Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From old qualifying exam: Let $E$ be the union of the two coordinate axes, i.e. $E = \{z=x+iy : xy=0\}$. Describe all entire functions satisfying $f(E) \subset E$.

I feel like the best approach is to consider the power series of $f$. My first approach was to write down constraints by considering the function applied to the real or imaginary axis. When I didn't get anywhere with this, I began thinking of the function geometrically: on $E$ it's allowed to scale by a real constant, and rotate by $k\pi/2$. But again, I couldn't see how to usefully translate this to produce information about the power series. Thanks!

As an example, $z^2$ has this property. In fact, so does $az^2+bz^4$ (with $a,b \in \mathbb{R}$) since each term maps the imaginary axis to the real axis, which ends up back on the real axis when added. A similar argument shows real odd polynomials work, too.

share|cite|improve this question
For example, any function of the form $f(z) = \sum_{n=0}^\infty c_n z^{2n}$, where the $c_n$ are real numbers converging to 0 fast enough that $f$ is entire, is in this class. So I don't know how tidy a characterization one will be able to get. – Greg Martin Jan 27 '12 at 20:50
Functions of the form $Ce^{i\theta}z^n$ where $\theta\in\left\{0,\frac{\pi}2,\pi,\frac{3\pi}2\right\}$, $C\in\mathbb R$ and $n\in\mathbb N$ seems to work (but are not the only one). – Davide Giraudo Jan 27 '12 at 20:52
So a guess would be: any entire function whose power series 1) has strictly even (and no constant term) or strictly odd terms and 2) has strictly real or strictly imaginary coefficients. Can't see why this would capture all such functions. – dls Jan 27 '12 at 21:37
up vote 5 down vote accepted

If $f(E)\subset E$, then $(f^2)(\mathbb{R})\subset\mathbb{R}$. Thus, the coefficients of the power series of $(f)^2$ (around $z=0$) are real. If $\{c_n\}$ are the Taylor coefficients of $f$ around $z=0$, then $$ \sum_{j=0}^nc_jc_{n-j}\in\mathbb{R},\quad n=0,1,2,\dots\tag1 $$ If $f(E)\subset E$, the same is true of $g(z)=f(i\,z)$. Since the coefficients of $g$ are $i^n\,c_n$, it folows that $$ i^n\,\sum_{j=0}^nc_jc_{n-j}\in\mathbb{R},\quad n=0,1,2,\dots\tag2 $$ From (1) and (2) we get $$ \sum_{j=0}^nc_jc_{n-j}=0\quad\text{if $n$ is odd.}\tag3 $$ Now it is not to difficult to see that the power series of $f$ must be one of $$ \sum_{n=0}^\infty c_nz^{2n},\quad i\,\sum_{n=0}^\infty c_nz^{2n},\quad \sum_{n=0}^\infty c_nz^{2n+1},\quad i\,\sum_{n=0}^\infty c_nz^{2n+1},\text{ with } c_n\in\mathbb{R}. $$

share|cite|improve this answer
I have to write out the details of the last implication; but I think $c_0=0$ in the even case. For instance, $z\mapsto1$ does not satisfy the original property. Thanks! – dls Jan 28 '12 at 3:49
@dls: How does it not satisfy? $\{1\} \subset E$, right? Also note that in the even case, the image of $E$ is either entirely real or entirely imaginary. – Niels Diepeveen Jan 28 '12 at 4:57
@NielsDiepeveen Doh! For some reason I was thinking $z\mapsto1+z$ in my brain... which makes no sense. – dls Jan 28 '12 at 5:29

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.