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Let $f: \mathbb{R} \to \mathbb{C}$ and suppose that we know $|f(\lambda)| = 1$ for all $\lambda$. Consider the Taylor series around $0$: $$ f(\lambda) = a + b\lambda + c\lambda^2 + \cdots. $$ Instead of calculating $c$ explicitly, can one find the second-order Taylor polynomial by computing the first-order Taylor polynomial and then normalizing? What if we also know that $f(0) = 1$? What other conditions are necessary for this to work? I have seen that it works if $a = 1$ and $c$ is real, or more generally if $\bar{a}c$ is real. Under what conditions can $\bar{a}c$ be shown to be real?

Background: A quantum mechanics problem was to find a certain probability (= $|\text{amplitude}|^2$) to lowest order in $\lambda$, and the example solution proceeded to find all amplitudes to order $\lambda$ and then normalizing them by dividing by the sum of all norm-squared amplitudes.


My thought process so far: Set $z = f(\lambda)$. $|z| = 1$ gives $$ 1 = \bar{z}z = \underbrace{\bar{a}a}_1 + \underbrace{(\bar{a}b + \bar{b}a)}_0\lambda + \underbrace{(\bar{a}c + \bar{b}b + \bar{c}a)}_0 \lambda^2 + \cdots. $$ Compare $z = a + b\lambda + c\lambda^2$ with $$ \frac{z^{(1)}}{|z^{(1)}|} = \frac{a + b\lambda}{\sqrt{\bar{a}a + (\bar{a}b + \bar{b}{a})\lambda + \bar{b}{b}\lambda^2}} = \frac{a + b\lambda}{\sqrt{1 + \bar{b}b\lambda^2}} = (a + b\lambda)\left(1 - \frac{\bar{b}b\lambda^2}{2} + \cdots\right) = a + b\lambda - \frac{a\bar{b}b}{2}\lambda^2 + \cdots. $$ Simplifying using $-a\bar{b}b = a(\bar{a}c + \bar{c}a) = c + a^2 \bar{c}$, we need to assume $a = f(0) = 1$ to make progress; then we get $$ \frac{z^{(1)}}{|z^{(1)}|} = a + b\lambda + \frac{1}{2}(c + \bar{c}) \lambda^2 + \cdots. $$ This method then works iff $\frac{1}{2}(c + \bar{c}) = c$, i.e. $c$ is real. More generally, it works whenever $\frac{1}{2}(c + a^2\bar{c}) = c$, i.e. $\bar{a}c = \bar{c}a$.

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