Problem 1806 (difficulty: 9/10)

Call an entire function \(\displaystyle f\) ``interesting'', if \(\displaystyle f(z)\) is real along the parabola \(\displaystyle \re z=(\im z)^2\).

(b) Prove that if \(\displaystyle f\) is an interesting function then \(\displaystyle f'(-3/4)=0\).

(a) Show an example for a non-constant interesting function.

CIIM 2014, Costa Rica

Solution:

(a) Such a function is

\(\displaystyle f(z) = \cos\left(2\pi\sqrt{\tfrac14-z}\right). \)

Since \(\displaystyle \cos z\) is even, the two branches of the square root give the same value. So, \(\displaystyle f\) is indeed an entire function. (The power series \(\displaystyle f(z)= \sum\limits_{k=0}^\infty \tfrac{(2\pi)^{2k}(z-\tfrac14)^k}{(2k)!}\) also converges everywhere.)

Now consider a point \(\displaystyle z=it+t^2\) of the parabola.

\(\displaystyle f(z) = \cos\left(2\pi\sqrt{\tfrac14-it-t^2}\right) = \cos\left(\pi - 2\pi it\right) = \cos (2\pi it) = \cosh (2\pi t) \in \RR. \)

(b) Assume that \(\displaystyle f\) is an arbitrary interesting function and let \(\displaystyle g(z)=f(z-z^2)\). If \(\displaystyle z=it\) with some real \(\displaystyle t\), then the point \(\displaystyle z-z^2=it+t^2\) lies on the parabola, therefore \(\displaystyle g(it)=f(it+t^2)\) is real. Hence, \(\displaystyle g(z)\) is real along the imaginary axis.

Consider the function \(\displaystyle g_1(z)=\overline{g(-\overline{z})}\). For every real \(\displaystyle t\) we have \(\displaystyle g_1(it) = \overline{g(-\overline{it})} = \overline{g(it)} = g(it)\), so, by the unicity theorem, \(\displaystyle g_1=g\). Hence, \(\displaystyle g(z)=\overline{g(-\overline{z})}\). Taking the derivative at \(\displaystyle z=\frac12\) we get

From the definition of \(\displaystyle g\) we have \(\displaystyle g'(z) = f'(z-z^2) (1-2z)\); in particular, for \(\displaystyle z=\pm\tfrac12\) this shows that

The relations (1) and (2) together prove that \(\displaystyle f'(-\tfrac34)=0\).

Give me another random problem!