Let $f(z)=|z|^2+z$.
- Find $z=z_1$ where $f(z)$ is not differentiable
- Does there exist analytic function $g:\Bbb{C} \to \Bbb{C}$ such that $f(z)+g(z)$ would be analytic in $\Bbb{C}$?
- Does there exist any function (not neceserilly analytic) $g:\Bbb{C} \to \Bbb{C}$ such that $f(z)+g(z)$ would be analytic in $\Bbb{C}$?
My attempt so far:
- $f(x+iy)=(x^2+y^2+x)+iy$. By Cauchy-Riemann equations $u_x=2x+1$, $v_y=1 \to$ $2x+1=1 \to$ $x=0$. $\quad v_x=0,-u_y=-2y \to y=0$ That means that given function is differentiable only at the point $(0,0)$. So, for example, it is not differentiable at for example $z_1=1$. Am I right? Or do I have to prove it by using limit to show that the functions derivate is infinite at $z_1$?
- To be honest, I guess that the answer is negative. But how should I prove it?
- Lets take $g(z)=-|z|^2$ which is not analytic (but differentiable at only one point $(x,y)=(0,0)$). Then $f(z)+g(z)=z$ which is analytic function.
You cannot find such a analytic function. If $g$ and $f+g$ are analytic so is $f=(f+g)-g$. But then $|z|^{2}=f(z)-z$ is analytic. But no (non-constant) real valued function is analytic. If $g(x)=-|z|^{2}$ then $f+g$ is analytic (but $g$ is not analytic). $f$ is not differentiable at any point other than $0$.