$i=\sqrt{-1}$
$\operatorname{Re}(z)+i\cdot\operatorname{Im}(z)=z$
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