I have to find the order of $f (z)=\sinh z -z;$ then I must show that $f $ has infinite zeros, and that the zeros different from $0$ have order $1$.
The order of $f $ is $1$; however I don't know how to continue. There is a theorem stating that $\mathfrak n (r)\le Cr^\rho$, where $\mathfrak n (r)$ is the number of zeros in the disk $D_r (0)$, $\rho$ is the order of the function and $C $ is a constant. However this is not enough to say that the zeros are infinitely many. How can I proceed? Thank you
The fact that $f$ has infinitely many zeros follows from the fact that $0$ is not a Picard exceptional value, i.e., that $f$ is not of the form $Pe^Q$ with polynomials $P$ and $Q$, but it can also be shown more directly using the argument principle, which even gives you a count of the zeros: If $R = [-2\pi k,2\pi k]^2$ denotes the square centered at zero, with side length $4\pi k$, where $k$ is a large positive integer, then by the argument principle the number of zeros of $f$ inside $R$, counted with multiplicity, is $N = \frac{1}{2\pi} \int_{\partial R} d \arg f(z)$. Writing $z=x+iy$, on the right boundary of $R$ you have that $|\sinh z| \gg |z|$, so $\arg f(z) \approx \arg \sinh z \approx y$, so the argument of $f(z)$ changes by approximately $4\pi k$. Similarly, on the left boundary you have $\arg f(z) \approx -y$, so the argument also changes by approximately $4\pi k$. On the top boundary the argument changes by $\approx -\pi$, and the same is true on the bottom boundary. Taking everything together you get that $N \approx \frac{8\pi k -2\pi}{2\pi} = 4k-1$. Since this has to be an integer, it will be equal to $4k-1$ (for $k$ large enough), so that this is the number of zeros inside $R$.