Question

What is ${\left[\frac{\sin \frac{\pi }{6}+i(1-\cos \frac{\pi }{6})}{\sin \frac{\pi }{6}-i(1-\cos \frac{\pi }{6})}\right]}^2$, where $i=\sqrt{-1}$ equal to?

Answer 1

We have,

${\left[\frac{\sin \frac{\pi }{6}+i(1-\cos \frac{\pi }{6})}{\sin \frac{\pi }{6}-i(1-\cos \frac{\pi }{6})}\right]}^2$

Now,

${\left[\frac{\frac{1}{2}+i(1-\frac{\sqrt{3}}{2})}{\frac{1}{2}-i(1-\frac{\sqrt{3}}{2})}\right]}^2$

$\Rightarrow \frac{{[\frac{1}{2}+i(1-\frac{\sqrt{3}}{2})]}^2}{{[\frac{1}{2}-i(1-\frac{\sqrt{3}}{2})]}^2}$

$\Rightarrow \frac{\frac{1}{4}+i^2{(1-\frac{\sqrt{3}}{2})}^2+2\times \frac{1}{2}i(1-\frac{\sqrt{3}}{2})}{\frac{1}{4}+i^2{(1-\frac{\sqrt{3}}{2})}^2-2\times \frac{1}{2}i(1-\frac{\sqrt{3}}{2})}\therefore i^2=-1$

$\Rightarrow \frac{\frac{1}{4}-1(1^2+\frac{3}{4}-2\times \frac{\sqrt{3}}{2})+i(1-\frac{\sqrt{3}}{2})}{\frac{1}{4}-1(1^2+\frac{3}{4}-2\times \frac{\sqrt{3}}{2})-i(1-\frac{\sqrt{3}}{2})}$

$\Rightarrow \frac{\frac{1-4-3+4\sqrt{3}+2i(2-\sqrt{3})}{4}}{\frac{1-4-3+4\sqrt{3}-2i(2-\sqrt{3})}{4}}$

$\Rightarrow \frac{-6+4\sqrt{3}+2i(2-\sqrt{3})}{-6+4\sqrt{3}-2i(2-\sqrt{3})}\therefore i=\sqrt{-1}$

$\Rightarrow \frac{-6+4\sqrt{3}+2\sqrt{-1}(2-\sqrt{3})}{-6+4\sqrt{3}-2\sqrt{-1}(2-\sqrt{3})}$

Hence, this is the answer.