Question

this is a hard question pls answer tho

Answer 1

Answer:

A

Step-by-step explanation:

Given,

           i = √( - 1 )

      ⇒ i^2 = ( √-1 )^2

      ⇒ i^2 = - 1

Here,

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

   Divide as well as multiply by 3 + 2i:

$\implies\dfrac{(8-i)(3+2i)}{(3-2i)(3+2i)}\\\\\\\implies\dfrac{26+16i-3i+</p><p>-2i^2}{(3)^2-(2i)^2}$

In the last step : ( 8 - i )( 3 + 2i ) = 24 + 16i - 3i - 2i^2

     ( 3 - 2i )( 3 + 2i ) = ( 3 )^2 - ( 2i )^2

$\implies\dfrac{24+13i-2i^2}{9-4(i)^2}$

     As discussed above, i^2 = - 1

$\implies\dfrac{24+13i-2(-1)}{9-4(-1)}\\\\\\\implies\dfrac{24+13i+2}{9+4}\\\\\\\implies\dfrac{24+13i+2}{13}\\\\\\\implies\dfrac{26+13i}{13}$

⇒ 2 + i

Now,

         2 + i = a + bi

      ⇒ a = 2    ;    bi = i

      ⇒ a = 2    ; b = 1

  Required value of a is 2.

         Option A