Let f(z) be analytic with a zero of order k

[Jamin2112]

Let f(z) be analytic with a zero of order k ...

Homework Statement

Let f(z) be analytic with a zero of order k at z₀. Show that f '(z) has a zero of order k-1 at z₀.

Homework Equations

f(z) has a zero of order k at z₀ if Ʃc_n(z-z₀)ⁿ (here n goes from k to ∞, and k ≠ 0)

The Attempt at a Solution

Well, we of course f '(z) = kc_k(z-z₀)^k-1 + (k+1)c_k+1(z-z₀)^k + ... ,

which looks to me like the definition of a function with a zero of order k at z₀. ?

 

[Dick]

It looks to me like f'(z) has a zero of order k-1 at z0. Why do you think it's order k?

 

[Jamin2112]

It looks to me like f'(z) has a zero of order k-1 at z0. Why do you think it's order k?

If it had order k-1, then c_k-1 would not equal zero.

At least according to http://chanarchive.org/content/63_x/3693517/1267659070786.jpg

A function f(z) analytic in D_r(z₀) has a zero of order k at the point z=z₀ iff its Taylor series given by f(z) = ∑c_n(z-z₀)ⁿ has c₀ = c₁ = ... = c_k-1 = 0 and c_k ≠ 0

 

[Dick]

Homework Equations

f(z) has a zero of order k at z₀ if Ʃc_n(z-z₀)ⁿ (here n goes from k to ∞, and k ≠ 0)

Perhaps this definition is faulty. It should have read

f(z) has a zero of order k at z₀ if Ʃc_n(z-z₀)ⁿ (here n goes from k to ∞, and $c_k$ ≠ 0)

The order of a zero is the index of the smallest nonzero coefficient in the power series expansion.

 

[Dick]

Ah. In f '(z) = kck(z-z0)k-1 + (k+1)ck+1(z-z0)k + ..., "c_(k-1)" is the coefficient of (x-x0)^(k-1). That's k*c_k. When they say c_k, it means different things in f(z) and f'(z). That's an imaginative way of being confused!

 

[Jamin2112]

Perhaps this definition is faulty. It should have read

f(z) has a zero of order k at z₀ if Ʃc_n(z-z₀)ⁿ (here n goes from k to ∞, and $c_k$ ≠ 0)

The order of a zero is the index of the smallest nonzero coefficient in the power series expansion.

I see what the problem was. lol

 

[Jamin2112]

Another question:

Locate the poles of the following functions and determine their orders.

(z⁶ + 1)^-1



z^-5sin(z)



(z²sin(z))^-1




I'm not sure where to start with these. I haven't bought my textbook yet, and Google isn't giving me any good results.

 

[Dick]

I'm not sure where to start with these. I haven't bought my textbook yet, and Google isn't giving me any good results.

That's pretty weak. Look up the definition of what a pole is. Now start with the first one 1/(z^6+1). If it's going to have a pole at a point z then z^6+1=0, right? Where can that happen?

 

[Jamin2112]

That's pretty weak. Look up the definition of what a pole is. Now start with the first one 1/(z^6+1). If it's going to have a pole at a point z then z^6+1=0, right? Where can that happen?

z = i

for one

 

[Dick]

z = i

for one

Ok. Yes, that's one. You are supposed to be doing complex analysis here. The question of where x^6+1=0 isn't complex analysis. It's complex arithmetic. You should have covered this before. It's de Moivre. What are all six of the solutions?

 

[Jamin2112]

It's de Moivre.

We never learned that, but according to Wikipedia ...

(cos(x) + i sin(x))ⁿ = cos(nx) + i sin(nx).

Not sure exactly how to invoke it.

z⁶ = r⁶e^i(6ø) and e^iπ = -1, so ...

Is that on the right track?

 

[Dick]

We never learned that, but according to Wikipedia ...

(cos(x) + i sin(x))ⁿ = cos(nx) + i sin(nx).

Not sure exactly how to invoke it.

z⁶ = r⁶e^i(6ø) and e^iπ = -1, so ...

Is that on the right track?

I would complain to your teachers they never taught you that, if you dare. I wouldn't. I think they probably did and you forgot. You already know i works. That's because i=exp(i*pi/2) and i^6=exp(i*pi/2)^2=exp(6*i*pi/2)=exp(3*i*pi)=(-1). Doesn't this ring kind of a bell that angles might have something to do with this? Hint: exp(i*pi/6) also works.