Question

`(8-i)/(3-2i)` If the expression above is rewritten in the form a+bi, where a and b are real numbers, what is the value of a? (Note:
`i=√(-1)`

Answer 1

a = 2

b = 1

Explanation:

`(8-i)/(3-2i)`

Write the fraction in this form:

`(a+bi)/(c+di)\:=\:(\left(c-di\right)\left(a+bi\right))/(\left(c-di\right)\left(c+di\right))=\:(\left(ac+bd\right)+\left(bc-ad\right)i)/(c^2+d^2)`

`(\left(8(3)+-1(-2)\right)+\left(-1(3)-8(-2)\right)i)/(3^2+-2^2)`

Evaluate.

`(26+13i)/(13)`

Factor the numerator.

`(13\left(2+i\right))/(13)`

`2+1i`

Answer 2

a = 2 , b = 1

Explanation:

`(8-i)/(3-2i)*(3+2i)/(3+2i)`

=>
`((8-i)(3+2i))/(9+4)`

=>
`(24+13i-2i^2)/(13)`

=>
`(26+13i)/(13)`

Comparing it with a+bi

a = 26/13 , b = 13/13

a = 2, b = 1