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exactly like how ${x^2=-1}$ is impossible in $\mathbb{R}$

is there any equation that is impossible in $\mathbb{C}$ and how to deal with ?

tc1729
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Mostafa 36a2
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    Any polynomial equation of positive degree with complex coefficients is solvable in $;\Bbb C;$ . This is the Fundamental Theorem of Algebra (FTA). Is this what you meant? – DonAntonio Jan 04 '14 at 20:24
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    The equation $e^z=0$ has no solutions in $\mathbb C$. – Git Gud Jan 04 '14 at 20:24
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    $\mathbb{C}$ is algebraically closed: That is, every non-constant polynomial with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$. –  Jan 04 '14 at 20:24
  • $x=x+1$ is still impossible... – Gina Jan 04 '14 at 20:28
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    @Gina your polynomial is of degree $0$ once you collect like terms. – Kevin Carlson Jan 04 '14 at 20:34
  • @KevinCarlson:uh, the asker is asking for a an equality that is impossible to solve. I assume that "to solve" at least means that there is a variable in there. Of course, $x=x+1$ and $0=1$ is equivalent in the sense that both have no solutions; but not equivalence in the formal sense: one have variable $x$, the other one don't. – Gina Jan 04 '14 at 20:36
  • exactly , i'm asking about something that may rake us to third dimension ${real,imagine,something else}$ – Mostafa 36a2 Jan 04 '14 at 20:40
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    @Mostafa36a2:There is the quaternion and the octonion. You might want to look at this: http://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29 ; in other word, the higher you go, the more nice algebraic property get broken. If you want to increase the dimension, it have to be infinite dimensional if it were to be like $\mathbb{C}$ at all. – Gina Jan 04 '14 at 21:02
  • Are there any more interesting answers than $e^z = 0$, perhaps also involving $e^z$? – Stefan Smith Jan 05 '14 at 04:21

2 Answers2

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Sure, you can't solve equations like

$$xy-yx=1$$

in the complex plane. But if we extend the complex plane into quaternions then this equation does have a solution. The basis elements are usually written as $1,i,j,k$ so just like how $2+3i$ is a complex number, we can have $1+2i+3j+4k$ as a quaternion number. The reason we need four basis elements is because you can think of quaternions as two complex planes put together. There is no three-dimensional analogue in between because there is no meaningful way for us to extend the complex division into three dimensions. From two we go to four and then from four we have to go to eight (octonions) and then that's it.

Here is another answer of mine explaining this more in detail.

FYI, the above equations can be solved in quaternions because multiplication is not commutative anymore. So in general $xy\neq yx$ so we can find two numbers $x,y$ which can solve that equation above.

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Fixed Point
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  • so ${xy-yx=1}$ is not ${0=1}$ ! they redefined multiplication ? What set are ${x,y}$ are from ? – Mostafa 36a2 Jan 05 '14 at 06:21
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    $x,y$ are quaternions. Multiplication hasn't really been redefined. It has just been extended to all of quaternions but it is not in general commutative anymore. The real numbers and the complex numbers are subsets of quaternions so when we restrict quaternion multiplication to the reals, it becomes the usual real multiplication which is commutative. – Fixed Point Jan 05 '14 at 07:25
  • Ok , so is there any equation in complex which solution require this extension to quaternions because it has no solution in complex ? that's what I actually ask for – Mostafa 36a2 Jan 05 '14 at 08:09
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    That's what I answered. The equation $xy-yx=i$ has no solution in the complex numbers. But in quaternions there is a solution. – Fixed Point Jan 05 '14 at 08:46
  • Now I understand , in ${R}$ the equation ${x^2=-1}$ has no solutions where x is real number , but if x is complex there is a solution , same argument ${xy-yx=i}$ has no solution if ${x,y}$ are in ${C}$ but if they are quanternions , then the equation has a solution , Thank you very much. – Mostafa 36a2 Jan 05 '14 at 08:51
  • Glad to be of help. – Fixed Point Jan 05 '14 at 09:03
  • This is unclear. Your first answer is $xy - yx = 1$ then in the comments you write $xy - yx = i$. Which one is correct and why? – Heidegger Mar 26 '24 at 19:00
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    @MartinandFriends They are both correct. The $i$ was a typo. I meant to write $1$. But it is a moot point because any nonzero number on the right side will work as a counterexample. The thing is that in quarternions, multiplication is not commutative anymore so $xy$ doesn't have to be equal to $yx$. It is commutative for some $x$ and $y$ but not for all $x$ and $y$. So $xy-yx=1$ has a solution in the quaternions. In the complex plane, multiplication is always commutative so $xy-yx=1$ will never have a solution. $xy-yx$ is always zero in the complex plane. – Fixed Point Mar 27 '24 at 22:32
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    Also note here that $xy-yx=1$ is a polynomial with all complex coefficients but it still has no solution in the complex plane. This is because it is a bivariate polynomial, a polynomial with two variables. All univariate complex polynomials will have a solution in the complex plane. – Fixed Point Mar 27 '24 at 22:35
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All non-constant polynomials with coefficients in $\mathbb{C}$ have a root in $\mathbb{C}$, so polynomial equations are solvable. There are, of course, equations that aren't polynomials that have no solutions, e.g.

$$e^{z} = 0$$

$$\frac{1}{z} = 0$$

and so on.

  • those are limits ! aren't they ? I'm asking for something would take us to third dimension ${x+iy+?z}$ it seems there isn't – Mostafa 36a2 Jan 04 '14 at 20:42