ & =\dfrac{1}{2}\left( -\sqrt{3}+i \right)+\dfrac{1}{2}\left( -\sqrt{3}-i \right) \\\ & =-\sqrt{3} \end{aligned}$
Now, we will change this value of $z$ to the general form of complex number, so that we could observe the real and imaginary parts of $z$and thus could check the options, The general form of writing complex number is $z=a+ib$ , We will convert $z$ into this form and observe it
$\begin{aligned} & z=-\sqrt{3} \\\ & z=-\sqrt{3}+0i \\\ \end{aligned}$
$\Rightarrow R\left( z \right)=-\sqrt{3}$ and $I\left( z \right)=0$
Hence, $R\left( z \right)<0$ and $I\left( z \right)=0$
(a) is not correct as $R\left( z \right)<0$
(b) satisfies that $R\left( z \right)<0$ but , $I\left( z \right)$ is not greater than 0 , it could be correct if it says $I\left( z \right)\ge 0$
(c) is not correct

So, the correct answer is “Option d”.

Note: The alternative for this question is to expand $z={{\left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \right)}^{5}}$ directly with the binomial expansion formula and then proceed further as same . The binomial expansion formula is used to expand large powers and is given by
${{\left( a+b \right)}^{n}}=\left( \begin{matrix} n \\\ 0 \\\ \end{matrix} \right){{a}^{n}}{{b}^{0}}+\left( \begin{matrix} n \\\ 1 \\\ \end{matrix} \right){{a}^{n-1}}{{b}^{1}}+\left( \begin{matrix} n \\\ 2 \\\ \end{matrix} \right){{a}^{n-2}}{{b}^{2}}+...+\left( \begin{matrix} n \\\ n-1 \\\ \end{matrix} \right){{a}^{1}}{{b}^{n-1}}+\left( \begin{matrix} n \\\ n \\\ \end{matrix} \right){{a}^{0}}{{b}^{n}}$