Thank you for the response. The theorems you cite are what I had in mind.

Here is the specific thing on which I am working:

z^a + c*z = w where c and w are fixed nonzero complex numbers, preferably chosen generically so that we can avoid special cases.

My goal: find all infinitely many solutions z=F(a,c,w,k), where k indexes the solutions.

z is an analytic function of complex variable 'a' as long as we avoid the essential singularity at a=0.

The case with c=0 is already known. We already all infinitely many solution,

z=G(a,w,k) of z^a = w. Writing w=exp(u+(v+2*pi*k)*i) for real numbers x and y, we have

G(a,w,k) = exp((u+(v+2*pi*k)*i) / a ) . Writing a = x+y*i for real numbers, x and y we can break apart the real and imaginary parts of G(a,w,k).

But, here is the problem. Knowing all infinitely many solutions, G(a,w,k) of z^a=w, we can compute the Taylor series for z as a function of 'a' around any nonzero complex number, p. The radius of convergence of that series will be the distance from p to the essential singularity, a=0, in other words, abs(p). So, let's created two different Taylor series expansions for z as a function of 'a' with two real centers, say, p and q.

Since we know a priori that z=w^(1/a) - a multivalued function - we can compute the Taylor coefficients, d^m z/d a^m = P(m,1/a)*w^1/a where P(m,t) is a polynomial in the variable, t, with integer coefficients. Hence, when we set a = p, our chosen center of the Taylor series, we get d^m z/ d a^m at a=p = P(m,1/p)*p^(1/a) where, P(m,1/p) is single-valued, but p^1/a depends upon an index, k, in other words, p^1/a = G(a,p,k)

So, the k-th Taylor series for z as a function of 'a' centered at a=p is

sum of 1/m! * P(m,1/p)*G(a,p,k) * ( a-p)^m, indexed by k

Similarly, the Taylor series for z as a function of 'a' centered at a=q is

sum of 1/m! * P(m,1/q)*G(a,q,k')*(a-q)^m, indexed by k'

If we choose the centers p & q of these two infinite collections of Taylor series close enough, such that the abs(p - q) is less than the radius of convergence of either, these two Taylor series must coincide, i.e. they must be equal for all complex values of 'a' in some sufficiently small disk situated between p and q

But, which k of the first infinite collection of Taylor series matches up with which k' of the second infinite collection of Taylor series?

Now, apparently, the G(a,p,k) and G(a,q,k') factor out of each series, separately. But, nevertheless, G(a,p,k)*(a power series in 'a' that does not depend upon k) still has to equal G(a,q,k')*(a power series in 'a' that does not depend upon k'). So, we still have to match two power series which each depend upon which "root" of z^a = w we are taking.

This is what confuses me. It becomes more complicated with z^a + c*z=w.