Minima and maxima of $|f(z)|=|\overline{z}(z-2)-2Re(z)|$

by user561763   Last Updated May 14, 2018 16:20 PM


Now I thought that: $z*$ is a local min. (max.) for $|f|$$\Leftrightarrow$ $z*$ is a local min. (max.) for $|f|^2$. (How can I reason that?)

And $|f(z)|^2=(x(x-4)+y^2)^2+y^4.$ So $x=y=0$ and $x=4,y=0$ should be global minima.

But how can I find the local minima? And what about the local and global maxima? Thank you in advance!

Answers 2

Use that $$|f(z)|=\sqrt{(x(x-4)+y^2)^2+4y^2}$$

Dr. Sonnhard Graubner
Dr. Sonnhard Graubner
May 14, 2018 15:26 PM

To answer your first question. If $f$ is nonnegative function, $f$ and $f^2$ have the same local extrema because $x\mapsto x^2$ is strictly increasing on $[0,+\infty)$.

Now, $|f(z)|^2$ should actually be $(x(x-4)+y^2)^2+\color{blue}{4y^2}$, not what you wrote.

To find extrema, consider $g\colon\mathbb R^2\to\mathbb R$, $g(x,y) = (x^2-4x+y^2)^2 + 4y^2$. You can verify that $$\nabla g(x,y) = \left(2 (2 x-4) \left(x^2-4x+y^2\right),4 y \left((x^2-4x+y^2\right)+8 y\right).$$

Local extrema happen at stationary points, i.e. at the points where $\nabla g(x,y) = 0$.

To check whether these points are local extrema, use Hessian

$$H(x,y) = \left(\begin{array}{cc} 2 (2 x-4)^2+4 \left(x^2-4x+y^2\right) & 4 (2 x-4) y \\ 4 (2 x-4) y & 8 y^2+4 \left(x^2-4x+y^2\right)+8 \\ \end{array}\right).$$

Remember that positive definite Hessian implies local minimum, negative definite implies local maximum and indefinite implies saddle point.

Global extrema happen either at critical points or on the boundary of the domain. Since you are looking at whole $\mathbb C$ (presumably, you didn't write), you should consider how $g$ behaves at infinity, just note that unlike with $\mathbb R$, $\mathbb C$ has infinitely many directions to go at infinity. In general, I'd try using polar coordinates and see what happens on the boundary of the ball $B(0,R)$ when $R$ goes to infinity.

However, in this case it's a bit easier. First of all, $|f(z)|^2\geq 0$, so if you can find a point such that it is equal to $0$, you've found yourself global minimum (there's more than one point at which global minimum is achieved - but you've noticed it yourself). And for global maximum, what happens if $y\to\infty$?

May 14, 2018 15:55 PM

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