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

1 Answer
Aug 24, 2016

Answer:

#-3/13+2/13 i#

Explanation:

The aim here is to obtain a real value on the denominator of the fraction.
This is achieved by multiplying the numerator and denominator by the #color(blue)"complex conjugate"# of the denominator.

Given a complex number #z=x ± yi then the complex conjugate is.

#bar(z)=x∓yi#

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

The conjugate of #2-3i" is " 2+3i#

and #(2-3i)(2+3i)=4-9i^2=4+9=13" a real number"#

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

We must multiply numerator/denominator by 2+3i

#rArri/(2-3i)xx(2+3i)/(2+3i)=(i(2+3i))/((2-3i)(2+3i))#

#=(2i+3i^2)/13=(-3+2i)/13=-3/13+2/13i#