To simplify a number with a complex denominator, we need to utilize the complex conjugate. For any complex number #a+bi#, its complex conjugate, #bar (a+bi)#, is the complex number #a-bi#. The conjugate is remarkably handy in turning complex denominators into real ones, which, in this case, will allow us to break apart our fraction into a real and imaginary part of the form #a+bi#.
When we multiply #a+bi# by #bar (a+bi)#, we can remember the fact that the particular case of the binomials #(a+b)(a-b)# has the product #a^2-b^2#. Since #i# has the special property that #i^2=-1#, #(a+bi)(a-bi)# actually ends up giving us #a^2+b^2#; this is especially convenient, as the imaginary part is completely eliminated in the product.
Turning out attention back to the problem, we'd like to turn the denominator into a real number, but we can't change the value of our number as a whole, so we'll have to multiply it by some form of 1. To get our denominator #2-3i# into the right form, we'll choose the fraction #bar(2-3i)/bar(2-3i)#, or #(2+3i)/(2+3i)#.
Multiplying that out, we get:
#(7i)/(2-3i)*(2+3i)/(2+3i)=(7i(2+3i))/((2-3i)(2+3i))#
The denominator, as mentioned before, becomes #2^2+3^2=13#, and distributing the #7i# to the #2# and the #3i# in the numerator gets us #14i-21#. Altogether, we have
#(14i-21)/13#
which we can break apart into
#(14i)/13-21/13#
and rearrange into the form #a+bi#:
#-21/13+14/13i#
