# On the Zeros of the Differential Polynomial \(\varphi(z)f^{2}(z)f^{\prime}(z)^{2}-1\)
## 2 Lemmas
In the following, we divide into two Subcases:  
Subcase 3.1. Let \(1\leq l\leq l_{1}\). Then  
\[ \begin{array}{llll}\omega(\frac{1}{f},z_{0})+\omega(\frac{1}{h},z_{0})&=& \omega(\frac{1}{fh},z_{0})\leq\omega(\frac{1}{fh},z_{0})+\omega(f,z_{0})\\ &\leq&\omega(\frac{1}{fh},z_{0})+\omega(\varphi,z_{0})+\omega(\frac{1}{\varphi },z_{0}).\end{array} \]  
Subcase 3.2. Let \(l>l_{1}\geq 0\). Using the Laurent series of \(\frac{h}{f^{3}\varphi}=2[\frac{f^{\prime}f^{\prime\prime}}{f^{2}}+(\frac{f^{ \prime}}{f})^{3}]+\frac{\varphi^{\prime}}{\varphi}\cdot(\frac{f^{\prime}}{f})^{2}\) at the point \(z_{0}\), we obtain the coefficient of \((z-z_{0})^{-3}\):  
\[a_{-3}=-2l^{2}(l+1)-2l^{3}-l^{2}(l_{1}+1)<0.\]  
Thus \(\omega(\frac{\partial}{h},z_{0})=0\). Therefore,  
\[ \begin{array}{llll}\omega(\frac{1}{f},z_{0})+\omega(\frac{1}{h})&=&\omega( \frac{\beta_{1}^{3}\varphi}{h}\cdot\frac{1}{f^{3}}\cdot\frac{1}{\varphi},z_{0} )\\ &\leq&\omega(\frac{\beta_{2}^{3}\varphi}{h}\cdot z_{0})+3\omega(\frac{1}{f},z_{ 0})+\omega(\frac{1}{\varphi},z_{0})=\omega(\frac{1}{\varphi},z_{0}).\end{array} \]Hence the inequality (16) holds. This completes the proof of Lemma 4. Under the hypotheses of Theorem 3.2, if \(z_{0}\in\mathbb{C}\) and \(G(z_{0})=0\), then  
\[\omega(\phi,z_{0})\leq 2\omega(\varphi,z_{0})+\omega(\frac{1}{\varphi},z_{0}), \tag{17}\]  
\[\omega(\frac{1}{F},z_{0})\leq\omega(\frac{1}{h},z_{0})+2\omega(\varphi,z_{0})+ \omega(\frac{1}{\varphi},z_{0}). \tag{18}\]  
Proof. : First, we prove the following inequality  
\[\overline{\omega}(\frac{1}{F},z_{0})\leq\omega(\varphi,z_{0})+\omega(\frac{1} {\varphi},z_{0}). \tag{19}\]  
Obviously, if \(F(z_{0})\
eq 0\), then the inequality (19) holds. Now let \(\omega(\frac{1}{F},z_{0})=l(l\geq 1)\) and \(\varphi(z_{0})\
eq 0,\infty\). Since \(\varphi(z_{0})f^{2}(z_{0})f^{\prime 2}(z_{0})=F(z_{0})+1=1\), then \(f(z_{0})\
eq 0,\infty\). Thus, \(z_{0}\) is a zero of \(h(z)=\frac{F^{\prime}(z)}{f(z)}\) with multiplicity \(l-1\) (if \(l=1\) and \(h(z_{0})\
eq 0,\infty\)). Using the Laurent series of \(G(z)\) at the point \(z_{0}\), we obtain the coefficient of \((z-z_{0})^{-2}\):  
\[b_{-2}=20l^{2}-24l-39l(l-1)+18(l-1)^{2}+18(l-1)=-l(l+3)<0.\]  
It contradicts with \(G(z_{0})=0\). Hence \(z_{0}\) is a zero or a pole of \(\varphi(z_{0})\). This implies that the inequality (19) holds. In order to prove (17), we will divide two Cases. Case 1. \(f(z_{0})\
eq 0\). Suppose that \(F(z_{0})\
eq\infty\). By (19), we have  
\[\omega(\phi,z_{0}) = \omega(\frac{1}{f}\cdot\frac{F^{\prime}}{F},z_{0})\leq\omega( \frac{1}{f},z_{0})+\omega(\frac{F^{\prime}}{F},z_{0})\] \[= \overline{\omega}(\frac{1}{F},z_{0})\leq\omega(\varphi,z_{0})+ \omega(\frac{1}{\varphi},z_{0}).\]  
Suppose that \(F(z_{0})=\infty\). If \(f(z_{0})=\infty\) and \(\frac{1}{f}\cdot\frac{F^{\prime}}{F}\
eq\infty(z=z_{0})\), then  
\[\omega(\phi,z_{0})=\omega(\frac{1}{f}\cdot\frac{F^{\prime}}{F},z_{0})=0.\]  
If \(f(z_{0})\
eq\infty\), since \(\varphi(z_{0})f^{2}(z_{0})f^{\prime 2}(z_{0})=F(z_{0})+1=\infty\), then we get  
\[\omega(\varphi,z_{0})\geq 1.\]  
Therefore,  
\[\omega(\phi,z_{0})\leq\omega(\frac{1}{f},z_{0})+\omega(\frac{F^{\prime}}{F},z _{0})=1\leq\omega(\varphi,z_{0}).\]  
Case 2. \(f(z_{0})=0\). Suppose that \(\omega(\frac{1}{f},z_{0})>\omega(\varphi,z_{0})\). Then \(F(z_{0})=\varphi(z_{0})f^{2}(z_{0})f^{\prime 2}(z_{0})-1=-1\) and \(\frac{\varphi^{\prime}}{\varphi}f\
eq\infty\)\((z=z_{0})\), therefore  
\[\omega(\frac{1}{F},z_{0})=0,\]  
\[\omega(\frac{1}{F},z_{0})\leq\omega(\frac{1}{h},z_{0})+2\omega(\varphi,z_{0})+ \omega(\frac{1}{\varphi},z_{0}). \tag{18}\]0Thus, \(\omega(\phi,z_{0})=\omega(\frac{1}{F}\cdot\frac{h}{\varphi}\cdot\varphi,z_{0})\leq \omega(\frac{1}{F},z_{0})+\omega(\frac{h}{\varphi},z_{0})+\omega(\varphi,z_{0})= \omega(\varphi,z_{0})\). Suppose that \(1\leq\omega(\frac{1}{f},z_{0})\leq\omega(\varphi,z_{0})\). Then  
\[\omega(\frac{1}{F},z_{0})\leq\omega(\frac{1}{h},z_{0})+2\omega(\varphi,z_{0})+ \omega(\frac{1}{\varphi},z_{0}). \tag{18}\]1  
Therefore, the inequality (17) holds.