# Why complex graphs are 4D?

Note that I'm trying to study complex numbers online (Can't find much good stuff..). So if I'm correct to solve a function like this $f(x) = x^2+1$ we need a 4D graph. Or 2 2D graphs basically. A x-axis,x-imaginary-axis and a y-axis,y-imaginary-axis. My question is. "Why?!" Can't we just integrate those 2 imaginary axis as one like with the classical xyz system with z = imaginary axis. Then couldn't we work in the 3D space between those 2 planes?

• $\newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}\newcommand{\i}[0]{\mathrm{i}}\newcommand{\kume}[1]{\mathbb{#1}}\newcommand{\bold}[1]{\textbf{#1}}\newcommand{\italic}[1]{\textit{#1}}\newcommand{\kumedigerBETA}[1]{\rm #1\!#1}$$x= x_{\text{real}}, \space y = y_{\text{real}}, z = x_{\text{imaginary}}$ Now where is $y_\text{imaginary}$? We need $1$ dimension more. – MCCCS Sep 19 '17 at 16:19
• The dimension of the graph is (dimension of domain) + (dimension of range) = 2 + 2 = 4. – MPW Sep 19 '17 at 16:28
• Just image a cube. It's corner is 0 and that corner's 3 sides are the x, y and imaginary axis. So one face of the cube it's a x,imaginary and on the other face it's y,imaginary. Why can't we just have the imaginary axis be the same but depending on the face we see it from change it's name (x or y). – alienCY Sep 19 '17 at 16:48
• Because if you specify the "x-real", "x-imaginary" and "y-real" coordinates for a point of such cube, then "y-imaginary" coordinate becomes determined (and equal to "x-imaginary"). – colt_browning Sep 19 '17 at 20:12