Can we write a complex number z like this Re(z)+Im(z)? I mean without the i. Re(z) and Im(z) would be real numbers right?
 A: for $z\in \mathbb{C}$, we can write $z=a+b\mathrm{i}$. In this case $a=\mathrm{Re}(z)$, and $b=\mathrm{Im}(z)$.
A: Let a complex number $ z $ be: $ z = a + ib $; then we define $ \text{Re}(z) = a $ and $ \text{Im}(z) = b $. 
As you see $ \text{Re}(z) $  and $ \text{Im}(z) $ are real numbers and thus the real number $ z_2 = \text{Re}(z) + \text{Im}(z) $ is different from $z$.
A: No, but you can write it $z = \operatorname{Re}(z) + i \operatorname{Im}(z)$. (Note that $\operatorname{Im}(z)$ is a real number.)
A: for $z\in \mathbb{C}$, we can write $z=a+b\mathrm{i}$. In this case $a=\mathrm{Re}(z)$, and $b=\mathrm{Im}(z)$.
Re(z) and Im(z) would be real numbers right?
Yes.
$\forall \, z \in \mathbb{C}, \exists! \, (a,b) \in \mathbb{R}^2, z = a + b\imath$
And you have:
Re(z) = a
Im(z) = b
Thus, Re(z) and Im(z) are real numbers.
Can we write a complex number z like this Re(z)+Im(z)?
Yes.
You can write a complex number z such as z = Re(z) + Im(z).
$z = 0$ works, for example.
Let's take $z \in \mathbb{C}, \exists! \, (a,b) \in \mathbb{R}^2, z = a + b\imath$.
$z = Re(z) + Im(z)$
$\Leftrightarrow a + b\imath = Re(z) + Im(z)$
$\Leftrightarrow a + b\imath = a + b$
$\Leftrightarrow b = 0$  
Thus, $z = Re(z) + Im(z) \Leftrightarrow z \in \mathbb{R}$. In fact, all the real numbers works.
