Question

1-underroot 3iota÷1+underroot3iota rationlise it please

Answer 1

$\frac{1 - \sqrt{3} i}{1 + \sqrt{3} i} \\ \\ = \frac{1 - \sqrt{3} i}{1 + \sqrt{3} i} \times \frac{1 - \sqrt{3} i}{1 - \sqrt{3} i}$

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

$= \frac{ 1 - 3 - 2 \sqrt{3}i }{1 + 3} \\ \\ = \frac{ - 2 - 2 \sqrt{3}i }{4}$

$= - \frac{ 1 + \sqrt{3}i }{2}$