Mathematics - Complex Number Question with Solution | TestHub

Given,

${\left(\frac{1+\sin \frac{2\mathrm{\pi }}{9}+i\cos \frac{2\mathrm{\pi }}{9}}{1+\sin \frac{2\mathrm{\pi }}{9}-i\cos \frac{2\mathrm{\pi }}{9}}\right)}^3$

Now let $z=\sin \frac{2\mathrm{\pi }}{9}+i\cos \frac{2\mathrm{\pi }}{9}$,

So, $\overline{z}=\sin \frac{2\mathrm{\pi }}{9}-i\cos \frac{2\mathrm{\pi }}{9}=\frac{1}{z}$

So, ${\left(\frac{1+\sin \frac{2\mathrm{\pi }}{9}+i\cos \frac{2\mathrm{\pi }}{9}}{1+\sin \frac{2\mathrm{\pi }}{9}-i\cos \frac{2\mathrm{\pi }}{9}}\right)}^3$

$={\left(\frac{1+z}{1+\overline{z}}\right)}^3$

$={\left(\frac{1+z}{1+{\displaystyle \frac{1}{z}}}\right)}^3$

$=z^3{\left(\frac{1+z}{1+{\displaystyle z}}\right)}^3$

$=z^3$

$={\left(\sin \frac{2\mathrm{\pi }}{9}+i\cos \frac{2\mathrm{\pi }}{9}\right)}^3$

$=i^3{\left(\cos \frac{2\mathrm{\pi }}{9}-i\sin \frac{2\mathrm{\pi }}{9}\right)}^3$

$=-i\left(\cos \left(3\times \frac{2\mathrm{\pi }}{9}\right)-i\sin \left(3\times \frac{2\mathrm{\pi }}{9}\right)\right)$

$=-i\left(\cos \frac{2\mathrm{\pi }}{3}-i\sin \frac{2\mathrm{\pi }}{3}\right)$

$=-i\left(\frac{-1}{2}-i\frac{\sqrt{3}}{2}\right)$

$=-\frac{1}{2}\left(\sqrt{3}-i\right)$