# On the Zeros of the Differential Polynomial \(\varphi(z)f^{2}(z)f^{\prime}(z)^{2}-1\)
## 2 Lemmas  
In order to prove our result, we need the following lemma. **Lemma 1**. : (see [15]) _Let \(f\) be a non-constant meromorphic function in the complex plane, let \(Q_{1}[f],Q_{2}[f]\) be quasi-differential polynomials in \(f\), satisfying \(f^{n}Q_{1}[f]=Q_{2}[f].\) If the total degree of \(Q_{2}\) is inferior or equal to \(n\), then_  
\[m(r,Q_{1}[f])=S(r,f).\]  
**Notations:**  
\[F(z)=\varphi(z)f^{2}(z)(f^{\prime}(z))^{2}-1, \tag{6}\]  
\[h(z)=\frac{F^{\prime}(z)}{f(z)}=2\varphi(z)[f(z)f^{\prime}(z)f^{\prime\prime}( z)+(f^{\prime}(z))^{3}]+\varphi^{\prime}(z)f(z)(f^{\prime}(z))^{2}, \tag{7}\]  
\[\phi(z)=\frac{h(z)}{F(z)}=\frac{1}{f(z)}\cdot\frac{F^{\prime}(z)}{F(z)}, \tag{8}\]  
\[ \begin{array}{ll}G(z)&=20(\frac{F^{\prime}(z)}{F(z)})^{2}+24(\frac{F^{ \prime}(z)}{F(z)})^{\prime}-39\frac{F^{\prime}(z)}{F(z)}\frac{h^{\prime}(z)}{ h(z)}+18(\frac{h^{\prime}(z)}{h(z)})^{2}-18(\frac{h^{\prime}(z)}{h(z)})^{ \prime}\\ &+\frac{15}{4}\frac{\varphi^{\prime}(z)}{\varphi(z)}\frac{F^{\prime}(z)}{F(z) }-\frac{9}{4}\frac{\varphi^{\prime}(z)}{\varphi(z)}\frac{h^{\prime}(z)}{h(z)} -9(\frac{\varphi^{\prime}(z)}{\varphi(z)})^{\prime}-\frac{15}{8}(\frac{ \varphi^{\prime}(z)}{\varphi(z)})^{2}.\end{array}  \tag{9}\]  
**Lemma 2**. : _Let \(f(z)\) be a transcendental meromorphic function and let \(\varphi(z)(\
ot\equiv 0)\) be a small function. Then \(\varphi f^{2}f^{\prime 2}\) is not equivalent to a constant._  
Proof. : Suppose \(\varphi f^{2}f^{\prime 2}\equiv C\), where \(C\) is a constant. Obviously, \(C\
eq 0\). Then  
\[\frac{1}{f^{4}}\equiv\frac{\varphi}{C}(\frac{f^{\prime}}{f})^{2},\quad\frac{1 }{f^{2}f^{\prime 2}}\equiv C\varphi.\]  
Therefore,  
\[ \begin{array}{lll}m(r,\frac{1}{f})&\leq&\frac{1}{4}m(r,\frac{1}{C}\varphi( \frac{f^{\prime}}{f}))^{2}\\ &\leq&\frac{1}{4}m(r,\varphi)+\frac{1}{4}m(r,(\frac{f^{\prime}}{f})^{2})+O(1) =S(r,f),\\ N(r,\frac{1}{f})&\leq&N(r,\frac{1}{f^{2}f^{\prime 2}})\\ &=&N(r,\frac{1}{C}\varphi)=S(r,f).\end{array} \]  
From the above, we have \(T(r,f)=S(r,f)\). It is a contradiction. Hence the proof of Lemma 2 is completed. Hence we have  
\[ \begin{array}{ll}4N(r,\frac{1}{f})-N_{00}(r,\frac{1}{(\varphi f^{2}f^{\prime 2})^{ \prime}})&\leq 2N_{1})(r,\frac{1}{f})+\overline{N}_{1})(r,\frac{1}{f})+3 \overline{N}_{(2}(r,\frac{1}{f})+N(r,\varphi^{2})\\ &=2N_{1})(r,\frac{1}{f})+\overline{N}(r,\frac{1}{f})+2\overline{N}_{(2}(r, \frac{1}{f})+S(r,f).\end{array}  \tag{15}\]  
Combining (12)-(15), we have  
\[ \begin{array}{llll}4T(r,f)&\leq&\overline{N}(r,f)+3\overline{N}(r,\frac{1}{ f})+\overline{N}(r,\frac{1}{\varphi f^{2}f^{\prime 2}-1})-N_{0}(r,\frac{1}{( \varphi f^{2}f^{\prime 2})^{\prime}}).\end{array} \]  
This completes the proof of the Lemma 3. **Lemma 4**. : _Under the hypotheses of Theorem 6, for any \(z_{0}\in\mathbb{C}\), we have_  
\[\omega(\frac{1}{f},z_{0})+\omega(\frac{1}{h},z_{0})\leq\omega(\frac{1}{fh},z_{ 0})+\omega(\varphi,z_{0})+\omega(\frac{1}{\varphi},z_{0}). \tag{16}\]  
Proof. : This proof is divided into three Cases:  
Case 1. \(f(z_{0})\
eq 0,\infty\). If \(h(z_{0})\
eq 0\), then \(\omega(\frac{1}{f},z_{0})+\omega(\frac{1}{h},z_{0})=0\). If \(h(z_{0})=0\), from (2), then we get  
\[\omega(\frac{1}{f},z_{0})+\omega(\frac{1}{h},z_{0})=\omega(\frac{1}{fh},z_{0}).\]  
Hence the inequality (16) holds. Case 2. \(f(z_{0})=0\). By \(\omega(\frac{\varphi}{\varphi},z_{0})\leq 1\), we have  
\[\frac{h}{\varphi}=2[ff^{\prime}f^{\prime\prime}+(f^{\prime})^{3}]+\frac{ \varphi^{\prime}}{\varphi}f(f^{\prime})^{2}\
eq\infty(z=z_{0}).\]  
From this, (1) and (2), we have  
\[ \begin{array}{llll}\omega(\frac{1}{f},z_{0})+\omega(\frac{1}{h},z_{0})& \leq&\omega(\frac{1}{f},z_{0})+\omega(\frac{\varphi}{h},z_{0})+\omega(\frac{1 }{\varphi},z_{0})\\ &=&\omega(\frac{1}{f},\frac{\varphi}{h},z_{0})+\omega(\frac{1}{\varphi},z_{0} )\\ &\leq&\omega(\frac{1}{fh},z_{0})+\omega(\varphi,z_{0})+\omega(\frac{1}{\varphi },z_{0}).\end{array} \]  
Hence the inequality (16) holds. Case 3. \(f(z_{0})=\infty\). Suppose \(l=\omega(f,z_{0}),l_{1}=\max\{\omega(\varphi,z_{0}),\omega(\frac{1}{\varphi},z_{0})\}\).