Show that if $\ p(z) = 1 + 2z + 3z^2 + 4z^3 + \cdots + nz^{n-1}$, then all zeros of $\ p(z)$ are inside the disk of radius one centered at the origin in $\mathbb{C}$. Consider the polynomial $\ q(z) = (1-z)p(z)$
I got to $\ q(z) = 1 + z + z^2 + z^3 + z^4 +\cdots + (-nz^n) $ and really don't know how to continue. I would appreciate an orientation.
We have $p(x)=1+2x+3x^2\cdots nx^{n-1}$
Now we define a polynomial $P(x)=1+x+x^2+\cdots x^n$
It is clear to see $P'(x)=p(x)$
We know the roots of $P$ are nth roots of unity and by applying Gauss Lucas we get then all roots of $p$ lie within the regular n-gon circumscribed by the unit circle