Let $f$ be holomorphic on connected open set $D$. Prove $u^4+v^4$ being constant implies $f=u+iv$ being constant.

Question

Differentiating $u^4+v^4$, we have $$ 4u^3u_x + 4v^3v_x = 0 $$ $$ 4u^3u_y + 4v^3v_y = 0 $$

Then we have $$u^3v^3 u_x v_y = u^3v^3 u_y v_x.$$ If $uv\neq 0$, then we can apply Cauchy-Riemann condition to conclude $u,v$ are constant. However, what if $uv=0$? I struggle to proceed with this case.

Answer 1

Apply Cauchy-Riemann equation to the two equations you have got, we have $$4u^3u_x-4v^3u_y=0$$ $$4v^3u_x+4u^3u_y=0$$ which is an linear equation of $u_x,u_y$ with the determinant of coefficient matrix $16(u^6+v^6)$.

Therefore, if the determinant is $0$ at some point in $D$, then $u^4+v^4\equiv 0$, which implies $f\equiv 0$, if the determinant is not $0$ at every point in $D$, then the equation has unique solution $u_x=u_y=0$, which implies $f$ is constant.

Answer 2

If $uv=0$ it means that at least one of them is identically constant (being $D$ connected). Say $v\equiv0$. Then use CR equations to get that $u$ is constant. So $f$ is.