Determining local extrema of $|f(x)|$ for $f(z)=\overline{z}(z-2)-2Re(z)$, $z=x+iy \in \mathbb{C}$ $f(z)=\overline{z}(z-2)-2Re(z)$, $z=x+iy \in \mathbb{C}$
I want to determine all local extrema of $|f(x)|$.
$|f(z)|=|\overline{z}(z-2)-2Re(z)|=|(x-iy)(x-2+iy)-2x| =|x^2+y^2-4x+i2y|\le|x^2|+|y^2|+|-4x|+|i2y|=x^2+y^2+4x+2y$. But that doesn't get me any further.
I know that $|.|$ is always $\ge0$. Does this mean that $0$ is a global and therefore local minimum? How can I prove this rigorously? And what about the local maxima?
 A: We have $|f(z)|=x^2+y^2-4x+2iy$, hence
$|f(z)|^2=(x^2+y^2-4x)^2+4y^2$.
Now observe: $|f|$ has a local max. (min) at $z_0  \iff |f|^2$ has a local max. (min) at $z_0 $.
Can you proceed ?
Remark: we have $|f| \ge 0$ and $|f(0)|=0$, hence $|f|$ has at $0$ a global minimum.
A: We have
$$
|f(z)|^2 = ((x(x-4)+y^2)^2+4y^2
$$
so we have two  global minima for
$$
y = 0\\
y^2+x(x-4) = 0
$$
or $x = 0, y = 0$  and $x = 4, y= 0$
and a local maximum for $x = 2, y = 0$ because
$\min y^2+x(x-4) = -4\rightarrow |f(z)|^2 = 16$
Attached a level local map. In red the global minima and in green the local maximum.

A: As you have already written $f(z)=x^2-4x+y^2-2iy$ so $|f(z)|=(x^2-4x+y^2)^2+4y^2$ which we can write as $[(x-2)^2+y^2-4]^2+4y^2$. Now writing $x-2=t$, we will have:
$$|f(z)|=[t^2+y^2-4]^2+4y^2$$
Now if $t^2+y^2>4$ then obviously there won't be an extremum (increasing (decreasing) $|y|$ or $|t|$ will increase (decrease) $|f(z)|$). If $t^2+y^2=4$ then increasing or decreasing $t$ (or $|t|$) will definitely increase $|f(z)|$. But for $y$ increasing $|y|$ will definitely increase $|f(z)|$ but dereasing $|y|$ will decrease $|f(z)|$ if $|z|>0$.
A third case remains but it is similar. 
I guess you can fill the gaps and correct the obvious (silly) mistakes if any.
