Analytic complex function with $u=v^2$

Question

I must prove that if $f = u+iv$ is analytic with $v=u^2$ then $f$ is constant. My attempt is $$f(x,y) = u(x,y)+iv(x,y) = u(x,y)+iu(x,y)^2$$ and write $$\frac{\partial v}{\partial y} = \frac{\partial }{\partial y} (u(x,y)^2) = 2u(x,y) \cdot u_y$$ $$\frac{\partial v}{\partial x} = \frac{\partial}{\partial x} (u(x,y)^2) = 2u(x,y) \cdot u_x$$ f is analytic so we can use the Cauchy - Riemann equations $$\frac{\partial u }{\partial x} = 2u(x,y) \frac{\partial u}{\partial y}$$ $$\frac{\partial u}{\partial y} = -2u(x,y)\frac{\partial u}{\partial x}$$ so $$\frac{\partial u}{\partial x} = -4u(x,y)^2\frac{\partial u}{\partial x} \implies v(x,y) = -\frac{1}{4}$$ is that right? I feel like I'm missing a lot of things. Any help is appreciated!

Answer 1

The image of the holomorphic f lies on the parabola u=$v^2$ and hence the image omits more than one point in the complex plane. Thus the function is a constant by Picard’s theorem.