Let $f(z)=e^z-z$

I want to check $f(z)$ is finite order.

And how to show that $f(z)$ has infinitely many zeros and that each zero is simple.

Dfn: an entire function f is finite order if $\exists$ integer $k \gt 0$ and $R\gt 0$ such that for all $|z|\ge R$, we have $|f(z)| \le e^{|z|^k}$

Theorem: suppose $f$ is an entire function of finite order. Then either f has infinite many zeros or $f(z)=Q(z)e^{P(z)}$ where $Q$ and $P$ are polynomials.


Well, from the definition it follows that $f(z)$ is of order $1.$ (since $z$ is of order $1,$ and you should show that the sum of two functions of order $1$ is of order $1.$)

For the second, to show that $f$ has infinitely many zeros, you need to show that it is not of the form $Q(z) \exp(P(z)).$ If it were, $P(z)$ would have to be a linear polynomial (by comparing order) of form $P(z) = a + z.$ It should be easy to check that this is not possible -- if it were, $z$ would grow exponentially).

For the last question, a zero is multiple if it is also a zero of the derivative. The derivative of $f$ is $\exp z - 1,$ so $z$ is a zero of both the function and the derivative if $\exp z - z = \exp z - 1 = 0.$ Is that possible?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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