Question 149993
The idea behind simplifying expressions containing complex numbers in their denominator, such as you have here, is to change the denominator into a "real" number.
You do this by multiplying the top and bottom of your expression by the complex conjugate of the denominator.
The conjugate of a complex number (a+bi) is simply (a-bi), so, in your problem,...
{{{(6-i)/(2+3i)}}} The denominator is the complex number (2+3i) and its conjugate is (2-3i), so let's proceed...
{{{(6-i)highlight((2-3i))/(2+3i)highlight((2-3i))}}} ...as you can see, we are really multiplying your original expression by {{{(2-3i)/(2-3i) = 1}}} and so its value does not change.
Let's perform the indicated multiplication using the FOIL method.
{{{((6*2)+(6*(-3i))+(2(-i))+((-2)(-3i)))/((2*2)+(2*(-3i))+(2*3i)+((3i)(-3i)))}}} Simplifying all of this and recalling that {{{i^2 = -1}}}, we get:
{{{(12-18i-2i+3i^2)/(4-6i+6i-9i^2)}}}={{{(9-20i)/13}}}