A real value of x satisfies the equation `(3-4ix)/(3+4ix)=alpha-ibeta(alpha,beta in R)`, if `alpha^2+beta^2=`A. ` x = 2n +1`B. `x = 4n`C. `x = 2n`D. `x = 4n + 1`

A real value of x satisfies the equation `(3-4ix)/(3+4ix)=alpha-ibeta(

1.	A real value of x satisfies the equation `(3-4ix)/(3+4ix)=alpha-ibeta(alpha,beta in R)`, if `alpha^2+beta^2=`A. ` x = 2n +1`B. `x = 4n`C. `x = 2n`D. `x = 4n + 1`
Answer» Correct Answer - A Given equation, ` ((3 - 4 ix)/(3 + 4 ix )) = alpha - i beta ( alpha , beta in R ) ` `rArr [(3 - 4 ix)/(3 + 4 ix )] = alpha - i beta` Now, ` (alpha -ibeta ) =((3 -4ix)(3 - 4ix))/((3 + 4ix)(3 - 4 ix) )=(9+ 26i^(2) x^(2) - 24ix)/(9 - 16i^(2)x^(2))` `rArr (alpha -ibeta ) =(9+ 26i^(2) x^(2) - 24ix)/(9 - 16i^(2)x^(2))` `rArr alpha - ibeta = (9- 16x^(2))/(9 + 16x^(2) - i24x) /(9 + 16x^(2)` `rArr alpha - ibeta = (9 - 16x^(2))/(9 + 16x^(2)) - (i24x)/(9 + 16x^(2)) ...(i)` `:." " alpha + ibeta = (9 - 16x^(2))/(9 + 16x^(2)) - (i24x)/(9 + 16x^(2)) ...(ii)` So, `(alpha - ibeta)(alpha + i beta)= ((9 - 16x^(2))/(9 + 16x^(2))^(2)) - ((i24x)/(9 + 16x^(2)))^(2)` `alpha^(2) + beta^(2) = (81+ 256x^(4) - 288x^(2) + 576x^(2))/((9 + 16x^(2))^(2))` `(( 9 + 16x^(2))^(2))/(9 + 16x^(2) )^(2) = 1`