Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of complex numbers with $a_1 \ne 0$. Moreover we assume that $\sum na_n$ converges absolutely, and that $\sum_{n = 2}^\infty n \mid a_n \mid \leq \mid a_1\mid$. Prove that $f$ is one-to-one on the open unit disk (ie.$\{z \in \mathbb{C}, \mid z \mid < 1 \}$), where $f(x) = \sum_{n = 0}^\infty a_nx^n$.
Here is what I've done :
We have $f'(x) = a_1 + \sum_{n = 2}^{\infty} a_nnx^{n-1}$. Moreover, since we have $$\mid \sum_{ n= 2}^\infty a_nnx^{n-1} \mid \leq \sum_{ n = 2}^\infty \mid a_n \mid \mid nx^{n-1} \mid \leq \sum_{ n = 2}^\infty n \mid a_n \mid \leq \mid a_1 \mid$$
Hence we have (we can assume $a_1\geq 0$ since when $a_1 < 0$ it is the same argument in reverse) :
$$-a_1 \leq \sum_{n = 2}^ \infty a_nnx^{n-1} \leq a_1$$
So we have $0 \leq f'(x) \leq 2a_1$ (if $a_1 < 0$ then we will get $-2a_1 \leq f'(x) \leq 0$).
Thus to prove that $f$ is one-to-one it is sufficient to prove that $f'(x) \ne 0$ for all $x$. But this clearly is true, because to have $f'(x) = 0$ for some $x$ we need
$$\sum_{n = 2}^\infty n \mid a_n \mid =\mid a_ 1 \mid$$ Yet since we clearly have
$$\mid \sum_{n = 2}^\infty a_nnx^{n-1} \mid \leq \sum_{n = 2}^\infty \mid a_n \mid \mid nx^{n-1} \mid < \sum_{n = 2}^\infty n \mid a_n \mid $$
So it's impossible to find such an $x$. So to conclude we get $f'(x) > 0$ ($f'(x) < 0$ if $a_1 < 0$) for all $x$ in the open unit disk, so $f$ is one-to-one.
I don't know if what I've done is correct, because my book is doing something far more complicated using a proof by contradiction which is not intuitive.