Question

If $z={{\left( \dfrac{\sqrt{3}}{2}+\dfrac{i}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}}{2}-\dfrac{i}{2} \right)}^{5}}$ , then
(a) Re(z) = 0
(b) Im(z) = 0
(c) $\operatorname{Re}(z)>0,\operatorname{Im}(z)>0$
(d) $\operatorname{Re}(z)>0,\operatorname{Im}(z)<0$

Answer 1

Hint: The complex number is of the form $z=x+iy$, where x is known as real part of complex number z and y is known as imaginary part of complex number z and you can use the formula ${{e}^{i\theta }}=\cos \theta +i\sin \theta $.

Complete step-by-step answer:
Let us consider the given expression
 $z={{\left( \dfrac{\sqrt{3}}{2}+\dfrac{i}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}}{2}-\dfrac{i}{2} \right)}^{5}}$
We know that $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}$ and $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ and we replace this values in the given expression,
$z={{\left( \cos \left( \dfrac{\pi }{6} \right)+i\sin \left( \dfrac{\pi }{6} \right) \right)}^{5}}+{{\left( \cos \left( \dfrac{\pi }{6} \right)-i\sin \left( \dfrac{\pi }{6} \right) \right)}^{5}}$
We know that the Euler’s formula for the complex number, $\cos \theta +i\sin \theta ={{e}^{i\theta }}$
$z={{\left( {{e}^{i\dfrac{\pi }{6}}} \right)}^{5}}+{{\left( {{e}^{-i\dfrac{\pi }{6}}} \right)}^{5}}$
By using the indices formula ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$ and we have
\[z={{e}^{i\dfrac{5\pi }{6}}}+{{e}^{-i\dfrac{5\pi }{6}}}\]
Again, applying the Euler’s formula of a complex number in the reserve order ${{e}^{i\theta }}=\cos \theta +i\sin \theta $
\[z=\left[ \cos \left( \dfrac{5\pi }{6} \right)+i\sin \left( \dfrac{5\pi }{6} \right) \right]+\left[ \cos \left( \dfrac{5\pi }{6} \right)-i\sin \left( \dfrac{5\pi }{6} \right) \right]\]
Rearrange the term $\dfrac{5\pi }{6}=\pi -\dfrac{\pi }{6}$ in the above expression, we get
\[z=\left[ \cos \left( \pi -\dfrac{\pi }{6} \right)+i\sin \left( \pi -\dfrac{\pi }{6} \right) \right]+\left[ \cos \left( \pi -\dfrac{\pi }{6} \right)-i\sin \left( \pi -\dfrac{\pi }{6} \right) \right]\]
Applying the allied angle formulas $\cos \left( \pi -\dfrac{\pi }{6} \right)=-\cos \dfrac{\pi }{6}$ and $\sin \left( \pi -\dfrac{\pi }{6} \right)=\sin \dfrac{\pi }{6}$, we get
\[z=\left[ -\cos \left( \dfrac{\pi }{6} \right)+i\sin \left( \dfrac{\pi }{6} \right) \right]+\left[ -\cos \left( \dfrac{\pi }{6} \right)-i\sin \left( \dfrac{\pi }{6} \right) \right]\]
Rearranging the terms, we get
\[z=-\cos \left( \dfrac{\pi }{6} \right)+i\sin \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right)-i\sin \left( \dfrac{\pi }{6} \right)\]
Cancelling the term $i\sin \left( \dfrac{\pi }{6} \right)$ on the right side, we get
\[z=-\cos \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right)\]
\[z=-2\cos \left( \dfrac{\pi }{6} \right)\]
We have \[\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\] and replace this value in the above expression, we get
\[z=-2\times \dfrac{\sqrt{3}}{2}\]
\[z=-\sqrt{3}\]
It can be also written as
\[z=-\sqrt{3}+i0\]
Now comparing this complex number with the general complex number $z=x+iy$, we get
$x=-\sqrt{3}$ and $y=0$
Hence the imaginary part of the given expression is zero.
Therefore, the correct option for the given question is option (b).

Note: Alternatively, the question is solved as follows
The polar form the complex number is $z=r\left( \cos \theta +i\sin \theta \right)$ where ${{r}^{2}}=\dfrac{3}{4}+\dfrac{1}{4}=1$ and $\theta =\dfrac{\pi }{6}$. Therefore$z={{r}^{5}}\left[ {{e}^{i5\theta }}+{{e}^{-i5\theta }} \right]$, put r = 1 and by using De-Moivre’s theorem. This gives $z=2\cos 5\theta $. Hence it is a real number that means the imaginary part in the complex number is zero.