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. $$\bar{a}c = \bar{c}a$$.