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

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!

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 $$-\triangle f\leq 0$$ everywhere. You need to use the Cauchy-Riemann-equations for this.
Since an analytic function is smooth, we can use that $$f\in C^2$$ is subharmonic iff. $$-\triangle f\leq 0$$ everywhere.