How do I determine order of growth of periodic function? (sin)

Question

I am self-reading Levin's book, on Entire functions. There is a task that needs to be verified by the reader, and I cannot reproduce it. It states:

"Verify that $sin(Az)$ is of order $p = 1$ and type $\sigma = |A|$"

My attempt was the following: $sin (Az)$ is a function:

1. periodic
2. limited by growth, with a maximum result value equal to 1.

Thus, $$ p = \left[ \limsup_{r\to\infty} \frac{log(log(M_f(r)))}{log(r)} \right] $$

As $ log(log(M_f(r)))$ is limited, as sinus is limited, thus we are getting

$$ p = \left[ \limsup_{r\to\infty} \frac{limited}{log(r)} \right] $$ and $log(r)$ is -> to $\infty$ as $r\to\infty$ Thus $p = 0$, and

$$ \sigma = \left[ \limsup_{r\to\infty} \frac{log(M_f(r))}{r^p} \right] $$

Where $M_f(r) = 1$(maximum value of sinus), $log(1) = 0$ which results in $$ \sigma = \frac{0}{\infty} = 0 $$

Where had I taken it wrong? Should I somehow try to use "wonderful limit"? What is general rule to estimate growth of periodic functions?

By the way, I was thinking of converting the sin into the Taylor series, but in this case, I feel the temptation of putting to $p = \infty$, which nonetheless verifies anything.

Answer 1

Well,

1. It is required to rewrite $sin(z)$, like that: $sin(z) = \frac{ e^{iz} - e^{-iz} }{2i}$
2. I have found the complete solution:

P.S. Also, worth checking out this question & accepted answer