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$i=\sqrt{-1}$

$\operatorname{Re}(z)+i\cdot\operatorname{Im}(z)=z$

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

$\operatorname{Im}(z)=-i(z-\operatorname{Re}(z))$

$\operatorname{Re}(z^{2})=\operatorname{Re}^{2}(z)-\operatorname{Im}^{2}(z)$

$\operatorname{Re}(z^{2})=\operatorname{Re}^{2}(z)+(z-\operatorname{Re}(z))^{2}$

$\operatorname{Re}(z^{2})=2\operatorname{Re}^{2}(z)+z^{2}-2z\operatorname{Re}(z)$ 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

  • 8
    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.2012-01-16
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    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.2012-01-16
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    Perhaps There is another kind of number group that we have not known yet.2012-01-16
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    Nope. If you allowed $\Re(z) = \sqrt{-1}$ it would not longer be "the real part of $z$". It simply has no solution.2012-01-16
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    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$2012-01-16
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    If $X$ in the equation $\Re(X)^2=-1$ is a complex number, then your equation has no solution. So you suggest that perhaps $X$ is not complex, but a new type of number. In which case, I think you are asking, "how can the domain of the function $\Re$ be extended to include the value $-1$ in its range?" If you represent $\Re(X)$ as a power series of complex numbers, your sum will be either a complex number or $\infty$, neither of which work.2012-01-16
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    I think you've misunderstood the construction of complex numbers and suchlike. Either you mean that $\text{Re}(z)^2 = -1$, which has no solutions because $\text{Re}(z)$ is real; or you mean that $z=a+ib$ (with $a,b$ not necessarily real) and $a^2=-1$, in which case $a=\pm i$ (say $i$ for now) so that $z=(b+1)i$. But I get the feeling you didn't mean either of these, in which case your problem is ill-posed.2012-01-16

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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: $$\Re(z) = \frac{1}{2}(z + \bar{z}).$$

  • 0
    How can we prove that $\Re(z)$ is not analytic? It has many Function equations as shown above.2012-01-16
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    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.)2012-01-16
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    I understood what you mean.2012-01-17
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    Even simpler(?) write z as rexp(itheta) and note that the derivitve depends on theta (e.g. the "limit" isn't unique - so doesn't exist).2012-01-17
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    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 $$\partial_{x}(\Re(z))=\partial_{x} (x)$$ $$\partial_{x}(\Re(g(x,y)))=\Re'(g(x,y)) \partial_{x}{(g(x,y))}=$$2012-01-17
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    $$=\Re'(z) \partial_{x}{(x+iy)} =(a_1+a_22z+a_33z^{2}\dots).(1)=\partial_{x} {(x)} =1 \qquad (1)$$ $$\Re'(z)=(a_1+a_22z+a_33z^{2}\dots) =1$$ Secondly, Derivate by y both side $$\partial_{y}(\Re(z))=\partial_{y} (x)$$ $$\partial_{y}(\Re(g(x,y)))=\Re'(g(x,y)) \partial_{y}{(g(x,y))}=$$ $$\Re'(z) \partial_{y}{(x+iy)} =(a_1+a_22z+a_33z^{2}\dots).(i)=\partial_{y} {(x)} =0 \qquad (2)$$ $$\Re'(z) =(a_1+a_22z+a_33z^{2}\dots)=0/i=0 $$ There is contradiction between (1) result and (2) result because $\Re'(z)=1$ according to (1) result but $\text{}$$\Re'(z)=0$ according to (2) result.2012-01-17
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    Analytic functions have to have only one derivative function result. So it is impossible to define $\Re(z)$ in power series as $a_0+a_1z+a_2z^{2}+\dots$ If we summarize our results, $\Re(x) = \sqrt{-1}$ we cannot define x by using power series . Any function that is defined in power series can be solved by complex numbers. (a+ib) Maybe methods is still waiting for us to discover how to use the $\Re(x) = \sqrt{-1}$ in different applications of mathematics2012-01-17
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    Where is the contradiction between (1) and (2)? You consider the function $F:z\mapsto\Re(z)$ and you note that $\partial_xF(z)\ne\partial_yF(z)$... And so what? One also has $\partial_xG(z)\ne\partial_yG(z)$ for $G:z\mapsto z$ and for $G:z\mapsto z^2$, although these are both power series in $z$.2012-01-17
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    Please consider the defination of $\partial_{x} {[F(g(x,y))]}= F'(g(x,y)).\partial_{x} {g(x,y)}$ This is general defination of function derivatives. I used it2012-01-17
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    @Mathlover: Consistency between the two partial derivatives in the complex plane (one along the real axis, the other along the imaginary axis) is exactly what is required by the Cauchy-Riemann equations.2012-01-17
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    I understood what you mean, I just used contradiction method to prove that impossible to write power series of $\Re(z)$,if we firstly accept that we have $\Re(z)=a_0+a_1z+a_2z^{2}+a_3z^{3}\dots$ thanks alot for your advices2012-01-17
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    @Didier Piau: thank you for answer Please let me show that results (1) and (2) are equal for $G(z)=z$ $z=x+iy$ $G(z)=z$ firstly derivate both side by x $$\partial_{x}(G(z))=\partial_{x}(z)$ $\partial_{x}(z)=\partial_{x} (x+iy)=1\qquad (1)$$ $$\partial_{x}(G(z))=G'(z) \partial_{x}{(z)}=G'(z)\partial_{x} (x+iy)=G'(z)=1$$ now derivate both side by y $$\partial_{y}(G(z))=\partial_{y}(z)$$ $$\partial_{y}(z)=\partial_{y} (x+iy)=i\qquad (2)$$ $$\partial_{y}(G(z))=G'(z) \partial_{y}{(z)}=G'(z)\partial_{y} (x+iy)=G'(z).i=i$$ $$G'(z)=1$$2012-01-17
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    As you see if we select G(z)=z my test shows that result (1) and result (2) are equal, if so G(z) is analytic function. this test can be done for any analytic function such as $G(z)=z^2$ or $G(z)=e^{z} $ and result (1) and result (2) will be equal. Thanks for advice2012-01-17
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    @Mathlover: **You must use comments to respond to other comments**. Unless you are posting something that is an actual answer to the question, *you should not post it as an answer*.2012-01-17