# Prove that the function $xy(x^2-y^2)$ cannot have absolute extrema…

In the complex plane, set $z=x+iy$. How to prove that the function $xy(x^2-y^2)$ cannot have local maximum or minimum in $|z|<1$? I suspect that the maximum modulus principle plays a role but don't know how to make use of this idea

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Not exactly a modulus (this is not even nonnegative), but you can see it as the imaginary part of some simple holomorphic function, and as such it is harmonic. –  D. Thomine Dec 9 '11 at 23:20
@ D.Thomine Yes, I see this, it is the imaginary part of $z^4/4$ or something, but now how is the fact that it is harmonic going to help me? –  QuArK21343 Dec 9 '11 at 23:28
Because harmonic functions also satisfy a maximum principle (en.wikipedia.org/wiki/Harmonic_function). By the way, yes, $z^4/4$ works. –  D. Thomine Dec 9 '11 at 23:32
and I believe this could be done by the usual, pedestrian way (find critical points, etc) –  Glougloubarbaki Dec 9 '11 at 23:33

You don’t need complex functions at all. Just convert to polar coordinates:

$$xy(x^2-y^2)=r^4\cos\theta\sin\theta(\cos^2\theta-\sin^2\theta)=\frac12r^4\sin(2\theta)\cos(2\theta)=\frac14r^4\sin(4\theta)\,,$$ which obviously has neither a maximum nor a minimum in the open unit disk centred at the origin. Along $\theta=\frac12\pi$ it increases towards $1$ as you move away from the origin, and along $\theta=-\frac12\pi$ it decreases towards $-1$, and it’s along those rays (among others) that it attains its maximum and minimum on circles centred at the origin.

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Two parts:

It is the imaginary part of $\frac{z^4}{4}$.

This means that it is a harmonic function, and harmonic functions have a max/min principle.

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Oh - I see that while I was typing this up, the same answer came up in the comments. D. Thomine said the same thing. –  mixedmath Dec 9 '11 at 23:36