Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top



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

share|cite|improve this question
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: $$\Re(z) = \frac{1}{2}(z + \bar{z}).$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.