Question

Express in the form of a+ib in 5-√3i÷4+2√3i

Answer 1

5-√3i÷4+2√3i
={(5-√3i)/(4+2√3i)} X{ (4-2√3i)/(4-2√3i)}
={(5-√3i)(4-2√3i)}/( 16 + 12 )
= (20 - 10√3i -4√3i +6i^2)/28
= (14 - 14√3i )/28
= 1/2 - √3/2i

Answer 2

Answer:

Required answer is ($\frac{1}{2} - \frac{ \sqrt{3} }{2} i$)

Step-by-step explanation:

Here given,

$\frac{5 - \sqrt{3}i }{4 + 2 \sqrt{3} i}$

We want to convert it into a+ib form.Where a is real part and b is imaginary part.

Now,in given fraction denominator is( $4 + 2 \sqrt{3} i$)and numerator is ( $5 - \sqrt{3} i$)

We are multiplying denominator and numerator by $(4 - 2 \sqrt{3} i)$and get,

$\frac{(5 - \sqrt{3}i )(4 - 2 \sqrt{3} i)}{(4 + 2 \sqrt{3} i)(4 - 2 \sqrt{3} i)}$

$= \frac{20 - 10 \sqrt{3}i - 4 \sqrt{3}i + 6 {i}^{2} }{ {4}^{2} - {(2 \sqrt{3} i)}^{2} } = \frac{20 - 14 \sqrt{3} i - 6}{16 + 12} = \frac{14 - 14 \sqrt{3} i}{28} = \frac{14}{28} - \frac{14 \sqrt{3} i}{28} = \frac{1}{2} - \frac{ \sqrt{3} }{2} i$

So,we are comparing ($\frac{1}{2} - \frac{ \sqrt{3} }{2} i$) with( $a + bi$) and get $a = \frac{1}{2}$and

$b = - \frac{ \sqrt{3} }{2}$

Here applied formula is ${i}^{2} = - 1$