I could use some help with this problem:
Suppose that f is analytic on a domain D and has a zero of order m at z(subscript 0) in D. Show that:
1.) f' has a zero of order m - 1 at z(subscript 0), and
2.) f^2 has a zero of order 2m at z(subscript 0)
I could use some help with this problem:
Suppose that f is analytic on a domain D and has a zero of order m at z(subscript 0) in D. Show that:
1.) f' has a zero of order m - 1 at z(subscript 0), and
2.) f^2 has a zero of order 2m at z(subscript 0)
Here is an attempt.
We can write,
f(z) = (z-z_0)^m * g(z) for all z in D.
Then,
f'(z)=m(z-z_0)^{m-1} *g(z) + (z-z_0)^m*g'(z)
f'(z)=(z-z_0)^{m-1}*[mg(z)+(z-z_0)*g'(z)]
We need to show that the second function part does not attain a zero at z_0. Indeed! Substitute and see:
mg(z_0)+0*g'(z_0)=mg(z_0)!=0
Because m!=0 and g(z_0)!=0.
Thus, all the zero's belong to the first factor (z-z_0)^{m-1}.
Which is of order m-1.