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

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  • $\begingroup$ @PatrickStevens Then $x+y$ isn't real. $\endgroup$ – wythagoras Aug 21 '16 at 10:37
  • $\begingroup$ @PatrickStevens But he required that $x+y=xy\in\Bbb R$.. $\endgroup$ – BigbearZzz Aug 21 '16 at 10:37
  • $\begingroup$ Sorry all. I wasn't paying attention, clearly. $\endgroup$ – Patrick Stevens Aug 21 '16 at 10:38
  • $\begingroup$ The answer is yes; the essential idea is that $xy = x+y \in \mathbb{R}$ forces either that $x,y \in \mathbb{R}$, or that $x$ and $y$ are complex conjugates. $\endgroup$ – Drew N Aug 22 '16 at 2:22
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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 $a \in \mathbb{R}$.

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  • $\begingroup$ You do need $x,y\geq0$, for your last statement. $\endgroup$ – wythagoras Aug 21 '16 at 10:54
  • $\begingroup$ @6005: Even with $a>0$, you still need $x,y\not\le 0$. Consider $x=y=-1\in\mathbb R$ and $a=1/2>0$. Then no matter where you do the branch cut, either $x^a=y^a=i$ or $x^a=y^a=-i$. Basically, the equation $\overline x^a = \overline{x^a}$ is nor true for all $x$ if $a$ is not integer. $\endgroup$ – celtschk Aug 21 '16 at 11:06
  • $\begingroup$ As a second conjecture, I believe the casework depends on the value of $w$. That is, in fact $r = 0$ if and only if $0< w < 4$, and $x \in \mathbb{R}$ if and only if $w \geq 4$ ($w = 0$ is excluded by assumption). $\endgroup$ – Drew N Aug 21 '16 at 11:28
  • $\begingroup$ @celtschk Thanks for your valuable feedback. I believe I fixed all the problems and provided all the necessary caveats. $\endgroup$ – 6005 Aug 21 '16 at 11:48
  • $\begingroup$ @DrewN your second conjecture is correct. $\endgroup$ – 6005 Aug 21 '16 at 11:55
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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 $w>0$.

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  • $\begingroup$ Thanks for your feedback on my answer earlier. In your answer, note that $(\overline{x})^w = \overline{x^w}$ is only true for a particular branch of complex log, and will never be true for negative real $x$. However, you can guarantee it for all $x$ which aren't negative real by picking the relevant branch of complex log to define your exponential. $\endgroup$ – 6005 Aug 21 '16 at 11:53
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$x+y$ 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

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  • $\begingroup$ Actually, $x = \overline{(\overline{x})^w}$ is a bit more complicated; it's not sufficient that $w$ is real. We have to pick a definition of complex exponential that allows for the symmetry that you appeal to. But regardless of which definition we take (branch of complex log), $x^w = \overline{(\overline{x})^w}$ will be false for $x$ negative real. Consider $w = \tfrac12$. $\endgroup$ – 6005 Aug 21 '16 at 11:51
  • $\begingroup$ Yeah. That is indeed true. I'll add in my answer to look at yours for a more detailed explanation on when/why this works. $\endgroup$ – b00n heT Aug 21 '16 at 12:17

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