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If $\operatorname{Re}^{2}(x)=-1$, what is $x$?

$x$ cannot be defined in complex number as $(a+ib)$. { $a$ and $b$ are real numbers }

Let's try to find out $x$ by using function equations and power series




$\operatorname{Re}(z^{2})=2\operatorname{Re}^{2}(z)+z^{2}-2z\operatorname{Re}(z)$ That is function equation for real part function. We can obtain many such relation using similar method for $\operatorname{Re}(z^{n})$.

Also, $\operatorname{Re}(z_1+z_2)=\operatorname{Re}(z_1)+\operatorname{Re}(z_2)$.

it seems that $\operatorname{Re}(z)$ has a lot of relation as function equations. But I could not get it as power series ($a_0+a_1z+a_2z^{2}+\cdots$)

Does anybody know how to find $\operatorname{Re}(z)$ function in series of $z$?

If we can find it, we would define $x$ as new number group.

Thanks for help

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If your equation is saying that the square of the real part of $x$ is $-1$, then it's nonsense. For the square of the real part of $x$ to be $-1$, the real part of $x$ would have to be $\pm\sqrt{-1}$, which isn't real, so it can't be the real part of anything. – Gerry Myerson Jan 16 '12 at 10:56
I'd like to note that this problem is fundamentally different from the problem of "what's $\sqrt{-1}$?" that spawned $i$ in the first place. The statement "$\Re(z)^2=-1$" takes place after $\mathbb{C}$ has been discovered and defined, and it's defined so that nothing in $\mathbb{R}$ squared is negative, which is why $\Re(z)^2=-1$ has no solution. – tomcuchta Jan 16 '12 at 11:01
Perhaps There is another kind of number group that we have not known yet. – Mathlover Jan 16 '12 at 11:03
Nope. If you allowed $\Re(z) = \sqrt{-1}$ it would not longer be "the real part of $z$". It simply has no solution. – tomcuchta Jan 16 '12 at 11:04
If we can express Re(z) as power series or integral formula (I dont know if it is possible or not) we can define X in a equation as i defined in $Z^{2}+1=0$ – Mathlover Jan 16 '12 at 11:09
up vote 5 down vote accepted

There is no power series for the real part of $z$, because $\Re(z)$ is nowhere analytic. The best you can hope for is to express $\Re(z)$ as a sum of its analytic and anti-analytic pieces, in terms of which it is simple: $$\Re(z) = \frac{1}{2}(z + \bar{z}).$$

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How can we prove that $\Re(z)$ is not analytic? It has many Function equations as shown above. – Mathlover Jan 16 '12 at 13:31
You can verify that it does not satisfy the Cauchy-Riemann equations. If you write $\Re(z) = \Re(x + iy) = u(x, y) + iv(x, y)$, then you have $u(x, y) = x$ and $v(x, y) = 0$, so $\partial_{x}{u} = 1$, but $\partial_{y}{v} = 0 \neq 1$. (It does satisfy the other CR equation.) – mjqxxxx Jan 16 '12 at 15:46
I understood what you mean. – Mathlover Jan 17 '12 at 12:10
Thank you mjqxxxx a lot for answers to understand the problem. I have noticed another way to show that it is impossible to define $\Re(z)$ as Power Series in Z. $\Re(z)=\Re(x+iy)=x$ We assume that $$\Re(z)=a_0+a_1z+a_2z^{2}+a_3z^{3}\dots$$ Need to use general defination of $$\partial_{x} {[F(g(x,y))]}= F'(g(x,y)).\partial_{x} {g(x,y)}$$ and $$\partial_{y} {[F(g(x,y))]}= F'(g(x,y)).\partial_{y} {g(x,y)}$$Let's define $g(x,y)=x+iy=z$ and $F(x)=\Re(x)$ First of all, Derivate by x both side $$\partial_{x}(\Re(z))=\partial_{x} (x)$$ $$\partial_{x}(\Re(g(x,y)))=\Re'(g(x,y)) \partial_{x}{(g(x,y))}=$$ – Mathlover Jan 17 '12 at 13:03
@Mathlover: Consistency between the two partial derivatives in the complex plane (one along the real axis, the other along the imaginary axis) is exactly what is required by the Cauchy-Riemann equations. – mjqxxxx Jan 17 '12 at 14:39

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