<s>[INST] Extrema of $f(z)=\Big |\bar z(z-2)-2\Re(z) \Big|$ for $z \in \mathbb{C}$ I want to find extrema of $f(z)=\Big |\bar z(z-2)-2\Re(z) \Big|$ for $z \in \mathbb{C}$.
$f(z)=\Big |\bar z(z-2)-2\Re(z) \Big|=\Big| x(x-4)+y^2+i2y\Big|$
Then I defined $g(x,y):=f(z)^2=(x^2-4x+y^2)^2+4y^2$. Because $g$ is strictly monotone the extrema of $f$ and $g$ should be the same.
Now for finding the extrema I looked at the derivatives of $g$.
$g_x(x,y)=2(x^2-4x+y^2)(2x-4)$ and $g_y(x,y)=2(x^2-4x+y^2)2y+8y$. Then I tried to determine where $g_x$ and $g_y$ vanish.
$g_x(x,y)=0 \Leftrightarrow x_1=2$, $x_2=2-\sqrt{4-y^2},x_3=2+\sqrt{4-y^2}$
$g_y(x,y)=0 \Leftrightarrow y_1=0$, $y_2=\sqrt{-x^2+4x-2},y_3=-\sqrt{-x^2+4x-2}$
N0w, I need help with determing the critical points. $(2,0)$ should be one but how can I determine the others?
 [/INST] Suppose that $(x,y)$ is a critical point of $g$. Since $g_x(x,y)=2(x^2-4x+y^2)(2x-4)$ you have two possibilities:


*

*$x=2$. In that case, $0=g_y(x,y)=2(2^2-4\times2+y^2)2y+8y=4y^3-8y$ and therefore $y=0$ or $y=\pm\sqrt2$.

*$x^2-4x+y^2=0$. Then $0=g_y(x,y)=8y$. So, $y=0$ and therefore $x=0$ or $x=4$.


Therefore, the critical points are $(2,0)$, $\left(2,\pm\sqrt2\right)$, $(0,0)$, and $(4,0)$.
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