Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
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 ( 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

2 Answers 2

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.