$x+y=xy=w \in \mathbb{R}^+$. 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, $w \in (0,4)$.)
 A: Yes.
Since $x + y \in \mathbb{R}$,
$y = \overline{x} + r$ for some $r \in \mathbb{R}$.
Then $xy = |x|^2 + xr \in \mathbb{R}$, implying that either $r = 0$ or $x \in \mathbb{R}$. Then we do casework:


*

*If $r = 0$, then $y = \overline{x}$; this leads to 
$$x^w + y^w = x^w + \overline{x^w} \in \mathbb{R}.$$
Warning: for this to work, we had to pick the standard branch of the complex logarithm, specifically, the one undefined on the nonpositive real line, whose imaginary part is between $-\pi$ and $\pi$. Once we define $z^w := e^{w \ln z}$, ${(\overline{x})}^w = \overline{x^w}$ is true for this branch of $\ln$ (as $x$ is not nonpositive real), but might not be true for another branch.
This warning does not come into play when $w$ is an integer. But take, for example, $x = 1 + \frac{1 + i}{\sqrt{2}}$, $y = 1 + \frac{1 - i}{\sqrt{2}}$. Then $x + y = xy = 2 + \sqrt{2}$. If we picked a different branch of the complex logarithm, then we could have $x^w + y^w$ not real.

*On the other hand, if $x \in \mathbb{R}$, then $y \in \mathbb{R}$, so $x^w + y^w \in \mathbb{R}$. Since $x + y = xy > 0$, $x,y$ must both be positive, so we have no trouble with a negative base of the exponent.

Note there was nothing special about $w$: we could have reached the stronger conclusion that $x^a + y^a \in \mathbb{R}$ for all $a \in \mathbb{R}$.
A: 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 $w>0$.
A: $x+y$ real implies that $\Im(x)=-\Im(y)$. Thus if $x=a+bi$ then $y=c-bi$.
$xy$ real implies that, because $xy=ac+b^2+ib(c-a)$, $b=0\lor c=a$.
If $b=0$ the result is trivial as $x,y\in\mathbb{R}$.
If $c=a$ then $x=\overline{y}$. But then clearly 
$$x^w+y^w=x^w+\overline{x}^w=\overline{x^w+\overline{x}^w}=\overline{x^w+y^w},$$
where the middle equality holds by symmetry.
But then it must be real, as the only complex numbers satisfying $z=\bar{z}$ are real.
Note: for why/when we are allowed to write $x^w=\overline{\bar{x}^w}$ refer to  @6005's answer
