I was just playing around with some equations the other day and losing interest I started writing down what were mostly "random" equations at first. But, I realized that there's something special about them, they don't have solutions! At least, as far as I could tell. On the surface it seems there should be solutions to them, seeing they look simple; at any rate, couldn't we just "define" objects which would satisfy these equations? Similar to what was done when complex numbers were introduced? I can't say; anywho, here is a sample of what I have:

$$ \tag{1} \sqrt{ix} = -1, \sqrt{ix} = -i, \sqrt{x} = -1 - i, \\ \sqrt{ix} = -1 - i, \sqrt{-ix} = -1 - i, \sqrt{-ix} = i - 1, \\ \sqrt{-ix} = -i $$

What do you think? Can we get any solutions? Can we define solutions? Thanks in advance.


I have to mention that I'm looking for solutions that would "force" the LHS of the equations to equal the RHS. I'm not an expert so I'm not sure if this is the right question to ask, but that's why I have it here :)

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    $\begingroup$ Write $x=re^{i\theta}$, $i=e^{i\pi/2}$ and substitute. $\endgroup$ – dls Apr 29 '12 at 18:35
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    $\begingroup$ All of these have solutions in complex numbers. A boring fact is that all equations which dont simplify to 0=1 have solutions in C. $\endgroup$ – TROLLHUNTER Apr 29 '12 at 18:35
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    $\begingroup$ "...all equations which don't simplify to $0=1$ have solutions in $\bf C$." How about $e^x=0$? $\endgroup$ – Gerry Myerson Apr 30 '12 at 5:51
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    $\begingroup$ @GerryMyerson Multiply both sides by $e^{-x}$ ;) $\endgroup$ – TROLLHUNTER Apr 30 '12 at 13:46
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    $\begingroup$ @Holowitz, you can't multiply by $e^{-x}$ unless you know that's not tantamount to dividing by zero, which means you have to know independently that $e^x=0$ has no solution. $\endgroup$ – Gerry Myerson May 1 '12 at 0:20

The question shows that you are rigorously thinking about these concepts, which is good. Let's take a simple example which is similar in concept to those equations you've mentioned:

$$\sqrt x = -1$$

Because it is convenient, a generally agreed upon custom of mathematics is to denote a principal square root of a number by the $\sqrt{}$ symbol. For complex numbers, this is defined as follows (see wikipedia):

Let $x=ze^{i\varphi}$ where $\varphi$ is the principal angle ($\varphi\in(-\pi,\pi]$), then $\sqrt x = \sqrt z e^{i \varphi/2}=\sqrt z [\cos \tfrac \varphi 2 + i\sin \tfrac \varphi 2]$.

From this definition, we can see that the real part of $\sqrt x$ must be positive. Thus the equation $\sqrt x = -1$ has no solution in the complex numbers, when $\sqrt {}$ is taken to be the principal root. If we wanted to define a new number field for which $\sqrt x = -1$ has a solution, say $x=j$, then we would quickly run into trouble, as squaring both sides leads to the conclusion that $j=1$. Thus new rules would have to be created regarding how to treat the multiplication of square roots in our new number system, and very likely such a system would not end up being consistent.

In addition, such an extension does not yield as useful a result as the complex numbers, since we find that $x=1$ is a solution if we consider the multivalued square root function instead of the principal square root.

The complex numbers were introduced in order to create a field in which all algebraic equations had solutions. A good place to continue your investigation would be this thread: http://www.physicsforums.com/archive/index.php/t-315935.html

A small comment about why I dropped the $1=0$ approach: Now, although some equations may be able to be transformed to $1=0$, and thus the original equation was false within the given field, this is not reason in and of itself to dissuade one from an extension of the number system, such as the complex were to the reals. This was pointed out to me by Gerry Myerson in the comments above. To see why, consider a situation where the only number system is the reals, as it was before complex numbers were introduced. Now one may write an equation such as $x^2+1=0$ and use the following logic: $x^2>0 \implies x^2+1>1 \implies x^2+1\ne0$ and now, dividing our original equation by $x^2+1$ is valid, since $x^2+1\ne0$, and hence we have $1=0$.

So what we have shown is that, in $\mathbb R$, $x^2+1=0$ has no solution, but this did not prevent the extension of the reals to the complex numbers and so it should not serve as a reason to prevent an extension based on the equation $\sqrt x = -1$. Similarly to the way that the complex numbers made the inequality $x^2>0$ not true, our new number system would make the inequality $\sqrt x > 0$ not true.

  • $\begingroup$ That clears up everything, thanks! I had a feeling there was some mistake with my assumptions. $\endgroup$ – ThisIsNotAnId Apr 30 '12 at 23:41
  • $\begingroup$ I was going through the answer again, and "$\sqrt{x} = -1\ ... \implies \sqrt{1} = -1$" caught my eye. This is valid under the assumption that we are not considering the principal branch of the square root function; however, this is not the case with (1) in the OP. Though, I think what you stated in the paragraph prior, "... $\sqrt{x} = -1$" has no solution, addresses the question at any rate. On a more informal note, doesn't it make you wonder? Seems pretty suspect to me that a construction of this sort seems allowable, but doesn't admit a solution still. $\endgroup$ – ThisIsNotAnId May 1 '12 at 4:54
  • $\begingroup$ Actually, I think you're conclusion about the equation entailing a contradiction seems to be on mark, though, for a different reason than one concerning arithmetic. If we examine closely my OP, we find that I have demanded we be able to find solutions to equations of the form "$\sqrt{x} = -1$" while at the same time looking at the principal branch of the square root function. This is a contradiction, by definition. We could only arrive at the situation "1=0" if there was no contradiction as the one that was inherent to begin with. I hope this is right. $\endgroup$ – ThisIsNotAnId May 1 '12 at 5:52
  • $\begingroup$ @ThisIsNotAnId Look carefully at the step that falls between the statement $\sqrt x = -1$ and $\sqrt 1 = -1$, where (by squaring both sides) we arrive at the conclusion that $x=1$. Thus, we are just substituting to arrive at $\sqrt 1 = -1$. You are correct, in that this is already false, and the proof could have stopped there, but I just continued on to reach a more obviously false statement. It is a thought provoking question. $\endgroup$ – process91 May 1 '12 at 14:20
  • $\begingroup$ @ThisIsNotAnId A modification to my answer may be forthcoming, pending a response from Gerry... stay tuned. $\endgroup$ – process91 May 2 '12 at 14:11

Let's just look at one of the equations (the rest of them can be handled similarly), $$\sqrt{i x} = -1.$$ Squaring both sides we find $i x = 1$, so $x = -i$. Checking with the original equation we find $$\sqrt{i(-i)} = \sqrt{1}.$$ If we interpret $\sqrt{}$ as the principal root, then $\sqrt{1} = 1$, and $x=-i$ is not a solution. However, if we interpret $\sqrt{}$ as the multivalued root then there are two roots of $1$, $1$ and $-1$, since $1^2 = (-1)^2 = 1$. In this case $x=-i$ is a solution.

The square root is multivalued. This is something we see even with real numbers!

  • $\begingroup$ I was just hoping that there may be some way of "forcing" the LHS of the equation to equal -1. I'm no analysis expert, so I may be completely off about this after all. Maybe I ought to edit the OP to reflect this. But, thanks for the response. $\endgroup$ – ThisIsNotAnId Apr 30 '12 at 22:09
  • $\begingroup$ @oenamen I hate to disagree, but there really isn't a solution if you use the expression $\sqrt{}$ with it's standard meaning, since $\sqrt{ix}>0$. If $x=-i$ is a solution, then we can simplify the expression to $1=-1$, which is obviously not true. $\endgroup$ – process91 Apr 30 '12 at 23:05
  • $\begingroup$ @process91: I agree it is confusing to use the notation $\sqrt{}$ for the multivalued root. I will edit the answer. $\endgroup$ – user26872 Apr 30 '12 at 23:09
  • $\begingroup$ @ThisIsNotAnId: Glad to help. As I clarified in my answer, if we interpret the root as the multivalued root these equations have solutions in $\mathbb{C}$. For example, $1^{1/2} = e^{i k \pi}$, where $k=0,1$, is the typical definition of the multivalued root of $1\in\mathbb{C}$. $k$ tells us which branch we're on. $k=0$ is the principal branch. For $k=1$ this "forces" the LHS of the equation of interest to be $-1$. If you consider only the principal branch you only know half of the function $z^{1/2}$. $\endgroup$ – user26872 May 1 '12 at 0:36
  • $\begingroup$ @ThisIsNotAnId: See a diagram of the Riemann surface showing the two branches of the square root here. $\endgroup$ – user26872 May 1 '12 at 0:36

If your $\sqrt{}$ denotes the principal branch of the square root, you are correct in that this never has a negative real part, and when its real part is $0$ the imaginary part is nonnegative.


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