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Does this equation have a solution in the complex domain?

$$ \sqrt{x+3} = 3 + \sqrt{x} $$

Squaring both sides gives $\sqrt{x}=-1$, which suggests the solution to be $x=1$. But:

  1. How can the complex domain solution be a purely real number?
  2. Does this have a link to the multivalued nature of the function ($\sqrt{x}$) in the complex domain?
  3. What is the correct way to deal with such equations involving multivalued functions in the complex domain?

Thanks in advance.

Mohammad Ali
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  • When you write $\sqrt{z}$ for a complex number $z$, you need to specify what "branch" you are using. Otherwise, you are interpreting your equality in the equation as a set identity. –  May 03 '22 at 22:03
  • Where is this equation from? Do you make it up? Or is it from some exercise? That sort of information allows a more appropriate way to write an answer. –  May 03 '22 at 22:05
  • Ok, thanks for your comment. Can you refer me to some detailed resource? I am not an expert in complex analysis. – Mohammad Ali May 03 '22 at 22:05
  • Indeed, I confronted it on the web, someone asked about it in a math help page. – Mohammad Ali May 03 '22 at 22:08
  • "which suggests the solution to be $x=1$". Why? This is incorrect. –  May 03 '22 at 22:14
  • What "detailed resource" are you looking for? You may begin with https://en.wikipedia.org/wiki/Square_root#Square_roots_of_negative_and_complex_numbers –  May 03 '22 at 22:15
  • Don't we have that by squaring both sides? – Mohammad Ali May 03 '22 at 22:16
  • I mean some resource to read about the "branch" you mentioned above. – Mohammad Ali May 03 '22 at 22:16
  • https://en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number –  May 03 '22 at 22:20
  • OK, I will read the wiki page. But, why can't we say that sqrt(x) = -1 suggests that x = 1? I didn't get this point. Could you explain? – Mohammad Ali May 03 '22 at 22:22
  • In complex analysis, square root is a multivalued function: the square roots of $1$ are $-1$ and $1$. So $\sqrt{1+3} = 3 + \sqrt{1}$ is correct in the sense that there are values of the square roots that make this true. It just looks strange because we usually think of the square root of a positive real as the positive square root of that real, not the negative one. – Robert Israel May 03 '22 at 22:34
  • @RobertIsrael Then, x = 1 is a correct solution? – Mohammad Ali May 03 '22 at 23:00

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