Find the number of roots $f(z)=z^{10}+10z+9$ in $D(0,1)$. I want to find $g(z)$ s.t. $|f(z)-g(z)|<|g(z)|$, but I cannot. Any hint is appreciated.
Asked
Active
Viewed 452 times
1 Answers
4
First, we factor by $z+1$ to get $f(z)=(z+1)(z^9-z^8+z^7+\dots-z^2+z+9)$. Let $F(z):=z^9-z^8+z^7+\dots-z^2+z+9$ and $G(z)=9$. Then for $F$ of modulus strictly smaller than $1$, $|F(z)-G(z)|\leqslant 9|z| \lt |G(z)|$. thus for each positive $\delta$, we can find the number of zeros of $f$ on $B(0,1-\delta)$.
Davide Giraudo
- 172,925
-
@Sanchez You can post your alternative approach as an answer. – Davide Giraudo Nov 30 '12 at 23:13
-
actually let me delete the last comment, as I hope that OP can figure it out himself. – Nov 30 '12 at 23:14
-
Thank you very much, @Davide. I think you are right and there is no root inside the region. Only when we remove $(x+1)$, things get clearer. Also, thank Sanchez's comments. – Sam Nov 30 '12 at 23:20
-
1How about $z=-1$ again? – Ma Ming May 03 '13 at 13:56
-
Davide, in "$F$ of modulus $1$," $F$ should be replaced with $z$. How does your last inequality imply that $\vert F(z)-g(z)\vert<9$? – The Substitute Feb 09 '15 at 10:40
-
1Also, in your last inequality, what if $z=-1$? – The Substitute Feb 09 '15 at 10:47
-
This answer is incorrect. As Ma Ming and The Substitute have pointed out, $F(z)$ still has a zero at $z=-1$, and so you cannot apply Rouche's Theorem on this boundary. – Travis Bemrose Apr 03 '15 at 09:36
-
@TravisBemrose Actually, we have to apply it on $B(0,1-\delta)$; I should have been more careful. – Davide Giraudo Apr 03 '15 at 09:40
-
@TheSubstitute and Travis Thanks for your remarks. – Davide Giraudo Apr 03 '15 at 09:41
-
Ah. Thanks. I kept trying to apply the theorem to $\frac{f(z)}{(z+1)^2}$ on either $B(0,1)$ or $B(0,1-\delta)$. – Travis Bemrose Apr 03 '15 at 09:43