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This question is from Conway Complex Analysis, page 287, exercise 9(a).

My attempt: Write the product as $\underset{n}\prod(1-\frac{z}{b^n})$, where $b=1/a$. First note that this entire function has genus $0$ since $\underset{n}\sum \frac{1}{|b|^n}<\infty$ for $|b|>1$. I would like to imitate the argument in this post: Order of the entire function: $\prod\limits_{n=1}^{\infty} \left(1-\frac{z}{n^k}\right)$ to calculate the order$$\lambda=\underset{r\rightarrow\infty}\limsup\dfrac{\log\log M(r)}{\log r}.$$

But the problem is that $b$ might not be real. Hence the maximum function $M(r)$ is not quite easy to determine (if $b$ is real and positive, then the maximum is attained at $z=-r$ on each circle $\{|z|=r\}$). So I wounder if there is a way to determine explicitly the maximum function $M(r)$ for this particular case.

kzkzkzz
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1 Answers1

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The idea is to consider the power series of this product. Using q-Pochhammer symbol (link in Somos's comment) the coefficient of $z^n$ is $$c_n=\dfrac{(-1)^na^{n(n-1)/2}}{\prod_{m=1}^n(1-a^{m})}$$

Then we can compute the order by $\lambda=1/\alpha$, where

$$\alpha=\underset{n\rightarrow \infty}\liminf\dfrac{-\log|c_n|}{n\log n}=\infty$$.

Hence the order is $0$.

Remark: for the case of $\prod(1-\frac{z}{n^k})$, it has order $1/k$. When $k$ gets bigger, the order gets smaller. In our case, the zeros grow exponentially, and thus its order must be smaller than all $1/k$.

kzkzkzz
  • 331