If $(u(x,y))^2+u(x,y)v(x,y)$ has a local maximum or minimum in $D$, then $f$ must be constant?












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Let $f(z) = u(x,y)+iv(x,y)$ be an analytic function on a connected open set $D$ with $u(x,y)$ and $v(x,y)$ being the real and imaginary parts of $f(z)$, respectively. If $(u(x,y))^2+u(x,y)v(x,y)$ has a local maximum or minimum in $D$, then $f(z)$ must be a constant.




I am preparing for an upcoming final in complex analysis, and this question was given as a practice problem. The solution given seems very tedious and I suspect this can be proven with a simple contradiction. Since $D$ is open and connected, maybe we can assume $f$ is not constant and apply the open mapping theorem to arrive at a contradiction? Maybe apply the maximum modulus principle?



My apologies for the lack of work, I am not too sure how to attempt the problem. Any help would be much appreciated!






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    Let $f(z) = u(x,y)+iv(x,y)$ be an analytic function on a connected open set $D$ with $u(x,y)$ and $v(x,y)$ being the real and imaginary parts of $f(z)$, respectively. If $(u(x,y))^2+u(x,y)v(x,y)$ has a local maximum or minimum in $D$, then $f(z)$ must be a constant.




    I am preparing for an upcoming final in complex analysis, and this question was given as a practice problem. The solution given seems very tedious and I suspect this can be proven with a simple contradiction. Since $D$ is open and connected, maybe we can assume $f$ is not constant and apply the open mapping theorem to arrive at a contradiction? Maybe apply the maximum modulus principle?



    My apologies for the lack of work, I am not too sure how to attempt the problem. Any help would be much appreciated!






    complex-analysis







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      Let $f(z) = u(x,y)+iv(x,y)$ be an analytic function on a connected open set $D$ with $u(x,y)$ and $v(x,y)$ being the real and imaginary parts of $f(z)$, respectively. If $(u(x,y))^2+u(x,y)v(x,y)$ has a local maximum or minimum in $D$, then $f(z)$ must be a constant.




      I am preparing for an upcoming final in complex analysis, and this question was given as a practice problem. The solution given seems very tedious and I suspect this can be proven with a simple contradiction. Since $D$ is open and connected, maybe we can assume $f$ is not constant and apply the open mapping theorem to arrive at a contradiction? Maybe apply the maximum modulus principle?



      My apologies for the lack of work, I am not too sure how to attempt the problem. Any help would be much appreciated!






      complex-analysis







      share|cite|improve this question














      Let $f(z) = u(x,y)+iv(x,y)$ be an analytic function on a connected open set $D$ with $u(x,y)$ and $v(x,y)$ being the real and imaginary parts of $f(z)$, respectively. If $(u(x,y))^2+u(x,y)v(x,y)$ has a local maximum or minimum in $D$, then $f(z)$ must be a constant.




      I am preparing for an upcoming final in complex analysis, and this question was given as a practice problem. The solution given seems very tedious and I suspect this can be proven with a simple contradiction. Since $D$ is open and connected, maybe we can assume $f$ is not constant and apply the open mapping theorem to arrive at a contradiction? Maybe apply the maximum modulus principle?



      My apologies for the lack of work, I am not too sure how to attempt the problem. Any help would be much appreciated!






      complex-analysis




      complex-analysis






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      asked Dec 13 '18 at 8:43









      The math godThe math god

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          I didn't think about it much, so I don't know if it helps for the answer, but we can show that $(u(x,y))^2+u(x,y)v(x,y)$ is constant.



          For seeing this you should first calculate the Laplace operator of $f$, i.e. $triangle f$ and see that $-triangle fleq 0$ everywhere. You need to use the Cauchy-Riemann-equations for this.



          Since an analytic function is smooth, we can use that $fin C^2$ is subharmonic iff. $-triangle fleq 0$ everywhere.



          You can now use the maximum principle for subharmonic functions.






          share|cite|improve this answer





















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            I didn't think about it much, so I don't know if it helps for the answer, but we can show that $(u(x,y))^2+u(x,y)v(x,y)$ is constant.



            For seeing this you should first calculate the Laplace operator of $f$, i.e. $triangle f$ and see that $-triangle fleq 0$ everywhere. You need to use the Cauchy-Riemann-equations for this.



            Since an analytic function is smooth, we can use that $fin C^2$ is subharmonic iff. $-triangle fleq 0$ everywhere.



            You can now use the maximum principle for subharmonic functions.






            share|cite|improve this answer


























              0














              I didn't think about it much, so I don't know if it helps for the answer, but we can show that $(u(x,y))^2+u(x,y)v(x,y)$ is constant.



              For seeing this you should first calculate the Laplace operator of $f$, i.e. $triangle f$ and see that $-triangle fleq 0$ everywhere. You need to use the Cauchy-Riemann-equations for this.



              Since an analytic function is smooth, we can use that $fin C^2$ is subharmonic iff. $-triangle fleq 0$ everywhere.



              You can now use the maximum principle for subharmonic functions.






              share|cite|improve this answer
























                0












                0








                0






                I didn't think about it much, so I don't know if it helps for the answer, but we can show that $(u(x,y))^2+u(x,y)v(x,y)$ is constant.



                For seeing this you should first calculate the Laplace operator of $f$, i.e. $triangle f$ and see that $-triangle fleq 0$ everywhere. You need to use the Cauchy-Riemann-equations for this.



                Since an analytic function is smooth, we can use that $fin C^2$ is subharmonic iff. $-triangle fleq 0$ everywhere.



                You can now use the maximum principle for subharmonic functions.






                share|cite|improve this answer












                I didn't think about it much, so I don't know if it helps for the answer, but we can show that $(u(x,y))^2+u(x,y)v(x,y)$ is constant.



                For seeing this you should first calculate the Laplace operator of $f$, i.e. $triangle f$ and see that $-triangle fleq 0$ everywhere. You need to use the Cauchy-Riemann-equations for this.



                Since an analytic function is smooth, we can use that $fin C^2$ is subharmonic iff. $-triangle fleq 0$ everywhere.



                You can now use the maximum principle for subharmonic functions.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 11:25









                Student7Student7

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