Question

find the multiplicative inverse of 1 - root 3i​

Answer 1

$\huge\mathbb{ANSWER:}$

Given :

• 1-√3i

To find :

• Multiplicative inverse of 1-√3i

Solution :

• As we know multiplicative inverse is reciprocal of number.

So, Multiplicative inverse of complex number 1-√3i is

$\rm \longmapsto \: mi(1 - \sqrt{3}i) = \frac{1}{1 - \sqrt{3}i }$

• Simplifying by rationalising denominator

$\rm \longmapsto \frac{1}{1 - \sqrt{3}i }$

$\rm \longmapsto \frac{1}{1 - \sqrt{3}i } \times \frac{1 - \sqrt{3} i}{1 - \sqrt{3} i}$

$\rm \longmapsto \frac{1 + \sqrt{3} i}{(1 - \sqrt{3} i)(1 + \sqrt{3}i) }$

• IDENTITY USED :

$\red{ \bf \: (a + b)(a - b) = {a}^{2} - {b}^{2} }$

$\rm \longmapsto \frac{1 + \sqrt{3} i}{(1) {}^{2} - (\sqrt{3} i) {}^{2} }$

$\rm \longmapsto \frac{1 + \sqrt{3} i}{(1) - (\sqrt{3}) {}^{2} (i) {}^{2} }$

• i = iota = √-1

$\rm \longmapsto \frac{1 + \sqrt{3} i}{(1) - 3 ( \sqrt{ - 1} ) {}^{2} }$

$\rm \longmapsto \frac{1 + \sqrt{3} i}{(1) - 3 ( - 1 ) }$

$\rm \longmapsto \frac{1 + \sqrt{3} i}{1 + 3 }$

$\rm \longmapsto \frac{1 + \sqrt{3} i}{4 }$

$\longmapsto \boxed{\purple{ \bf\: mi(1 - \sqrt{3}i) = \frac{1 + \sqrt{3} i}{4 } }}$

Answer 2

Step-by-step explanation:

69 is the right answer....