Let's find the global minimum of $y=x^2$

First we calculate its first derivative, and make it equal to zero. $y'=2x=0$; $x=0$

Then we check its second derivative, if it's positive then it is minimum. $y''=2$

$(0,0)$ turned out to be global minimum.

Why doesn't derivative show the complex minimum? $(i,-1)$ is below it and that derivation should show the minima are $(∞i,-∞)$ and $(-∞i,-∞)$.

How can I calculate complex minima of other functions?

  • 1
    $\begingroup$ The complex numbers are not ordered, there are no minima/maxima for complex valued functions. What do you mean by complex minimum? But for the same reason, why does derivative not show the "minimum" of $x^3$ but only gives the saddle point $x=0$ but not $(-1,-1)$? Because thats not what a minimum is. $\endgroup$
    – M. Winter
    May 31, 2017 at 16:33

2 Answers 2


I see several problems in your understanding.

First, for general complex values $z\in\Bbb C$ there is no meaning in asking for the minimum of $z^2$. For exmaple, is $$i<1\quad \text{or} \quad 1 <i\;?$$ Is $-i<-1$? The complex numbers are "2-dimensional" and you cannot order the plane. There is no minimum because there is no way to decide what is smaller.

The second thing is that you seem to have a wrong understanding of the term minimum itself. Just because $(0,0)$ is a minimum does not mean that there cannot be a value of the function below $0$. There is just no smaller value in some neighborhood of your minimum. If this shows you something (at all), then that $(0,0)$ is no global minimum here. You can ask why the derivative of $x^3$ will only give you $(0,0)$ (which is just a saddle point) but not $(-1,-1)$. That is because there are smaller values close to $-1$. It also will not give you $(-\infty,-\infty)$. So this is not a problem with complex numbers.


If a codomain is $\Bbb C$, then there is a nonsense to speak about the reasonable order. That is why we don't consider extrema of complex-valued functions.


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