$x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ real?

Question

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, $w \in (0,4)$.)

Answer 1

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.

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Note there was nothing special about $w$: we could have reached the stronger conclusion that $x^a + y^a \in \mathbb{R}$ for all $a \in \mathbb{R}$.

Answer 2

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 $w>0$.

Answer 3

$x+y$ 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