Suppose that f is analytic on a domain D and has a zero of order m at z_{0} in D.Show that a)f¨ has a zero of

Suppose that f is analytic on a domain D and has a zero of order m at z_{0} in D.Show that
a)f¨ has a zero of order m-1 at z_{0}
b)f^{2} has a zero of order 2m at z_{0}

By hypothesis, f(z) = (z - z₀)^m g(z), where g(z) is analytic at z₀ and g'(z₀) ≠ 0.

a) Using the product rule,
f '(z) = m(z - z₀)^(m-1) * g(z) + (z - z₀)^m * g'(z)
.......= (z - z₀)^(m-1) [m g(z) + (z - z₀) g'(z)].

Since m g(z) + (z - z₀) g'(z) is analytic at z₀ (since each part of it is), and
m g(z₀) + (z₀ - z₀) g'(z₀) = m g(z₀) ≠ 0, we conclude that f ' has a zero of order m-1 at z₀.

b) [f(z)]^2 = (z - z₀)^(2m) [g(z)]^2, where [g(z)]^2 is analytic at z₀ and [g'(z₀)]^2 ≠ 0.
Hence, f^2 has a zero of order 2m at z₀.

I hope this helps!