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A few months ago, I wrote down this set of simultaneous equations, hoping there might be a way to solve them:

$y+x^2=0$

$\frac{x^x}{2-\sqrt{y}}=\frac{1}{\sqrt{e^{\pi }}}$

I know that one solution to these equations is $x=i$ and $y=1$, but I’m struggling to prove if this is the only solution or if others might exist. I keep hitting a wall, especially with this expression that frequently comes up:

$ax^2+bx=c$

where $a,b$ and $c$ are constants. In the simplest case when $a=1$ and $b=0$, the equation reduces to $x^x=c$, and I’ve been able to solve that using the Lambert W function:

When $x^x=c$

$x=e^{W\left(ln\left(e\right)\right)}$

However, I’m not a big fan of the Lambert function; it often feels like a trick without general applicability, and I haven't managed to get further.

That’s why I’m reaching out to the StackExchange community. Could anyone help verify if $x=i$ and $y=1$ is indeed the only solution to the original system? And if possible, could someone suggest a method or tool for handling expressions like $ax^x+bx+c$ more generally?

Lastly, if the equations above are too complex or unsolvable, could they be modified without losing their essential properties to make them more approachable?

Thank you in advance for any insights!

Anne Bauval
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Ruchin Himasha
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    May be a typo somewhere. If $y=1$ the first equation gives $x^2=0$ – Claude Leibovici Nov 04 '24 at 11:14
  • @ClaudeLeibovici Thank you for pointing out. There were indeed to typos in. corrected them. – Ruchin Himasha Nov 04 '24 at 12:30
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    The thing about the Lambert function is it's just a name for a useful "inverse function" that let's you write down solutions of some equations which cannot otherwise be solved using other functions already learned. That's not unlike the square root function, an inverse function named so that you can solve $x^2=a$. Or the natural logarithm function, an inverse function named so that you can solve $e^x=a$. – Lee Mosher Nov 04 '24 at 12:59
  • Make sure your expressions are well-defined. $x^x$ can only make sense if $x$ is real positive. For example, $i^i$ doesn't exist (or you have to define it correctly). Similarly, $\sqrt{y}$ only makes sense when $y$ is real non-negative, or else you don't have a canonical way to define it. – Cactus Nov 04 '24 at 12:59
  • @Cactus x for this equation equals i. So if imaginary numbers are allowed should I add those restrictions? – Ruchin Himasha Nov 04 '24 at 13:10
  • If you allow non real complex numbers, you must make sure that your equation makes sens. What is $i^i$ ? You can define $i^n$ when $n$ is a non-negative integer by iteration of the product and $i^n$ when $n$ is a negative integer as $1/(i^{-n})$.

    If $a$ is a positive real number and $z$ any complex number, you can also define $a^z$ as $\exp(z\ln(a))$ where $\exp(w) = \sum_{n \geqslant 0} \frac{w^n}{n!}$ when $w$ is a complex number. However, there is no good way to extend powers to any couple of complex numbers $z^w$.

     – Cactus Nov 04 '24 at 13:39
  • You can make arbitrary choices for example, like saying that $(re^{i\theta})^z$ equals $r^ze^{i\theta z}$ with $\theta$ chosen in the interval $]-\pi,\pi]$. However, if you change the value of $\theta$ by a multiple of $2\pi$, it will change the result (unless $z$ is a relative integer). – Cactus Nov 04 '24 at 13:41
  • @Cactus There is a standard notion of complex exponentiation. It's called "the principal value" of $z^w$. It doesn't have all the nice properties that the real counterpart, but it's useful and there is universal consensus on which of values of the multivalued $\log$ use in the definition. – jjagmath Nov 05 '24 at 03:36
  • This complex plot shows infinite solutions for $x$. $ax^x+bx+c=0$ is partly covered here is – Тyma Gaidash Nov 05 '24 at 16:38

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