How do you divide #(4-9i)/(-6i)#?

1 Answer
Jul 25, 2016

Answer:

#3/2+2/3i#

Explanation:

We require to multiply the numerator and denominator of the fraction by the #color(blue)"complex conjugate"# of the denominator. This will eliminate the i from the denominator, leaving a real number.

If z = a ± bi then the #color(blue)"complex conjugate"# is

#color(red)(|bar(ul(color(white)(a/a)color(black)(barz=a∓bi)color(white)(a/a)|)))#
Note that the real part remains unchanged while the 'sign' of the imaginary part is reversed.

The complex number on the denominator is 0 - 6i , hence the conjugate is 0 + 6i = 6i

Multiply numerator and denominator by 6i

#(4-9i)/(-6i)xx(6i)/(6i)=(6i(4-9i))/(-36i^2)=(24i-54i^2)/(-36i^2)#

#color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(i^2=(sqrt(-1))^2=-1)color(white)(a/a)|)))#

#rArr(4-9i)/(-6i)=(24i+54)/36=54/36+24/36i=3/2+2/3i#