Analytic solution of the functional equation $V\left(z^d\right)=dz^{d-1} V(z)+\sum_{k=2}^d b_k z^{d-k}$ Here's my question: let $b_2, \dots, b_d \in \mathbb{C}$ ($d$ is an integer greater than $2$), and consider the functional equation
$$V\left(z^d\right)=dz^{d-1} V(z)+\left(b_2 z^{d-2} + b_3 z^{d-3}+\ldots + b_d\right)$$
In this article: http://www.math.harvard.edu/~ctm/papers/home/text/papers/wp/wp.pdf , page 33, McMullen claims that it is clear this functional equation admits a unique analytic solution on $\{z\in\mathbb C : |z|>1\}$, and that the solution is
$$V(z)=\frac{-z}{d} \sum_{k=2}^d \sum_{n=0}^{\infty} \frac{b_k z^{-k d^n}}{d^n}$$
Of course I believe him =) , but I've been staring at the equation and the solution for some time, and it's not clear to me that the solution is unique (or how he found the solution). All I can do is check it is indeed a solution. So:
Question 1: why is it unique?
Question 2: can you see a method on how the formula was obtained?
PS: I mentioned the article, but I believe there is no need to refer to it. This problem should be self contained.
Thanks in advance!
 A: Define $ \tilde V ( z ) = \frac { V ( z ) } z $ and rewrite the equation
$$ V \left ( z ^ d \right ) = d z ^ { d - 1 } V ( z ) + b _ 2 z ^ { d - 2 } + b _ 3 z ^ { d - 3 } + \dots + b _ d $$
as
$$ \tilde V ( z ) = \frac 1 d \tilde V \left ( z ^ d \right ) - \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k } } d \tag 0 \label 0 \text . $$
For any nonnegative integer $ n $, substituting $ z ^ { d ^ n } $ for $ z $ in \eqref{0}, you have
$$ \tilde V \left ( z ^ { d ^ n } \right ) = \frac 1 d \tilde V \left ( z ^ { d ^ { n + 1 } } \right ) - \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } d \text , $$
or equivalently
$$ \frac 1 { d ^ n } \tilde V \left ( z ^ { d ^ n } \right ) = \frac 1 { d ^ { n + 1 } } \tilde V \left ( z ^ { d ^ { n + 1 } } \right ) - \frac 1 d \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } { d ^ n } \tag 1 \label 1 \text . $$
Using \eqref{0} as the basis and \eqref{1} for the inductive step, you can prove
$$ \tilde V ( z ) = \frac 1 { d ^ { N + 1 } } \tilde V \left ( z ^ { d ^ { N + 1 } } \right ) - \frac 1 d \sum _ { n = 0 } ^ N \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } { d ^ n } \tag 2 \label 2 $$
by induction on the nonnegative integer $ N $. Rewriting \eqref{2} in terms of $ V $, you have
$$ V ( z ) = \frac { V \left ( z ^ { d ^ { N + 1 } } \right ) } { d ^ { N + 1 } z ^ { d ^ { N + 1 } - 1 } } - \frac z d \sum _ { n = 0 } ^ N \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k d ^ n } } { d ^ n } \text . \tag 3 \label 3 $$
The formula for the mentioned series is just the result of interchanging the sums and letting $ N $ go to infinity in the second term on the right-hand side of \eqref{3}. Now that we have an idea of what $ V ( z ) $ should look like, we can be more precise on the matter: define $ V : \bigl \{ z \in \mathbb C \bigm | | z | > 1 \bigr \} \to \mathbb C $ with $ V ( z ) = - \frac z d \sum \limits _ { k = 2 } ^ d \sum \limits _ { n = 0 } ^ \infty \frac { b _ k z ^ { - k d ^ n } } { d ^ n } $. Note that the series converges absolutely, for example by the root test. It's then straightforward to verify that $ V $ satisfies the functional equation, and as it's defined in terms of power series, it's analytic. Also, note that for this function, the first term on the right-hand side of \eqref{3} tends to $ 0 $ as $ N $ goes to infinity. This is not needed for the proof, but is a useful observation for seeing that $ V $ is in fact the result of taking a limit from both sides of \eqref{3}.
For proving uniqueness, consider another analytic function $ U $ satisfying the same functional equation. Define $ \tilde U ( z ) = \frac { U ( z ) } z $, and conclude
$$ \tilde U ( z ) = \frac 1 d \tilde U \left ( z ^ d \right ) - \sum _ { k = 2 } ^ d \frac { b _ k z ^ { - k } } d \tag 4 \label 4 \text . $$
Define $ F ( z ) = \tilde V ( z ) - \tilde U ( z ) $. Subtract \eqref{4} from \eqref{0} to get
$$ F ( z ) = \frac 1 d F \left ( z ^ d \right ) \text. \tag 5 \label 5 $$
By \eqref{5} and induction, we get
$$ F ( z ) = \frac 1 { d ^ n } F \left ( z ^ { d ^ n } \right ) \text , \tag 6 \label 6 $$
for any nonnegative integer $ n $. Letting $ a _ n = 2 \exp \left ( \frac { 2 \pi i } { d ^ n } \right ) $ and using \eqref{6} we get
$$ f ( a _ n ) = \frac 1 { d ^ n } F \left ( a _ n ^ { d ^ n } \right ) = \frac 1 { d ^ n } F \left ( 2 ^ { d ^ n } \right ) = F ( 2 ) \text . $$
Hence, $ F $ is an analytic function which is constant on the sequence $ ( a _ n ) _ { n = 0 } ^ \infty $, which has a limit point. Therefore, $ F $ must be constant. As the only constant function satisfying \eqref{5} is the zero function (since $ d > 2 $), we get $ \tilde U = \tilde V $, and $ U = V $.
