Laurent Expansion of a Composition (with the inverse)

Question

Suppose $f(z) = a_1z + a_0 + O(\frac{1}{z})$ as $|z| \rightarrow \infty$. Here $a_1 > 0$. Let $W(z) = z + \frac{1}{z}$.
How do I obtain that the composite function $L(z) = W \circ f^{-1}(z)$ has the expansion $\frac{z}{a_1} - \frac{a_0}{a_1} + O(\frac{1}{z})$ as $|z| \rightarrow \infty$

Answer 1

Suppose $f(z) = a_1z + a_0 + O(\frac{1}{z})$ as $|z| \rightarrow \infty$ with $a_1>0$. Consider $g(z) = \frac{1}{f(\frac{1}{z})}$. Then $g(0) = 0$ and $g'(0) \neq 0$. So $g$ is analytic near $0$ and has the form $g(z) = g_1z + g_2z^2 + O(z^3)$ near $0$. $$ \frac{1}{f\left(\frac{1}{z}\right)} = g(z) $$ $$ \therefore g_1 = \frac{1}{a_1} \text{ and } g_2 = -\frac{a_0}{a_1^2}$$ Now as the comment (above) suggests, for $g(z) = \frac{1}{f\left(\frac{1}{z}\right)}$ we have $g^{-1} = \frac{1}{f^{-1}\left(\frac{1}{z}\right)}$ and $g^{-1}$ is also analytic at $0$. So $g^{-1}(z)$ is of the form $b_1z + b_2z^2 + O\left(z^3\right)$ near $0$. $$ (g^{-1}(g(z)))' = 1 $$ $$ \left( \frac{1}{a_1} - \frac{2a_0}{a_1^2} (b_1z + b_2z^2)) + O(z^3)\right)\left( b_1 + 2b_2z + O(z^3)\right) = 1 $$ Which gives us $b_1 = a_1$ and $b_2 = a_0a_1$
Therefore we get that $\frac{1}{f^{-1}(\frac{1}{z})} = a_1z + a_0a_1z^2 + O(z^3)$.
Now since ord$_0 f^{-1}\left(\frac{1}{z}\right)) = -1$ we have $f^{-1}\left(\frac{1}{z}\right)$ is of the form $\frac{c_{-1}}{z} + c_o + c_1z + O(z^2)$ near $0$. Thus $$ \left( \frac{c_{-1}}{z} + c_0 + c_1z + O(z^2)\right)(a_1z + a_0a_1z^2 + O(z^3) = 1 $$ Which yields, $c_1 = \frac{1}{a_1}$ and $c_0 = -\frac{a_0}{a_1}$. So, $f^{-1}\left(\frac{1}{z}\right) = \frac{1}{a_1z} - \frac{a_0}{a_1} + O(z)$.

Now, $L(z) = W \circ f^{-1}(z) = f^{-1}(z) + \frac{1}{f^{-1}(z)}$. Both of these can be found by taking the reciprocals inside $f^{-1}\left(\frac{1}{z}\right)$ and $\frac{1}{f^{-1}\left(\frac{1}{z}\right)}$.