Prove that $\frac{\text{x+y}}{\text{r$^2$+xy}}\in \mathbb{R}$

Question

Let two complex numbers $\rm x$ and $\rm y$, satisfy $\rm |x|=|y|=r>0$ and $\rm r^2+xy \neq 0$.

Prove that $$\frac{x+y}{r^2+xy}\in \mathbb{R}$$

Answer 1

Since $$\bar xx=\bar yy=r^2,$$ $$u=\frac{x+y}{r^2+xy}=\frac{(r^2/\bar x)+(r^2/\bar y)}{r^2+xy}=\frac{r^2\bar y+r^2\bar x}{\bar x\bar y(r^2+xy)}=\frac{r^2(\bar x+\bar y)}{r^2(\bar x\bar y+r^2)}=\frac{\bar x+\bar y}{r^2+\bar x\bar y}=\bar u.$$

Answer 2

Let $\displaystyle x=r(\cos2A+i\sin2A), y=r(\cos2B+i\sin2B)$

Using Prosthaphaeresis Formulas
$\displaystyle x+y=r\left[\cos2A+\cos2B+i(\sin2A+\sin2B)\right]$ $\displaystyle\implies x+y=r\cdot2\cos(A-B)\underbrace{[\cos(A+B)+i\sin(A+B)]}\ \ \ \ (1)$

Now $\displaystyle xy=r^2[\cos(2A+2B)+i\sin(2A+2B)]$

$\displaystyle r^2+xy=r^2[1+\cos(2A+2B)+i\sin(2A+2B)]$

Using Double-Angle Formulas, $\displaystyle r^2+xy=r^2[2\cos^2(A+B)+i2\sin(A+B)\cos(A+B)]$

$\displaystyle\implies r^2+xy=r^2\cdot2\cos(A+B)\underbrace{[\cos(A+B)+i\sin(A+B)]}\ \ \ \ (2)$