Let $f:\mathbb{C} \to \mathbb{C}$ be defined by $$f(z)=(1-z)e^{\big( z+ \frac{z^2}{2} \big)}=1+ \sum_{n=1}^{\infty} a_nz^n.$$
Then, which of the following is FALSE?
$f'(z)=-z^2e^{\big( z+ \frac{z^2}{2} \big)}$
$a_1=a_2$
$a_n \in (-\infty,0]$
$\sum_{n=3}^{\infty} |a_n|<1$
My attempt:
Straight forward calculations yield that both options 1) and 2) are TRUE.We can see that $a_1=0,a_2=0$ and $a_3=-\frac{2}{6}.$ But that does not guarantee that $a_n \in (-\infty,0]$ for all $n.$
The answer key says that option 4) is FALSE, meaning that 4) is the right answer for this question.
I am stuck here and not able to go ahead. How to see this ? Please help and provide some hints. Thank you very much.
All Taylor coefficients of $$ f'(z)=-z^2e^{\big( z+ \frac{z^2}{2} \big)} $$ are zero or negative, which implies that the same is true for the Taylor coefficients of $f$, with the possible exception of the constant coefficient. This proves that (3) is true.
As already said in a comment, you can then use that $$ 0 = f(1) = 1 + \sum_{n=3}^\infty a_n = 1 - \sum_{n=3}^\infty |a_n| \, . $$