How do you divide #(2i)/(3-9i)#?

1 Answer
Sep 20, 2016

Answer:

#-1/5+1/15i#

Explanation:

Before we can divide, we require the denominator to be a real number.
This is achieved by multiplying the denominator by it's #color(blue)"conjugate"#

#"If "z=x±yi" is a complex number then it's conjugate is"#

#color(red)(bar(ul(|color(white)(a/a)color(black)(bar(z)=x∓yi)color(white)(a/a)|)))#

Note that the real part remains unchanged while the #color(red)"sign"# of the imaginary part is reversed.

Hence the conjugate of 3-9i is 3 + 9i

and if we multiply them.

#(3-9i)(3+9i)=9+27i-27i-81i^2=9-81i^2#

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

Then #9-81i^2=90larr" a real number"#

Since we have a fraction then the numerator/denominator must be multiplied by the complex conjugate.

#(2i)/(3-9i)=(2i(3+9i))/((3-9i)(3+9i))=(6i+18i^2)/90=(-18+6i)/90#

#=(-18)/90+6/90i=-1/5+1/15i#