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Answer 1

Answer :

• a = 2.
• b = 1.

SolutioN :

$\tt \rightarrow \: \dfrac{8 - i}{3 - 2i}$

$\tt \bullet \: \: \: \: \: (a + b)(a - b) = {a}^{2} - {b}^{2}.$

$\tt \rightarrow \: \dfrac{8 - i}{3 - 2i} \times \frac{3 + 2i}{3 +2 i}$

$\tt \rightarrow \: \dfrac{(8 - i)(3 + 2i)}{ {(3)}^{2} -{( 2i)}^{2} }$

$\tt \rightarrow \: \dfrac{24 + 16i - 3i - 2 {i}^{2} }{ 9 -{ 4i}^{2} }$

$\tt \bullet \: \: \: \: \: Note : \: i = \sqrt{ -1}.$

$\tt \rightarrow \: \dfrac{26 + 13i}{ 13 }$

⚝ Compare with a + bi form.

• a = 26 / 13 = 2.
• bi = 13i / 13 = 1.

Therefore, the value of a is 2 and b is 1.

Answer 2

Given :-

8 - i/3 - 2i. If the expression above is rewritten in the form of a + bi, where a and b are real number

To Find :-

Value of a

Solution :-

$\sf\dfrac{8-i}{3-2i}$

We may rationalize the denominator by using the algebric identity

(a + b)(a - b) = a² - b²

$\sf\dfrac{8-i}{3-2i}\times\dfrac{3-2i}{3-2i}$

$\sf\dfrac{(8-i)(3-2i)}{(3-2i)(3-2i)}$

$\sf \dfrac{24 + 16i - 3i - 2i^2}{(3)^2-(2i)^2}$

$\sf\dfrac{24+16i-3i-2i^2}{9 - 4i^2}$

Using i = √-1

$\sf \dfrac{24+16(\sqrt{-1}) - 3(\sqrt{-1}) - 2(\sqrt{-1})^2}{9 - 4(\sqrt{-1}^2)}$

$\sf \dfrac{24+\sqrt{-16} + \sqrt{3}+ 2(1)}{9 - 4(-1)}$

$\sf \dfrac{24 + \sqrt{-16} + \sqrt{3} +2}{9 - (-4)}$

$\sf \dfrac{26 + \sqrt{-16} + \sqrt{3} }{13}$

$\sf\dfrac{26 + \sqrt{-13}}{13}$

Using √-13 = 13i as 13 × √-1 = √-13

$\sf\dfrac{26+13i}{13}$

On comparing with given form i.e a + bi

We get

$\sf \dfrac{a+bi}{2}$

$\sf \dfrac{26+13i}{13}$

$\sf 2 = a$