Plotting $e^{i (x+y)}+e^{i (x+z)}+e^{i (z+y)}$ for x,y,z are real I want to plot on the complex plane the values of  $e^{i (x+y)}+e^{i (x+z)}+e^{i (z+y)}$ for x,y,z are real.
For example for $x=y=0$, we get a $1+2\cos(z)+i2\sin(z)$ i.e. a circle of radius 2 center at $1+i0$.
1)Then how about with more variables? Any ways to go about it. I can't afford mathematica. 
2)Will it cover the entire 3-disk (even though in a complicated way)?
3)Fixing $x=0$, we get $e^{iy}+e^{iz}+e^{i(y+z)}$. So as we vary z we get  circles of radius $r_{z}:=(1+e^{iz})$ rotating around the origin with centers on 1-circle ($e^{iz}$). At $z=\pi$ we just get point -1.
Thank you
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$The image of the mapping $f:\Reals^{3} \to \Cpx$ defined by
$$
f(x, y, z) = e^{i(x + y)} + e^{i(x + z)} + e^{i(y + z)}
$$
is the closed disk of radius $3$ centered at $0$.

To prove this, note that:


*

*The image of $f$ is invariant under rotations about $0$: If $\theta$ is real, then
$$
f(x + \tfrac{1}{2}\theta, y + \tfrac{1}{2}\theta, z + \tfrac{1}{2}\theta)
  = e^{i\theta}\, f(x, y, z).
$$

*Fixing two variables gives, for example,
$$
f(x_{0}, y_{0}, z) = e^{i(x_{0} + y_{0})} + (e^{ix_{0}} + e^{iy_{0}}) e^{iz},
$$
which parametrizes the circle of radius $|(e^{ix_{0}} + e^{iy_{0}})|$ centered at $e^{i(x_{0} + y_{0})}$. If $y_{0} = -x_{0}$ in addition, we have
$$
f(x_{0}, -x_{0}, z) = 1 + (e^{ix_{0}} + e^{-ix_{0}}) e^{iz}
  = 1 + 2(\cos x_{0}) e^{iz}.
$$
The radius takes every value between $0$ and $2$ (inclusive), so the image of $f$ contains the interval $[0, 3]$ on the real axis.
By rotation invariance, the image contains the closed disk of radius $3$.
By the triangle inequality, the image is contained in the closed disk of radius $3$.
