0
$\begingroup$

$i=\sqrt{-1}$

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

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

  • 0
    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

1 Answers 1

5

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
    @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