Consider $P(z) = z^4+4z^3+6z^2-4z+3$. How many zeros does $P(z)$ have within $|z-1|<1$.
I've looked at wolfram alpha and it appears the answer is two. I've tried many things to apply Rouche's Theorem to solve this, but nothing quite works.
This problem is from Stewart and Tall's text, page 267, number 23.
Thank you in advance for you help.
Thanks!
Ugly but here it is: with the substitution $z-1=w$ we need to find the roots of $p(w) = w^4+8w^3+24w^2+24w+10$ inside the unit circle. Guided by wolfram alpha as in the OP we suspect there are $2$ so we look to apply Rouche to $w^4+8w^3$ vs $24w^2+24w+10$ which has indeed two roots inside (complex conjugate so same modulus and product is less than $1$).
Hence we need to prove $|w^4+8w^3| < |24w^2+24w+10|$ for $|w|=1$ or $|w+8|<2|12w^2+12w+5|, |w|=1$. Squaring and using $w^2+\bar w^2=2\cos 2\theta=4x^2-2, w+\bar w=2\cos \theta =2x$ it comes down to showing an ugly quadratic has no roots inside $[-1,1]$ and it turns out to be a positive definite one as squaring requires us to prove:
$$65 +16x<4(313+408x+120(2x^2-1)), x \in [-1,1]$$
Quadratic then is $$960x^2+1616x+707$$ and the discriminant is $1616^2-4\times 960 \times 707=-103424$ so the quadratic is indeed positive definite and we are done!
Maybe there is a cleverer solution using that the product of the roots is $10$ and they are conjugated in pairs since there is no real root (easily seen by checking $-1<w<0, -1\le w \le -2, w<-2$ and distributing coefficients appropriately) and arguing that not all can be outside the unit circle, hence two must be in and two out