Proof that the integral $ frac{f'(x)}{f(x)}$ is zero












2















If $f$ is differentiable in a region $U$, $f'$ is continuous in $U$ and $|f(x)-1|<1$ in $U$, proof that the integral $frac{f'(z)}{f(z)}$ on any closed curve in $U$ is zero.




I know that if $f$ is differentiable then it is holomorphic. And if $F(z)=f'(z)$, $f$ holomorphic then the integral is zero. But I don't know how to use the hypothesis: $|f(x)-1|<1$






complex-analysis







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  • $frac{f'(z)}{f(z)}$ is the logarithmic derivative. What do you mean by integral?
    – Nodt Greenish
    Nov 25 '18 at 7:59










  • Also what do you think of the case $|f(z)-1|le 1$ ?
    – reuns
    Nov 25 '18 at 8:05


















2















If $f$ is differentiable in a region $U$, $f'$ is continuous in $U$ and $|f(x)-1|<1$ in $U$, proof that the integral $frac{f'(z)}{f(z)}$ on any closed curve in $U$ is zero.




I know that if $f$ is differentiable then it is holomorphic. And if $F(z)=f'(z)$, $f$ holomorphic then the integral is zero. But I don't know how to use the hypothesis: $|f(x)-1|<1$






complex-analysis







share|cite|improve this question
























  • $frac{f'(z)}{f(z)}$ is the logarithmic derivative. What do you mean by integral?
    – Nodt Greenish
    Nov 25 '18 at 7:59










  • Also what do you think of the case $|f(z)-1|le 1$ ?
    – reuns
    Nov 25 '18 at 8:05
















2












2








2








If $f$ is differentiable in a region $U$, $f'$ is continuous in $U$ and $|f(x)-1|<1$ in $U$, proof that the integral $frac{f'(z)}{f(z)}$ on any closed curve in $U$ is zero.




I know that if $f$ is differentiable then it is holomorphic. And if $F(z)=f'(z)$, $f$ holomorphic then the integral is zero. But I don't know how to use the hypothesis: $|f(x)-1|<1$






complex-analysis







share|cite|improve this question
















If $f$ is differentiable in a region $U$, $f'$ is continuous in $U$ and $|f(x)-1|<1$ in $U$, proof that the integral $frac{f'(z)}{f(z)}$ on any closed curve in $U$ is zero.




I know that if $f$ is differentiable then it is holomorphic. And if $F(z)=f'(z)$, $f$ holomorphic then the integral is zero. But I don't know how to use the hypothesis: $|f(x)-1|<1$






complex-analysis




complex-analysis






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 8:02









Mason

1,9591530




1,9591530










asked Nov 25 '18 at 7:52









Jazmín Jones

519




519












  • $frac{f'(z)}{f(z)}$ is the logarithmic derivative. What do you mean by integral?
    – Nodt Greenish
    Nov 25 '18 at 7:59










  • Also what do you think of the case $|f(z)-1|le 1$ ?
    – reuns
    Nov 25 '18 at 8:05




















  • $frac{f'(z)}{f(z)}$ is the logarithmic derivative. What do you mean by integral?
    – Nodt Greenish
    Nov 25 '18 at 7:59










  • Also what do you think of the case $|f(z)-1|le 1$ ?
    – reuns
    Nov 25 '18 at 8:05


















$frac{f'(z)}{f(z)}$ is the logarithmic derivative. What do you mean by integral?
– Nodt Greenish
Nov 25 '18 at 7:59




$frac{f'(z)}{f(z)}$ is the logarithmic derivative. What do you mean by integral?
– Nodt Greenish
Nov 25 '18 at 7:59












Also what do you think of the case $|f(z)-1|le 1$ ?
– reuns
Nov 25 '18 at 8:05






Also what do you think of the case $|f(z)-1|le 1$ ?
– reuns
Nov 25 '18 at 8:05












1 Answer
1






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2














The hypothesis that $|f(x)-1|<1$ on $U$ implies that $f(U)$ is contained in the right half plane, where the principal branch of the logarithm is analytic with derivative $(ln z)'=1/z$. By the chain rule, $(ln f(z))'=f'(z)/f(z)$, so use $F(z)=ln f(z)$ as in your question to get the result.






share|cite|improve this answer





















  • thanks again C:
    – Jazmín Jones
    Nov 25 '18 at 18:30











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














The hypothesis that $|f(x)-1|<1$ on $U$ implies that $f(U)$ is contained in the right half plane, where the principal branch of the logarithm is analytic with derivative $(ln z)'=1/z$. By the chain rule, $(ln f(z))'=f'(z)/f(z)$, so use $F(z)=ln f(z)$ as in your question to get the result.






share|cite|improve this answer





















  • thanks again C:
    – Jazmín Jones
    Nov 25 '18 at 18:30
















2














The hypothesis that $|f(x)-1|<1$ on $U$ implies that $f(U)$ is contained in the right half plane, where the principal branch of the logarithm is analytic with derivative $(ln z)'=1/z$. By the chain rule, $(ln f(z))'=f'(z)/f(z)$, so use $F(z)=ln f(z)$ as in your question to get the result.






share|cite|improve this answer





















  • thanks again C:
    – Jazmín Jones
    Nov 25 '18 at 18:30














2












2








2






The hypothesis that $|f(x)-1|<1$ on $U$ implies that $f(U)$ is contained in the right half plane, where the principal branch of the logarithm is analytic with derivative $(ln z)'=1/z$. By the chain rule, $(ln f(z))'=f'(z)/f(z)$, so use $F(z)=ln f(z)$ as in your question to get the result.






share|cite|improve this answer












The hypothesis that $|f(x)-1|<1$ on $U$ implies that $f(U)$ is contained in the right half plane, where the principal branch of the logarithm is analytic with derivative $(ln z)'=1/z$. By the chain rule, $(ln f(z))'=f'(z)/f(z)$, so use $F(z)=ln f(z)$ as in your question to get the result.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 8:01









Guacho Perez

3,88911131




3,88911131












  • thanks again C:
    – Jazmín Jones
    Nov 25 '18 at 18:30


















  • thanks again C:
    – Jazmín Jones
    Nov 25 '18 at 18:30
















thanks again C:
– Jazmín Jones
Nov 25 '18 at 18:30




thanks again C:
– Jazmín Jones
Nov 25 '18 at 18:30


















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