Test whether the following statements are TRUE or FALSE.
$1.$ Let, $f_j(j=1,2,...,n)$ is analytic in a domain $D$ such that $\sum_{j=1}^n|f_j(z)|^2$ is constant in $D$. Then each $f_j$ is costant throughout the domain $D$.
$2.$ If $u$ is a real valud function in a disc $D$ such that $u^{-1}+iu$ is analytic in $D$ then $u$ is constant throughout the disc.
$3.$ If $D$ is a domain which is symmetric about the real axis and if $f$ is differentiable at $a\in D$ as well as at $\bar a\in D$ , then $f(\bar z)$ is NOT differentiable at $a$.
Attempts:
$1.$ Let , $f_j(z)=u_j(x,y)+iv_j(x,y)$.
Now , $$\sum_{j=1}^n|f_j(z)|^2=\text{constant}.$$
$$\implies\sum_{j=1}^n\left(u_j^2+v_j^2\right)=\text{constant}.$$
$$\implies\left(u_j^2+v_j^2\right)=\text{constant}.$$
$$\implies u_j=\text{constant},v_j=\text{constant}$$
$$\implies f_j = \text{constant}.$$
$2.$ Let, $f(z)=u^{-1}+iu$ is analytic in $D$. So, $u^{-1}$ & $u$ satisfy Laplace equation.
So, $$\frac{\partial ^2}{\partial x^2}(u^{-1})+\frac{\partial ^2}{\partial y^2}(u^{-1})=0.$$
$$\implies \frac{\partial ^2}{\partial x\partial y}(u)-\frac{\partial ^2}{\partial x\partial y}(u^{-1})=0.$$ (using Cauchy-Riemann equation)
$$\implies \frac{\partial ^2}{\partial x\partial y}(u-u^{-1})=0.$$
$$\implies u-u^{-1}=\phi_1(x)+\phi_2(y).$$ not necessarily a constant , which contradicts the fact that if $u$ is constant then $u-u^{-1}$ is also constant. So this statement is false.
$3.$ I have no idea about this.
Please check my solutions & help for 3.
Edit : Please verify the problem 2. I'm confused with my solution and the posted answer.