Q:

Finding all values x and y of with complex numbers pt.2

Thank you for your help last time. This time, the problem is different. It is:
$$\frac{x+yi+2+3i}{2x+2iy-3} = i+2$$
x and y are assumed to be real numbers.
Basically my first trouble is knowing the conjugate of the denominator. Would it be $2x-2iy-3$ or $2x-2iy+3$? Secondly, I am unsure how to multiply the numerator by the denominator. I tried $(x+2)+i(y+3)\cdot (2x-3)+2iy$ or $(x+2)i(y+3) \cdot (2x+3) + 2iy $ depending on what the conjugate of the denominator is. Are either of these on the correct path?

A:

Write $2x+2iy-3$ as $(2x-3)+2yi$, so its conjugate is $(2x-3)-2yi$, but you don't need to multiply $(2x-3)+2yi$ by its conjugate. Just note that $$\frac{x+yi+2+3i}{2x+2iy-3} = i+2 \implies x+yi+2+3i=(2x+2iy-3)(i+2).$$
If we write every term in the form $a+bi$ we get $(x+2)+(y+3)i=((2x-3)+2yi)(2+i)$, then $(x+2)+(y+3)i=(4x-2y-6)+(2x+4y-3)i$. This lead us to the system:
\begin{cases}
x + 2 = 4x-2y-6 \\
y+3=2x+4y-3
\end{cases}
with solution $x=\frac{36}{13}$ and $y=\frac{2}{13}$.

