Order of $\frac{f}{g}$ An entire function is of finite order $\rho$ if $$\rho = \inf \{\lambda \geq 0  \ | \ \exists A, B > 0 \ s.t. \  |f(z)|\leq Ae^{B|z|^{\lambda}} \forall z \in \mathbb{C} \}$$
Prove that if $f$ and $g$ are entire functions of finite order  $\rho$, and $\frac{f}{g}$ is entire,then $\frac{f}{g}$ is of order $\leq \rho$.
Any hint ? 
 A: We will use the following equivalent definition: an entire function $f$ is of finite order $\rho$ if there exists $0\leq\rho<+\infty$ such that
$$\rho=\inf\ \{\lambda>0:\sup_{|z|=R}|f(z)|\in O(e^{rR^{\lambda}}), \ R\rightarrow +\infty\}$$ 
By definition of Big-O, this means that $f$ is of finite order $\rho$ if and only if $\forall \epsilon>0$ there exists a constant $C=C(\epsilon)$ such that
$$||f||_R\leq C\cdot\epsilon^{R^{\rho+\epsilon}}$$
definitely for $R$ sufficiently big, where $||f||_R$ is defined by
$$||f||_R:=\sup_{|z|=R}\{|f(z)|\}$$
Now take logarithm of both sides
$$\log ||f||_R\leq \tilde{C}\cdot R^{\rho+\epsilon}$$
for $R>>1$. Now by hypothesis, both $f$ and $g$ are of finite order. Then for every $\epsilon>0$ and $R$ big enough we have
$$\log ||fg||_R\leq\log||f||_R+\log||g||_R$$
$$\leq C_1\cdot R^{\rho+\epsilon}+C_2\cdot R^{\rho+\epsilon}$$
$$=(C_1+C_2)\cdot R^{\rho+\epsilon}$$
which proves that $fg$ is of finite order $\leq\rho$.
For the sum $f+g$ we have that:
$$||f+g||_R\leq||f||_R+||g||_R\leq C_1\cdot e^{R^{\rho+\epsilon}}+C_2\cdot e^{R^{\rho+\epsilon}}\leq 2C\cdot e^{R^{\rho+\epsilon}}$$
where $C:=\max(C_1,C_2)$, thus $f+g$ is of finite order $\leq \rho$.
Finally let's consider the quotient $f/g$. Using Hadamard factorization theorem for entire functions of finite order, we get
$$\displaystyle\frac{f(z)}{g(z)}=\frac{e^{g_1(z)}\cdot z^{m_1}\cdot\prod_{j=1}^{+\infty}E(\frac{z}{z_j^f},p_f)}{e^{g_2(z)}\cdot z^{m_2}\cdot\prod_{j=1}^{+\infty}E(\frac{z}{z_j^g},p_g)}$$
where: $g_1,g_2$ are polynomials with degree $\leq\rho$, $\{z_j^f\}$ is the sequence of zeros of $f$, $\{z_j^g\}$ the zeros sequence for $g$, $m_1=ord_0(f), m_2=ord_0(g),p_f$ is the genus of zeros of $f$, $p_g$ the genus of zeros of $g$.
Since by hypothesis $f/g$ is an entire function, the zeros of $g$ are a subset of zeros of $f$, so that $m_1\geq m_2$ and the infinite product in the denominator can be simplified:
$$\displaystyle\frac{f(z)}{g(z)}=e^{(g_1-g_2)(z)}\cdot z^{m_1-m_2}\cdot \prod E(\frac{z}{z_j},p)$$
where the Weierstrass product ranges over the zeros of $f$ which are not zeros of $g$. Hence we have
$$||\frac{f}{g}||_R\leq e^{C\cdot R^{\rho}}\cdot R^{m_1-m_2}\cdot \prod | E(\frac{z}{z_j},p)|$$
where $C$ is the leading coefficient of the polynomial $g_1-g_2$.
Since a canonical Weierstrass product has order of growth less or equal to the convergence exponent of the sequence of zeros, the right hand side of last inequality is definitely $\leq D\cdot e^{R^{\rho}}$, for suitable constants $D$ and $R$ big enough. This proves the finiteness of the order for the quotient $f/g$.
