How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| <...
$begingroup$
How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?
I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.
real-analysis abstract-algebra complex-analysis polynomials roots
asked Apr 23 '16 at 18:32
HappyHappy
561412
$endgroup$
add a comment |
$begingroup$
How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?
I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.
real-analysis abstract-algebra complex-analysis polynomials roots
asked Apr 23 '16 at 18:32
HappyHappy
561412
$endgroup$
add a comment |
$begingroup$
How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?
I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.
real-analysis abstract-algebra complex-analysis polynomials roots
asked Apr 23 '16 at 18:32
HappyHappy
561412
$endgroup$
How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?
I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.
real-analysis abstract-algebra complex-analysis polynomials roots
real-analysis abstract-algebra complex-analysis polynomials roots
asked Apr 23 '16 at 18:32
HappyHappy
561412
asked Apr 23 '16 at 18:32
HappyHappy
561412
asked Apr 23 '16 at 18:32
HappyHappy
561412
asked Apr 23 '16 at 18:32
HappyHappy
561412
asked Apr 23 '16 at 18:32
HappyHappy
561412
561412
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.
How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.
I will let you do the same with the circle of radius $2$.
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
$endgroup$
$begingroup$
I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
$endgroup$
– Happy
Apr 25 '16 at 16:52
$begingroup$
Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
$endgroup$
– Jennifer
Apr 25 '16 at 17:18
$begingroup$
That makes perfect sense! How does one know which $f$ to choose?
$endgroup$
– Happy
Apr 25 '16 at 17:24
$begingroup$
Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
$endgroup$
– Jennifer
Apr 25 '16 at 17:28
$begingroup$
I understand now. Thank you : )
$endgroup$
– Happy
Apr 25 '16 at 17:30
|
show 1 more comment
$begingroup$
I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.
answered Dec 21 '18 at 5:02
AMRAMR
1
$endgroup$
$begingroup$
Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
$endgroup$
– hardmath
Dec 21 '18 at 5:34
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.
How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.
I will let you do the same with the circle of radius $2$.
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
$endgroup$
$begingroup$
I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
$endgroup$
– Happy
Apr 25 '16 at 16:52
$begingroup$
Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
$endgroup$
– Jennifer
Apr 25 '16 at 17:18
$begingroup$
That makes perfect sense! How does one know which $f$ to choose?
$endgroup$
– Happy
Apr 25 '16 at 17:24
$begingroup$
Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
$endgroup$
– Jennifer
Apr 25 '16 at 17:28
$begingroup$
I understand now. Thank you : )
$endgroup$
– Happy
Apr 25 '16 at 17:30
|
show 1 more comment
$begingroup$
General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.
How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.
I will let you do the same with the circle of radius $2$.
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
$endgroup$
$begingroup$
I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
$endgroup$
– Happy
Apr 25 '16 at 16:52
$begingroup$
Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
$endgroup$
– Jennifer
Apr 25 '16 at 17:18
$begingroup$
That makes perfect sense! How does one know which $f$ to choose?
$endgroup$
– Happy
Apr 25 '16 at 17:24
$begingroup$
Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
$endgroup$
– Jennifer
Apr 25 '16 at 17:28
$begingroup$
I understand now. Thank you : )
$endgroup$
– Happy
Apr 25 '16 at 17:30
|
show 1 more comment
$begingroup$
General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.
How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.
I will let you do the same with the circle of radius $2$.
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
$endgroup$
General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.
How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.
I will let you do the same with the circle of radius $2$.
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
answered Apr 23 '16 at 18:44
JenniferJennifer
8,44721837
8,44721837
$begingroup$
I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
$endgroup$
– Happy
Apr 25 '16 at 16:52
$begingroup$
Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
$endgroup$
– Jennifer
Apr 25 '16 at 17:18
$begingroup$
That makes perfect sense! How does one know which $f$ to choose?
$endgroup$
– Happy
Apr 25 '16 at 17:24
$begingroup$
Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
$endgroup$
– Jennifer
Apr 25 '16 at 17:28
$begingroup$
I understand now. Thank you : )
$endgroup$
– Happy
Apr 25 '16 at 17:30
|
show 1 more comment
$begingroup$
I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
$endgroup$
– Happy
Apr 25 '16 at 16:52
$begingroup$
Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
$endgroup$
– Jennifer
Apr 25 '16 at 17:18
$begingroup$
That makes perfect sense! How does one know which $f$ to choose?
$endgroup$
– Happy
Apr 25 '16 at 17:24
$begingroup$
Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
$endgroup$
– Jennifer
Apr 25 '16 at 17:28
$begingroup$
I understand now. Thank you : )
$endgroup$
– Happy
Apr 25 '16 at 17:30
$begingroup$
I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
$endgroup$
– Happy
Apr 25 '16 at 16:52
$begingroup$
I did it and it says that there are $8$ zeros in the circle of radius $2$. So I get $8-2=6$, but I think that it is wrong.
$endgroup$
– Happy
Apr 25 '16 at 16:52
$begingroup$
Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
$endgroup$
– Jennifer
Apr 25 '16 at 17:18
$begingroup$
Hmm for the circle of radius $2$ I used $f=3z^7$, so I found $7$ zeros in this circle and so $5$ zeros in the annulus.
$endgroup$
– Jennifer
Apr 25 '16 at 17:18
$begingroup$
That makes perfect sense! How does one know which $f$ to choose?
$endgroup$
– Happy
Apr 25 '16 at 17:24
$begingroup$
That makes perfect sense! How does one know which $f$ to choose?
$endgroup$
– Happy
Apr 25 '16 at 17:24
$begingroup$
Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
$endgroup$
– Jennifer
Apr 25 '16 at 17:28
$begingroup$
Well I try to pick $f$ as a monome, and test with triangular inequality which one will verify $|g|<|f|$. For example here I knew that for $|z|=2$, I would have $|3z^7|>|z^8|$ so I knew it would be a good idea to try $f=3z^7$.
$endgroup$
– Jennifer
Apr 25 '16 at 17:28
$begingroup$
I understand now. Thank you : )
$endgroup$
– Happy
Apr 25 '16 at 17:30
$begingroup$
I understand now. Thank you : )
$endgroup$
– Happy
Apr 25 '16 at 17:30
|
show 1 more comment
$begingroup$
I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.
answered Dec 21 '18 at 5:02
AMRAMR
1
$endgroup$
$begingroup$
Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
$endgroup$
– hardmath
Dec 21 '18 at 5:34
add a comment |
$begingroup$
I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.
answered Dec 21 '18 at 5:02
AMRAMR
1
$endgroup$
$begingroup$
Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
$endgroup$
– hardmath
Dec 21 '18 at 5:34
add a comment |
$begingroup$
I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.
answered Dec 21 '18 at 5:02
AMRAMR
1
$endgroup$
I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.
answered Dec 21 '18 at 5:02
AMRAMR
1
answered Dec 21 '18 at 5:02
AMRAMR
1
answered Dec 21 '18 at 5:02
AMRAMR
1
answered Dec 21 '18 at 5:02
AMRAMR
1
1
$begingroup$
Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
$endgroup$
– hardmath
Dec 21 '18 at 5:34
add a comment |
$begingroup$
Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
$endgroup$
– hardmath
Dec 21 '18 at 5:34
$begingroup$
Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
$endgroup$
– hardmath
Dec 21 '18 at 5:34
$begingroup$
Apart from the subtraction $7-2=5$, I don't see any mathematical reasoning to follow. Surely there is more to the problem of locating roots inside/outside the annulus? This question has been around for a couple of years and already had an Accepted Answer, so it is important to read the previous response in order to highlight what new information you are adding.
$endgroup$
– hardmath
Dec 21 '18 at 5:34
add a comment |
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