Taylor series absolute value inequality This problem is driving me mad. Suppose you have that for $z$ near zero
$$\lambda(z)=1+a_nz^n+...$$
Where $a_n$ is non zero. I need to show this implies
$$|\lambda (z)|=1+Re(a_nz^n)+o(z^n)$$
I have tried showing that
$$\frac{|1+a_nz^n+O(z^{n+1})|-1-Re(a_nz^n)}{z^n}\rightarrow0$$
But I'm having difficulties...
Any tips?
 A: In what follows we will assume that $|z| < 1$ is small enough. We have
$$
\lambda(z) = 1 + a_n z^n + \sum\limits_{k=n+1}^\infty a_k z^k = 1+ a_n z^n + z^{n+1}g(z),
$$
where $g$ is bounded and analytic. From here we have
$$
\tag{1} |\lambda(z)|^2 = \lambda(z) \overline{\lambda(z)} = \left( 1 + a_n z^n + z^{n+1}g(z) \right) \left( 1 + \overline{a_n z^n} + \overline{z^{n+1}g(z)} \right) = \\ 1 + 2\mathrm{Re}(a_n z^n) + |a_n|^2 |z|^{2n} + z^{n+1}G(z),
$$
where we absorbed the rest of the terms in $G(z)$, which is bounded in the neighborhood of $0$. We now use the Taylor expansion of $(1+x)^\alpha$ with $\alpha>0$ and $x$ near the origin, which reads
$$
(1+x)^\alpha = 1 + \alpha x + O(x^2),
$$
hence with $\alpha = 1/2$, and taking as $x = 2 \mathrm{Re}(a_n z^n) + |a_n|^2 |z|^{2n} + z^{n+1}G(z)$ which is $O(|z|^{n})$ near the origin, from $(1)$ we get
$$
|\lambda(z)| = 1 + \frac 12 \left(2\mathrm{Re}(a_n z^n) + |a_n|^2 |z|^{2n} + z^{n+1}G(z) \right) + O(|z|^{2n}) = \\ 1 + \mathrm{Re}(a_n z^n)  + O(|z|^{n+1}),
$$
which gives the desired estimate on $|\lambda(z)|$.
