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Let f(z) be analytic function and $\forall z=x+iy\in\mathbb C, u_x=u_y$ ($u_x=\frac{\partial f}{\partial x},u_y=\frac{\partial f}{\partial y}$. Prove that $f(z)=az+b$.

I thought using Cauchy Riemann equations with the given equation: $$\begin{cases}u_x=u_y \\ u_x=v_y\\u_y=-v_x\end{cases}$$ $\Rightarrow u_x=-v_x\Rightarrow u(x,y)=v(x,y)+c(y)$ (where c(y) is a function of $\Im(z)$).As follows $$v_x=u_x=u_y=v_y + c_y(y)$$ and in C.R $u_y=-v_x\Rightarrow v_x=-v_x\Rightarrow v_x=0$ so also $v(x,y)$ is a function only of $y=\Im z$. So I got $f(x,y)=v(y)+c(y)+iv(y)=v(y)(1+i)+c(y)$. I can't find any way to prove that $c(y)\equiv c\in\mathbb C$ or $v(y)$ is a linear function. How can I do it?

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  • $\begingroup$ You for $v_x=0$. What does that imply about $u_x,u_y,v_y$? $\endgroup$ – N. S. Jul 9 '14 at 12:58
  • $\begingroup$ Ouch. So $u=c_1,v=c_2\quad (c_{1,2}\in\mathbb{C})$ ? but that's a constant function (not sure whether they asked for a linear function with incline $\neq 0$). Thanks. $\endgroup$ – user65985 Jul 9 '14 at 13:05
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From $u_x=u_y$ and $u_x=v_y$, you have $u_y=v_y$ which implies $$ u=v+2c_1(x). $$ Similarly from $u_x=u_y$ and $u_y=-v_x$, you have $u_x=-v_x$ which implies $$ u=-v+2c_2(y). $$ Thus $$ u=c_1(x)+c_2(y), v=c_2(y)-c_1(x). $$ Now use $u_x=u_y$ to get $c_1'(x)=c_2'(y)$. Since $c_1(x)$ and $c_2(y)$ are functions of $x$ and $y$, respectively, you have $$ c_1'(x)=c_2'(y)\equiv k $$ and hence $c_1(x)=kx+k_1$ and $c_2(y)=ky+k_2$. So $$ f(z)=u+iv=k(x+y)+ki(y-x)+b=(1-i)k(x+yi)+k_3=az+b $$ where $a=(1-i)k$ and $b=k_1+ik_2$.

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From Cauchy-Riemann we have $u_{xx} + u_{yy} = 0$, if now $u_x = u_y$, we have $$ u_{xx} = (u_x)_x = (u_y)_x = u_{xy} = (u_x)_y = u_{yy}$$ so $2u_{xx} = 0$, that is $u_{xx} = u_{yy} = u_{xy} = 0$. For $v$, we have $$ v_{xx} = (v_x)_x = -(v_y)_x = -v_{xy} = -(v_x)_y = v_{yy} $$ which again gives $v_{xx} = v_{xy} = v_{yy} = 0$ from Cauchy-Riemann. So both $u$ and $v$ are affine-linear, hence, $f$ is also. As $f$ is analytic, $f(z) = az+b$ for some $a,b \in \mathbb C$.

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  • $\begingroup$ why $u_{xx}+y_{yy}=0$? $\endgroup$ – user65985 Jul 9 '14 at 13:36

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