Entire function of finite order with deficient value

Question

There are some sufficient conditions for an entire function to have a finite deficient value e.g., if the order $\rho$ of an entire function $f$ is such that $2<\rho<+\infty$ with all but finitely many zeros of $f$ are real, then $0$ is a deficient value of $f$.

Question: Is there any sufficient condition for an entire function of finite order $1$ which has a finite deficient value?

In particular, does the function $f(z)=e^z+P(z)$ where $P(z)$ is a complex polynomial have a finite deficient value ?

Purpose: I want to know whether all the Fatou components of $e^z+P(z)$ are simply connected or not.

Any suggestions or comments are welcome.

Answer 1

The theorem you stated, and its various versions and generalizations, are the only simple sufficient conditions for $\delta(0)>0$. For example, if $f$ is entire of genus $1$, and zeros lie on a ray, then $\delta(0,f)>0$. (This covers some functions of order $1$. Your function is of genus $1$, but its zeros do not belong to a ray.)

Your function $f(z)=e^z+P(z)$ with non-constant $P$ has no finite deficient values, and one does not need any deep theorems to see this. Indeed, suppose that is has deficiency at some $a\in C$. WLOG we may assume that $a=0$. Then $g(z)=e^z/P(z)$ would have $\delta(0,g)=\delta(\infty,g)=1$, and $\delta(-1,g)>0$, which contradicts the deficiency relation.