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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 $f(z)=az+b$. 
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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?