I came across the following question on an exam:
Let $f(z) = \sin z – z \cos z$ be an analytic function on the entire complex plane. Prove that
$$f(z) = \frac{z^3}{3} \prod_{n=1}^\infty (1 – \frac{z}{\lambda_i})$$
where $\lambda_i$ is a sequence of nonzero complex numbers satisfying $\sum_{n=1}^\infty \frac{1}{\lambda_i^2} < \infty$
Well, $\sin$ and $\cos$ are genus 1 functions so that $f$ is as well, so it almost makes sense that we should have such a representation, since the zeroes of $f$ must have the requested growth constraint. However, this only gives us that
$$f(z) = e^{az + b} z^3\prod_{n=1}^\infty (1 – \frac{z}{\lambda_i})$$
So we still need to prove that $a = 0$ and $b = \log 1/3$.
I don't really know how to do this. I looked in Ahlfors at one example of computing this standard product representation of a genus 1 trigonometric function which was shown not to require the exponential factor, but it seemed very involved and ad hoc.
Any thoughts? Many thanks!
Best Answer
Hint:
From the product series in terms of $a,b$, $$\lim_{z\to 0}\frac{f(z)}{z^3}=e^b$$
The left hand side can be evaluated directly, and thus the value of $b$ is obtained.
Similarly,
$$-a=\lim_{z\to 0}\frac 1z \ln \bigg(e^b \frac{z^3}{f(z)}\prod^\infty_{i=1}\left(1-\frac z{\lambda_i}\right)\bigg)$$