Power series expansion of square root of z

[Isaac]

Find the power series expansion of sqrt(z) about z = 1 and find its radius
of convergence.

To start, well, I know sqrt(z), and any nth root of z, is analytic on the
domain of log(z). Does this mean that sqrt(z) about z = 1 is analytic in an
open disk of the same radius as where log(z) about z = 1 is analytic?

So, since log(z) is analytic of radius 1 about z = 1, then sqrt(z) and as
well z^(1/n) is analytic of radius 1 about z = 1?

Isaac

[Klueless]

Post by Isaac
To start, well, I know sqrt(z), and any nth root of z, is analytic on the
domain of log(z). Does this mean that sqrt(z) about z = 1 is analytic in an
open disk of the same radius as where log(z) about z = 1 is analytic?

Yes, because a composition of analytic functions is analytic
on the intersection of their domains. Since sqrt(z) = exp(1/2*log(z))
for z!=0, and exp is entire, what you say is correct.

Post by Isaac
So, since log(z) is analytic of radius 1 about z = 1, then sqrt(z) and as
well z^(1/n) is analytic of radius 1 about z = 1?

Yes.

[Zdislav V. Kovarik]

Post by Klueless

Post by Isaac
To start, well, I know sqrt(z), and any nth root of z, is analytic on the
domain of log(z). Does this mean that sqrt(z) about z = 1 is analytic in an
open disk of the same radius as where log(z) about z = 1 is analytic?

Yes, because a composition of analytic functions is analytic
on the intersection of their domains. Since sqrt(z) = exp(1/2*log(z))
for z!=0, and exp is entire, what you say is correct.

Post by Isaac
So, since log(z) is analytic of radius 1 about z = 1, then sqrt(z) and as
well z^(1/n) is analytic of radius 1 about z = 1?

Yes.

Caution: this is not how the composition behaves. The domains can be
actually disjoint, yet the composition can be well-defined. Why? The
intersection of the domains has nothing to do with it.

What matters is: for f(g(z)) = h(z), the range of g must have non-empty
intersection with the domain of f (assumed to be open). For pedants, this
would be a restriction of the composition.

Cheers, ZVK(Slavek).