Complex exercise to simplify given complex number as possible you can

Exercise:
Reduce this complex number $z=\frac{{{\left[ \sqrt{3}\left( \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right) \right]}^{4}}}{{{\left( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} \right)}^{2}}}$ as possible you can

Solution: Let ${{z}_{1}}=\sqrt{3}\left( \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right)\Leftrightarrow {{z}_{1}}=\sqrt{3}{{e}^{i\frac{\pi }{6}}}$

So ${{z}^{4}}_{1}={{\left( \sqrt{3}{{e}^{i\frac{\pi }{6}}} \right)}^{4}}=9{{e}^{i\frac{4\pi }{6}}}=9{{e}^{i\frac{2\pi }{3}}}$

Also put ${{z}_{2}}=\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}=\cos \left( \frac{-\pi }{4} \right)+i\sin \left( \frac{-\pi }{4} \right)={{e}^{-i\frac{\pi }{4}}}$

So ${{z}_{2}}^{2}={{e}^{-i\frac{2\pi }{4}}}={{e}^{-i \frac{\pi }{2}}}$

Hence $z=\frac{{{z}_{1}}^{4}}{{{z}_{2}}^{2}}=\frac{9{{e}^{i\frac{2\pi }{3}}}}{{{e}^{-i\frac{\pi }{2}}}}=9{{e}^{i\frac{2\pi }{3}}}\times {{e}^{i\frac{\pi }{2}}}=9{{e}^{i\frac{7\pi }{6}}}$ But $\frac{7\pi }{6}=\frac{12\pi -5\pi }{6}=2\pi -\frac{5\pi }{6}$ P.V angle

Thus $z=9{{e}^{i\frac{-5\pi }{6}}}=9\left( \cos \frac{-5\pi }{6}+i\sin \frac{-5\pi }{6} \right)=9\left( \cos \left( \frac{\pi }{6}-\pi  \right)+i\sin \left( \frac{\pi }{6}-\pi  \right) \right)=9\left( -\frac{\sqrt{3}}{2}-i\frac{1}{2} \right)$