How do you divide # (6-i) / (9-4i) #?
1 Answer
Jun 1, 2016
Explanation:
To divide a complex number by another , we require to
#color(blue)"rationalise the denominator"# ie. make it a real value.This is achieved by mutipling the numerator and denominator by the
#color(blue)"complex conjugate"# of the denominator.Given a complex number z = x ± iy then the conjugate is x ∓ iy
Note that the values of x and y remain unchanged but the 'sign' changes.
#rArr(x+iy)(x-iy)=x^2+y^2 " a real number"# We are also making use of the fact
#[i^2=(sqrt(-1))^2=-1]#
#rArr(6-i)/(9-4i)=(6-i)/(9-4i)xx(9+4i)/(9+4i)=((6-i)(9+4i))/((9-4i)(9+4i))# now expanding numerator/denominator
#rArr(54+15i-4i^2)/(81-16i^2)=(58+15i)/(81+16)=58/97+15/97 i#
