If $u+u^{-1}$ is analytic then $u$ is constant

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.

• In 1, the implication $\sum (u_i^2 + v_i^2) =$ constant implies $(u_i^2 + v_i^2) =$ constant for every $i$ is false or need further explanation. – Tlön Uqbar Orbis Tertius Mar 8 '15 at 8:23
• Ok...So, what will be it? – Empty Mar 9 '15 at 3:11
• Your last question lacks something : take any constant function, it's differentiable everywhere. An extra condition might be more interesting : either it be non constant, of $f$ analytic. – Tlön Uqbar Orbis Tertius Mar 9 '15 at 7:43

1. For the first one, take the laplacian $\Delta$ and use the fact that if $f$ is holomorphic, $\Delta|f|^2 = 4 |f'|^2$. Therefore, as the sum $\sum|f_i|^2$ is constant, you'll have $$0 = \Delta\left( |f_1|^2 + ... + |f_n|^2\right) = 4 \left(|f'_1|^2 + ... + |f'_n|^2 \right)$$ It is now clear that $|f'_i|^2 = 0$ for every $i$, showing that each of them is aconstant.

2. (someone please check this one) . It is a general fact that if $f$ is holomorphic and if $f = f_1 + if_2$, and if

$$J = \begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{pmatrix}$$ denotes its jacobian matrix, then its determinant $|J|$ is equal to $|f'|^2$. This is a consequence of the Cauchy-Riemann equations.

If $f = \frac{1}{u} + iu$ is holomorphic, then writing its jacobian matrix yelds

$$J = \begin{pmatrix} \frac{\partial u^{-1}}{\partial x} & \frac{\partial u^{-1}}{\partial y} \\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \end{pmatrix} = \begin{pmatrix} -\frac{\partial u}{\partial x}\frac{1}{u^2} & -\frac{\partial u}{\partial y}\frac{1}{u^2}\\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \end{pmatrix}$$

Its jacobien is clearly zero, so we must have $|f'|=0$ and $f$ must be constant, forcing its imaginary part $u$ to be a constant too.

3. The contant function equal to zero everywhere is differentiable everywhere.

• Where your doubt about 2.? – Empty Mar 9 '15 at 8:07
• well, I checked every part and it seems correct, but I always feel slightly uncomfortable with the links between jacobian and complex derivatives. – Tlön Uqbar Orbis Tertius Mar 9 '15 at 8:12