Minimum of $f(z) = \left|z^2+z+1\right|+\left|z^2-z+1\right|$ 
For $z\in\mathbb{C}$, calculate the minimum value of
$$
f(z) = \left|z^2+z+1\right|+\left|z^2-z+1\right|
$$

My Attempt:
Let $z= x+iy$. Then
$$
\begin{align}
z^2+z+1 &= (x+iy)^2+(x+iy)+1\\
&= (x^2-y^2+x+1)+i(2xy+y)
\end{align}
$$
and
$$
\begin{align}
z^2-z+1 &= (x+iy)^2-(x+iy)+1\\
&= (x^2-y^2-x+1)+i(2xy-y)
\end{align}
$$
Define $f:\mathbb{R}^2\rightarrow\mathbb{R}$ by
$$
f(x,y) = \sqrt{\big(x^2-y^2+x+1\big)^2+(2xy+y)^2}+\sqrt{\big(x^2-y^2-x+1\big)^2+(y-2xy)^2}
$$
Using the Triangle Inequality, we know that
$$
f(x,y) \geq 2\sqrt{\left(x^2-y^2+1\right)^2+y^2}
$$
How can the problem be solved from this point?
 A: We can write $f$ in the form
$$f(z)=2|z|\>\left(\left|J(z)-{1\over2}\right|+\left|J(z)+{1\over2}\right|\right)\ ,$$
where $J$ denotes the so-called Joukowski function:
$$J(z):={1\over2}\left(z+{1\over z}\right)\ .$$
It is well known that $J$ maps circles $|z|=r$ onto ellipses centered at $0$ with foci $\pm1$ and semiaxes $$a={1\over2}\left(r+{1\over r}\right),\quad b={1\over2}\left|r-{1\over r}\right|\ .$$ Looking at such an ellipse we see that the points minimizing $f$ are the points $J(\pm i r)$ on the imaginary axis. It follows that
$$g(r):=\min_{|z|=r} f(z)=f(ir)=2\sqrt{(r^2-1)^2+r^2}\ .$$
Now we have to minimize the right side over $r\geq0$. This leads to the equation
$4r^3-2r=0$ with the solutions $r_1=0$, $r_2={1\over\sqrt{2}}$. Since $g(0)=2$ and $g\bigl({1\over\sqrt{2}}\bigr)=\sqrt{3}$ we conclude that the global minimum of $f$ on the complex plane is $\sqrt{3}$, and this minimum is taken at the points $\pm{i\over\sqrt{2}}$.
A: Your analysis starts the right way. Substituting $z=x+iy$, you get as you write:
$$f(x,y)=\sqrt{(x^2-y^2+x+1)^2+(2xy+y)^2}+\sqrt{(x^2-y^2-x+1)^2+(y-2xy)^2}$$
The above now is a real multivariable function, so standard calculus techniques apply. In particular, the critical points are determined by looking first at the system:
$$\frac{df}{dx}=0$$
$$\frac{df}{dy}=0$$
This system has the solutions:
$$(x,y)=(0,0)$$
and
$$(x,y)=\left(0,\pm \frac{\sqrt{2}}{2}\right)$$
The null solution is rejected because it is not a global extremum and you are left with the second solution which after switching back to complex variables corresponds to:
$$\left(x=0,y=\pm\frac{i}{\sqrt{2}}\right)$$
as Cristian points out.
Here's the corresponding graph in two variables:

