I have the following problem:
If f is entire, find all g entire such that $|f(z)-g(z)|\leq |f(z)+g(z)|$ for every $z\in \mathbb{C}$
My initial thought was to write $f(x+iy)=u_1(x,y)+iv_1(x,y)$ and $g(x+iy)=u_2(x,y)+iv_2(x,y)$. Following the desired inequalities, I arrived at the following condition: $0\leq Re(f)Re(g)+Im(f)Im(g)$
However this doesn’t seem very intuitive to me, this condition doesn’t feel very enlightening. I was wondering if anyone could help me understand if this condition is the desired one and, if so, why?
If the set of zeroes of $f+g$ has an accumulation point, then $f+g$ is identically zero. Hence $g=-f$ and the inequality $|f-g|\le|f+g|$ implies that $f=g=0$.
Now suppose that all zeroes of $f+g$ are isolated points. Let $a$ be such a zero, so that $f+g$ is nonzero on a puntured open disc $A=\{z\in\mathbb C:0<|z-a|<r\}$. Therefore $h=(f-g)/(f+g)$ is analytic on $A$ and by the given inequality, it is also bounded on $A$. It follows from Great Picard Theorem that $a$ is not an essential singularity of $h$. Hence it is a pole. The boundedness of $h$ therefore further implies that $a$ is a removable singularity. Since this is true for every singularity of $h$, $h$ is extensible to an entire function. As it is bounded, it is a constant function, by Liouville's Theorem. Therefore $f-g=k(f+g)$ for some constant $k$ with modulus $\le1$. If $k=-1$, then $f=0$ and any entire function$g$ will satisfy the given inequality; if $k\ne-1$, then $g=\frac{1-k}{1+k}f$.
In summary, if $f=0$, any entire function $g$ will do; if $f$ is not the zero function, the solutions are given by $g=\frac{1-k}{1+k}f$ for some complex constant $k$ such that $k\ne-1$ and $|k|\le1$.