Taylor series absolute value inequality

Question

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?

Answer 1

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 $|\lambda(z)|$.