<s>[INST] 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?
 [/INST] 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:

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