If $f(z)$ and $g(z)$ are analytic, $g(1+i)=0$, is $h(z)=\frac{f(z)}{g(z)}$ analytic?

Question

1. Does there exist analytic functions $f(z)$ and $g(z)$, $g(1+i)=0$, such that $h(z)=\frac{f(z)}{g(z)}$ is also analytic in all $\Bbb{C}$.
2. Does there exist non-analytic functions $f(z)$ and $g(z)$ such that $h(z)=f(z)+g(z)$ is analytic in all $\Bbb{C}$?

Can anyone help me and give me ideas and other useful information that I can use to solve this problem? I am really doubt about this exercise.

First of all, if $g(1+i)=0$ and analytic in all $\Bbb{C}$ it follows (I guess) that $g(z)=0$ $\forall z \in \Bbb{C}$. That means that I need to construct $f(z)$ such that $h(z)=\frac{f(z)}{0}$ must exist and be finite. Does that mean that $f(z)\approx 0$?

Answer 1

I don't understand what you wrote about $g$. Do you believe that if $1+i$ is a zero of an analytic function $g$, then $g$ is the null function. What about $g(z)=z-(1+i)$ then?

Anyway, the answer is negative, because of $g(1+i)=0$, then $h$ is not even defined at $1+i$.

And if you define $f(z)=\overline z$ and $g(z)-\overline z$, then neither $f$ nor $g$ are analytic, but $f+g$ is the null function, which is analytic.