Use a Maclaurin series to solve the functional equation f (z^2) = z+f (z)-complex analysis

Use a Maclaurin series to solve the functional equation f (z^2) = z+f (z)
Where the series converges?

Assume that f(z) = Σ(n = 0 to ∞) a(n) z^n.

Substituting this into the functional equation yields
Σ(n = 0 to ∞) a(n) (z^2)^n = z + Σ(n = 0 to ∞) a(n) z^n
==> Σ(n = 0 to ∞) a(n) z^(2n) = a(0) + (a(1) + 1) z + Σ(n = 2 to ∞) a(n) z^n

*Note that the left hand side has no odd degree terms.
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Equate like coefficients:
Constant: a(0) = a(0)

z^1: 0 = a(1) + 1 ==> a(1) = -1

z^(2k+1) with k > 0: 0 = a(2k+1).

z^(2k) with k > 0: a(k) = a(2k).
Due to a(2k+1) = 0 for all k, we conclude that the only nonzero a(k) with k > 1
occur when the index is a power of 2:
-1 = a(1) = a(2) = a(4) = ... = a(2^j) for some non-negative integer j.

Hence, for any constant C = a(0):
f(z) = C + Σ(j = 0 to ∞) -1 * z^(2^j).
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Using the Root Test:
r = lim(j→∞) |-1 * z^(2^j)|^(1/j)
..= lim(j→∞) |z|^(2^j / j)
..= 0 if |z| < 1, and ∞ if |z| > 1 [since lim(j→∞) 2^j / j = ∞]

When z = -1 or 1, this series clearly diverges.
Since a series on the boundary converges on the whole boundary or with at most one exceptional point, we conclude that the series diverges whenever |z| = 1.

Thus, this series converges iff |z| < 1.
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I hope this helps!