Question

If $z={{\left( \dfrac{\sqrt{3}+i}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}-i}{2} \right)}^{5}}$ then which of the following option is correct:
(a) ${{R}_{e}}(z)=0$
(b) ${{I}_{m}}(z)=0$
(c) ${{R}_{e}}(z)=0$, ${{I}_{m}}(z)>0$
(d) ${{R}_{e}}(z)>0,{{I}_{m}}(z)<0$

Answer 1

Hint: First we solve the equation by opening brackets and then we have to rearrange some terms to make it in the form of a + ib. After doing this much then we will write the imaginary part and real part separately and find out the correct option.

Complete step-by-step answer:
First let’s try to find out the value of this one ${{\left( \dfrac{\sqrt{3}+i}{2} \right)}^{5}}$ .
The formula for ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we are going to use this formula for calculating the value of ${{z}^{2}}$, where z can any complex number.
Another formula that we are going to use is ${{i}^{2}}=-1$ ,
Now ${{\left( \dfrac{\sqrt{3}+i}{2} \right)}^{5}}$ will be,
$\begin{align}
  & =\dfrac{{{\left( \sqrt{3}+i \right)}^{5}}}{32} \\
 & =\dfrac{{{\left( \sqrt{3}+i \right)}^{2}}{{\left( \sqrt{3}+i \right)}^{2}}\left( \sqrt{3}+i \right)}{32} \\
\end{align}$
Now we will use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand,
$=\dfrac{\left( {{\left( \sqrt{3} \right)}^{2}}+{{i}^{2}}+2\sqrt{3}i \right)\left( {{\left( \sqrt{3} \right)}^{2}}+{{i}^{2}}+2\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32}$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$=\dfrac{{{\left( 2+2\sqrt{3}i \right)}^{2}}\left( \sqrt{3}+i \right)}{32}$
Now we will use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand,
$=\dfrac{\left( {{2}^{2}}+{{\left( 2\sqrt{3}i \right)}^{2}}+8\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32}$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$=\dfrac{\left( 4-12+8\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32}$
$\begin{align}
  & =\dfrac{\left( -8+8\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32} \\
 & =\dfrac{8\left( -1+\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32} \\
 & =\dfrac{8\left( -\sqrt{3}-i+3i+\sqrt{3}{{i}^{2}} \right)}{32} \\
\end{align}$
Now we know that ${{i}^{2}}=-1$ , using this we get,
$\begin{align}
  & =\dfrac{8\left( -2\sqrt{3}+2i \right)}{32} \\
 & =\dfrac{16\left( -\sqrt{3}+i \right)}{32} \\
\end{align}$
So, we have the first part now for the 2nd part we just have substitute i by –i in $\dfrac{16\left( -\sqrt{3}+i \right)}{32}$
${{\left( \dfrac{\sqrt{3}-i}{2} \right)}^{5}}$ , after substituting we get,
${{\left( \dfrac{\sqrt{3}-i}{2} \right)}^{5}}$= $\dfrac{16\left( -\sqrt{3}-i \right)}{32}$
Now we calculated all the values that are needed to write the given equation of z in the form of a + ib.
Therefore, the value of z becomes after substituting is:
$\begin{align}
  & \dfrac{16\left( -\sqrt{3}+i \right)}{32}+\dfrac{16\left( -\sqrt{3}-i \right)}{32} \\
 & z=\dfrac{-\sqrt{3}+i-\sqrt{3}-i}{2} \\
 & z=\dfrac{-2\sqrt{3}}{2} \\
 & z=-\sqrt{3} \\
\end{align}$
In a + ib, ‘a’ is the real part and ‘b’ is the imaginary part.
Now it is the form a + ib, now we can write the real part and the imaginary part.
${{R}_{e}}(z)$ = $-\sqrt{3}$
${{I}_{m}}\left( z \right)$ = 0
From this we can conclude that option (b) is correct.

Note: One can also solve this question by first writing all the powers of $\left( \sqrt{3}+i \right)$ up to 5 and then writing all the powers of $\left( \sqrt{3}-i \right)$ up to 5 and then substitute it in the given equation and then again we have to separate it imaginary part and real part to mark the correct option.