Question

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

Answer 1

$$\begin{gathered} \mathrm{for}\frac{\sqrt{3}}{2}+\frac{i}{2} \\ r=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\mathrm{and}\theta ={\tan }^{-1}\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}={\tan }^{-1}\frac{1}{\sqrt{3}} \\ \mathrm{i}.\mathrm{e}.\theta =\frac{\mathrm{\pi }}{6} \\ \mathrm{i}.\mathrm{e}.\frac{\sqrt{3}}{2}+\frac{i}{2}=r(\cos \theta +i\sin \theta ) \\ \mathrm{we}\mathrm{have}\theta =\frac{\mathrm{\pi }}{6}\mathrm{and}r=1 \\ \mathrm{and}\frac{\sqrt{3}}{2}-\frac{i}{2}=r\left(\cos \theta -i\sin \theta \right) \end{gathered}$$

$\begin{array}{rcl}\therefore z& =& {\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^5+{\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)}^5\\ & =& {\left[r\left(\cos \theta +i\sin \theta \right)\right]}^5+{\left[r(\cos \theta -i\sin \theta )\right]}^5\mathrm{where}r=1,\theta =\frac{\pi }{6}\\ & =& {\left(\cos \theta +i\sin \theta \right)}^5+{\left(\cos \theta -i\sin \theta \right)}^5\\ & =& \cos 5\theta +i\sin 5\theta +\cos 5\theta -i\sin 5\theta (\mathrm{By}\mathrm{De}-\mathrm{moivre}\mathrm{theorem})\\ & =& 2\cos 5\theta \end{array}$