Analytic Function II: Complex Calculation

[asqw121]

Let p(z)=A(z-z1)...(z-zn) where A and z1...zn are complex numbers and A not equal to 0 . Show that
P'(z)/P(z)=∑ (1/(z-zj)) z not equal z z1...zn

 

[Nono713]

Re: Analytic function

What have you tried? This isn't the place to dump homework and collect answers :(

 

[asqw121]

Re: Analytic function III

Already try hard but I really stuck on this question
Please help

 

[MarkFL]

Re: Analytic function III

Already try hard but I really stuck on this question
Please help

If you show what you have tried, then our helpers can help, otherwise they are left to do the problem for you, which is of little help to you as far as you being able to work this problem and others like it yourself.

 

[blue_raver22]

I would like to first clarify that p(z) is a polynomial function in the complex plane, and p'(z) is its derivative. With that in mind, let's proceed with the proof.

To begin with, we know that the derivative of a product of functions is given by the sum of the individual derivatives. Therefore, we can express p'(z) as:

p'(z) = A(z-z1)...(z-zn)' + A'(z-z1)...(z-zn) + ... + A(z-z1)...(z-zn-1)(z-zn)'

Now, let's focus on the last term of this expression, A(z-z1)...(z-zn-1)(z-zn)'. We can rewrite this as:

A(z-z1)...(z-zn-1)(z-zn)' = A(z-z1)...(z-zn-1)(1)

Notice that the only term that is affected by the derivative is (z-zn), which has a derivative of 1. Therefore, we can rewrite the entire expression of p'(z) as:

p'(z) = A(z-z1)...(z-zn)' + A'(z-z1)...(z-zn) + ... + A(z-z1)...(z-zn-1)(1)

Now, let's divide both sides by p(z):

p'(z)/p(z) = [A(z-z1)...(z-zn)'/p(z)] + [A'(z-z1)...(z-zn)/p(z)] + ... + [A(z-z1)...(z-zn-1)(1)/p(z)]

Next, we can use the quotient rule to evaluate the first term in the right-hand side:

A(z-z1)...(z-zn)'/p(z) = A'(z-z1)...(z-zn) + A(z-z1)...(z-zn-1)(z-zn)'/[A(z-z1)...(z-zn)]

Notice that the last term in this expression cancels out with the first term in p'(z)/p(z), leaving us with:

p'(z)/p(z) = A'(z-z1)...(z-zn) + ... + A(z-z1)...(z-zn-1)(1)/[A(z-z1)...(z-zn)]

Now,