Gauss-Lucas Theorem: Show Zeros of p(z) in Unit Disc

[wany]

Homework Statement

So I am told that p(z)=1+2z+3z^2+...+nz^(n-1). Now I need to use the Gauss-Lucas Theorem to show that all the zeroes of p(z) lie within the unit disc.

Homework Equations

Gauss-Lucas Theorem

The Attempt at a Solution

So I was thinking that I could integrate p(z) to get that q(z)=z+z^2+...+z^n+C.
Then I would need to show that the zeroes of this function lie within the unit circle then be able to apply the Gauss-Lucas Theorem. However, I am not sure how to do this, nor if the way I am going about is the right way.

 

[Dick]

There's really no need to put a +C in there. The convex hull of the roots of q(x) enclose the roots p(x) no matter what C is, since q'(x) is still p(x). Where are the roots of z+z^2+...+z^n?? Use that it's a geometric series. Sum it.

 

[wany]

So by the sum we have that z(1-z^n)/(1-z)
Then letting this equal 0 and multiplying both sides by 1-z we get that:
z(1-z^n)=0
So either z=0 or z^n=1.
Also we note that z cannot equal 1 or else we would be dividing by 0.
So looking at z^n=1 we can say that z^n=cis [n(theta)].
However, now I am not sure what to do.

 

[Dick]

So by the sum we have that z(1-z^n)/(1-z)
Then letting this equal 0 and multiplying both sides by 1-z we get that:
z(1-z^n)=0
So either z=0 or z^n=1.
Also we note that z cannot equal 1 or else we would be dividing by 0.
So looking at z^n=1 we can say that z^n=cis [n(theta)].
However, now I am not sure what to do.

You mean z=cis(2*pi*k/n) for some k. Then what is |z|? Where does z lie relative to the unit disc?

 

[wany]

So |z|=square root of (cos^2+sin^2)=1
Hence the modulus of z is 1, so z lies on the unit circle.
Note that z cannot equal 1 though, or else the previous equality fails.

 

[Dick]

So |z|=square root of (cos^2+sin^2)=1
Hence the modulus of z is 1, so z lies on the unit circle.
Note that z cannot equal 1 though, or else the previous equality fails.

Ok, so now you know where the roots of q(z) are. What does that tell you about the roots of p(z)?

 

[wany]

Well now we can just apply the Gauss Lucas Theorem to state that since the zeros of q(z) are on the unit disc, then the zeros of p(z) must be within the unit disc.

Thank you very much for your help.