Question

Rationalize the denominator of square root of negative 9 over open parentheses 4 minus 7 i close parentheses minus open parentheses 6 minus 6 i close parentheses.

Answer 1

The answer is
$\frac{-6i-3}{5}$.

Before we rationalize the denominator, we must simplify the numerator and denominator.  We evaluate the square root on the top, and combine like terms on the bottom:
$\frac{\sqrt{-9}}{(4-7i)-(6-6i)} \\ \\=\frac{\sqrt{9i^2}}{4-6-7i--6i} \\ \\=\frac{3i}{-2-i}$

To rationalize the denominator, we multiply by the conjugate.  The conjugate is found by making the imaginary term of the complex number, the i part, the opposite sign.  The conjugate of -2-i is -2+i:

$\frac{3i}{-2-i}\times \frac{-2+i}{-2+i} \\ \\=\frac{3i(-2+i)}{(-2-i)(-2+i)} \\ \\=\frac{3i\times -2 + 3i\times i}{-2\times -2 + -2\times i + -2\times -i + -i\times i} \\ \\=\frac{-6i+3i^2}{4-2i+2i-i^2} \\ \\=\frac{-6i+3(-1)}{4-i^2}=\frac{-6i-3}{4-(-1)}=\frac{-6i-3}{5}$