$x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ real? Question: For $x,y \in \mathbb{C}$, suppose $x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ necessarily real?
For instance, if $x+y=xy=3$, then one solution is $x = \frac{3 \pm i \sqrt{3}}{2}$, $y = \frac{3 \mp i \sqrt{3}}{2}$, but $x^3 + y^3 = 0$, which is real. 
I've checked this numerically for many values of $w$ that give complex $x$ and $y$ (namely, $w \in (0,4)$.)

Yes. Write $x=a+bi$, $y=c-di$, then clearly $b=d$ because $x+y$ is real. 
So   $x=a+bi$, $y=c-bi$. Then $xy=ac-abi+cbi+b^2$, so $a=c$ or $b=0$.


*

*If $a=c$, then $y = \overline{x}$, i.e. the complex conjugate of $x$.
So $x^w+y^w = x^w+(\overline{x})^w = x^w+\overline{x^w}$, which is
real.

*If $b=0$, $x$ and $y$ are real so $x^w+y^w$ is real as $w>0$.
