complex-analysis
Let $p(z) = c_0 + c_1z + c_2z^2 + \dots + c_nz^n$ where $0 \le c_0 \le c_1 \le \dots \le c_n$. I would like to show that all zeroes of this polynomial lie inside the unit disk by applying Rouche's theorem to the polynomial $(1-z)p(z)$. I'm not completely sure how to do this. Using the given, information I can deduce that
$$|(1 – z)p(z)| \le |c_nz^{n+1}|$$
on the unit circle, but this doesn't really match the assumptions of Rouche's theorem.
Help would be appreciated.
Two of the four complex roots are inside the unit circle centered at $1$. If you find these two pairs of conjugate roots, you can work out that $$ g(z)\approx(x^2+4.64751368404x+8.66302952977)(x^2-0.647513684039x+0.346299177406) $$ and that the roots are approximately $$ \{ -2.32375684201972 \pm 1.80642842895505 i,~ 0.32375684201972 \pm 0.491406842291623 i \}, $$ the former pair of which lie outside and the latter pair inside the given circle. However, I needed a numerical method to obtain the above and haven't used Rouche's theorem; I like your approach...I'll consider how to proceed...
In fact, $f(z)=2(4z^3+1)$ has roots at $-\frac1{\sqrt[3]{4}}\,e^{\frac{2\pi ik}{3}}$ for $k=0,\pm1$ along the circle of radius $2^{-\frac23}$, two of which are inside $|z-1|=1$ since $$\left|1+\frac1{\sqrt[3]{4}}\,e^{\pm\frac{2\pi i}{3}}\right|<1.$$ (For $k=\pm1$, the roots are on the side of a unit equilateral triangle stemming radially from the origin, which is thus inside the unit circle centered at $1$.)
Also, on the unit circle in question, if we let $z=1+e^{it}$ following @RobertIsrael's lead, then $$ \eqalign{ h(t)&= \left| f(z) \right|^2 =4\cdot \left| 4\left(1+e^{it}\right)^3+1 \right|^2 =4\cdot \left| 5+12e^{it}+12e^{2it}+4e^{3it} \right|^2 \\& =4\cdot \left[ \left(5+12\cos{t}+12\cos{2t}+4\cos{3t}\right)^2+4\, \left( 3\sin{t}+ 3\sin{2t}+ \sin{3t}\right)^2 \right] \\& =4\cdot \Big[ (5^2+12^2+12^2+4^2) + 2\cdot5\cdot4\, \left(3\cos{t}+3\cos{2t}+\cos{3t}\right) \Big.\\&\qquad\Big. +2\cdot4^2\cdot3\,\left(3\cos{(2t-t)}+\cos{(3t-t)}+\cos{(3t-2t)}\right) \Big] \qquad\text{(combining pairwise terms)} \\& =4\cdot \left[ 329 + 8\, \left(15\cos{t}+15\cos{2t}+5\cos{3t}\right) + 8\, \left(48\cos{t}+12\cos{2t}\right) \right] \\& =4\cdot \left[ 329 + 8\, \left(63\cos{t}+27\cos{2t}+5\cos{3t}\right) \right] } $$ by expanding the squares, combining pairwise terms and using $\cos(a-b)=$ $\cos a\cos b+$$\sin a\sin b$ with $a,b\in\{t,2t,3t\}$. Differentiating and using the double & triple angle formulas (or $\href{http://en.wikipedia.org/wiki/Chebyshev_polynomials#Examples}{U_1,~U_2}\in\mathbb{Z}[\cos t]$), $$ \eqalign{ h'(t)& =-96\, \left[ 21\sin{t}+18\sin{2t}+5\sin{3t} \right] \\&= -96\,\sin t\, \left[ 21 + 18 \left( 2\cos t \right) + 5 \left( 4 \cos^2t - 1 \right) \right] \\&= -96\,\sin t\, \left[ 20 \cos^2t + 36 \cos t + 16 \right] \\&= -384\,\sin t\, \left( \cos t + 1 \right) \left( 5 \cos t + 4 \right) } $$ we obtain the critical points and find the global minimum as he did in his comment, showing that $|f(z)|\ge\frac65$ on the circle in question.
So now you can use Rouché's theorem.
Best Answer
Similar to @DonAntonio's answer but uses Rouché's theorem. We have: $$ (1 - z)p(z) = c_0 + \sum_{k=1}^n (c_k - c_{k-1}) z^k - c_n z^{n+1} $$
Let $|z| = r > 1$, we have:
\begin{align} \left|(1 - z)p(z) - (-c_n z^{n+1})\right| &= \left| c_0 + \sum_{k=1}^n (c_k - c_{k-1}) z^k \right| \\ &< \left(c_0 + \sum_{k=1}^n (c_k - c_{k-1})\right)\left|-z^{n+1}\right| = \left|-c_n z^{n+1}\right| \end{align}
Thus, $(1 - z)p(z)$ and $-c_n z^{n+1}$ have the same number of zeros inside every circle $|z| = r$ for $r > 1$. But $-c_n z^{n+1}$ has $n+1$ zeros at $0$ and $(1-z)p(z)$ has $n+1$ zeros. It follows that all of the zeros of $(1-z)p(z)$ lie inside the circle $|z| = r$.
By letting $r \to 1$, we conclude that all zeros of $(1 - z)p(z)$ (and hence $p(z)$) lie in the closed unit disk. Notice that zeros can be on the unit circle as demonstrated by the polynomial $p(z) = 1 + z$.