# When are complex numbers insufficient?

Rationals can't solve $x^2=2$, and reals can't solve $x^2=-1$. Is there any problem that cannot be solved by complex numbers but can be solved by non-standard numbers?

Every polynomial with coefficients in $C$ can be solved by numbers in $C$, can every equation* be solved by numbers in $C$?

*(that can not be simplified to $1=0$)

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Your first Q's A depends on how you define a number since you already know C. Your 2nd Q: yes, because C is a field and if manipulations are also defined in this field. – puresky Nov 19 '11 at 15:53
Certainly not every equation has a solution in $\mathbb{C}$. As a trivial example, $1=2$, or to make it look more like an equation in a variable, $z+1-z=2$. As a less trivial example, consider $z \bar{z} = -1$. – Nate Eldredge Nov 19 '11 at 16:20
The equation $e^z=0$ cannot be "simplified" to $1=0$, but it doesn't have any solutions in $\mathbb{C}$. – Zev Chonoles Nov 19 '11 at 19:07
@ZevChonoles The equation $e^z=0$ can be "simplified" by multiplication with $e^{-z}$. Guess what you get... – Thomas Klimpel Nov 19 '11 at 19:50
@Thomas: That's true, but as long as we're playing that game, the statement that an equation $f(z)=0$ can't be solved is then equivalent to the statement that $1=0$, since $f$ has no zeros and is therefore invertible. Wouldn't you agree that there is at least content in the statement that there is no solution to $e^z=0$? – Zev Chonoles Nov 19 '11 at 20:18

As $\mathbb{C}$ is commutative, non-commutative problems cannot be solved in it.

E.g. $A \cdot B - B \cdot A = I.$

Equations like this are central to Quantum Mechanics and Lie Algebras.

However, matrices usually are non-commutative, so matrices over $\mathbb{C}$ or $\mathbb{R}$ are able to solve those equations. As can quaternions among others.

EDIT: You say that you are looking for non-standard numbers. May I suggest to have a look at Quaternions? They are the logical next step after $\mathbb{C}$. The Wikipedia entry might be a good starting point. Quaternions are a bit out of fashion, but theoretically and historically they are important.

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This is precisely 0=I – TROLLKILLER Nov 20 '11 at 18:41
@hyperion It's only 0=I when AB = BA. When something is non-commutative, one cannot assume that AB = BA anymore. – Sjoerd Nov 20 '11 at 19:52

The problem, "find a number strictly greater than zero but strictly less than every number $1/n$ with $n=1,2,3,\dots$" cannot be solved in the complex numbers, but can be solved in the non-standard numbers.

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I don't quite get this example, since there is no well-ordering of the complex numbers. – akkkk Dec 3 '12 at 11:13
Perhaps I should have put it this way: the problem can't be solved in the reals, and it doesn't help any to extend to the complex numbers, but it can be solved in the non-standard numbers. Or this way: find a number whose modulus is strictly greater than zero but.... – Gerry Myerson Dec 3 '12 at 11:36

I would like to respond to the question "Is there any problem that cannot be solved by complex numbers but can be solved by non-standard numbers?" Since one of the tags is "nonstandard analysis", I will interpret this as applying the the nonstandard numbers in that theory, namely the hyperreals. Back to the question: one important problem that the hyperreals allow one to solve, is relating the derivative of a function to ratios of infinitesimals, as it was done by Leibniz, a co-founder of infinitesimal calculus.

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