Question

convert-16÷1+iroot3 into a+ib form ​

Answer 1

Question :-

$\sf convert \: into \: (a + \iota \: b )\: form \:, to \: \frac{16}{1 + \iota \sqrt{3} }$

Answer :-

$\to \boxed{ 4 + \iota \:( - 4 \sqrt{3} )} \\ \\ \sf it \: is \: in \: the \: form \: of \: a + \iota \: b \\ \\ \sf here \: a = 4 \: \: and \: b = - 4 \sqrt{3} \:$

Solution :-

We have ,

$\to \: \frac{16}{1 + \iota \sqrt{3} } \\ \\ \sf \: rationalising \: by \: 1 - \iota \sqrt{3} \\ \\ \to \: \frac{16}{1 + \iota \sqrt{3} } \times \frac{ 1 - \iota\sqrt{3} }{ 1- \iota \sqrt{3} } \\ \\ \to \: \frac{16(1 - \iota \sqrt{3}) }{(1 + \iota \sqrt{3} )(1 - \iota \sqrt{3} )} \\ \\ \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} } \\ \therefore \\ \to \: \frac{16(1 - \iota \sqrt{3} )}{ {(1)}^{2} - {( \iota \sqrt{3} )}^{2} } \\ \\ \to \: \frac{16(1 - \iota \sqrt{3}) }{1 - 3 { \iota}^{2} } \\ \\ \because \: \iota = \sqrt{ - 1} \\ \therefore \: { \iota}^{2} = - 1 \\ \\ \to \: \frac{16(1 - \iota \sqrt{3} )}{1 - ( - 3)} \\ \\ \to \: \frac{16(1 - \iota \sqrt{3} )}{1 + 3} \\ \\ \to \: \frac{16(1 - \iota \sqrt{3}) }{4} \\ \\ \to \: 4(1 - \iota \sqrt{3} ) \\ \\ \to \boxed{ 4 + \iota \:( - 4 \sqrt{3} )} \\ \\ \sf it \: is \: in \: the \: form \: of \: a + \iota \: b \\ \\ \sf here \: a = 4 \: \: and \: b = - 4 \sqrt{3}$