Can we write a complex number z like this Re(z)+Im(z)?

Question

I mean without the i. Re(z) and Im(z) would be real numbers right?

No for the title, and yes. – Neeraj Kumar Nov 11 '15 at 10:23 — Neeraj Kumar, Nov 11 '15 at 10:23

score 16 · Accepted Answer · answered Nov 11 '15 at 10:25

16

No, but you can write it $z = \operatorname{Re}(z) + i \operatorname{Im}(z)$. (Note that $\operatorname{Im}(z)$ is a real number.)

answered Nov 11 '15 at 10:25

mrf

43,639

1

The map $z=x+iy→x+y$ is not one-to-one. But if you supplement it with a second map $z=x+iy→x-y$ then you can retrieve z from both. However, I see no advantages in proceeding along these lines. – Urgje Nov 11 '15 at 10:40
Are the $\Im$ and $\Re$ maps traditionally regarded as $\mathbb{C} \to \mathbb{R}$, or as $\mathbb{C} \to {x+0i| x \in \mathbb{R}}$. If the first, then when we write ${\Re}(z) + i {\Im}(z)$, the concatenation of $\Im(z)$ with $i$ is not really multiplication, right? – Ovi May 07 '19 at 23:05

R. Kasyfil · Answer 2 · 2015-11-11T11:02:51.027

7

for $z\in \mathbb{C}$, we can write $z=a+b\mathrm{i}$. In this case $a=\mathrm{Re}(z)$, and $b=\mathrm{Im}(z)$.

edited Nov 11 '15 at 11:02

answered Nov 11 '15 at 10:26

R. Kasyfil

106

score 3 · Answer 3 · answered Nov 11 '15 at 10:34

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 $z$.

score 0 · Answer 4 · answered Nov 11 '15 at 13:42

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.

$\forall \, z \in \mathbb{C}, \exists! \, (a,b) \in \mathbb{R}^2, z = a + b\imath$