Rouché for Polynomial $p(z)=z^7+z(z-3)^3+1$ around non-centered annulus.

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Let $p(z)=z^7+z(z-3)^3+1$. Find the numbers of zeros (including multiplicities) in $B_1(3)$.

I want to apply Rouché's Theorem for $p(z)$ and $g(z)=-z(z-3)^3$, but I don't know how to confine on $|z-3|=1$. The problem should be solved without using a calculator.

Advice appreciated.
Thanks

complex-analysis roots rouches-theorem

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edited Jul 19 at 3:15

Nosrati

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up vote
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favorite

Let $p(z)=z^7+z(z-3)^3+1$. Find the numbers of zeros (including multiplicities) in $B_1(3)$.

I want to apply Rouché's Theorem for $p(z)$ and $g(z)=-z(z-3)^3$, but I don't know how to confine on $|z-3|=1$. The problem should be solved without using a calculator.

Advice appreciated.
Thanks

complex-analysis roots rouches-theorem

share|cite|improve this question

edited Jul 19 at 3:15

Nosrati

19.6k41544

asked Jul 18 at 20:05

Clip

64

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $p(z)=z^7+z(z-3)^3+1$. Find the numbers of zeros (including multiplicities) in $B_1(3)$.

I want to apply Rouché's Theorem for $p(z)$ and $g(z)=-z(z-3)^3$, but I don't know how to confine on $|z-3|=1$. The problem should be solved without using a calculator.

Advice appreciated.
Thanks

complex-analysis roots rouches-theorem

share|cite|improve this question

edited Jul 19 at 3:15

Nosrati

19.6k41544

asked Jul 18 at 20:05

Clip

64

Let $p(z)=z^7+z(z-3)^3+1$. Find the numbers of zeros (including multiplicities) in $B_1(3)$.

I want to apply Rouché's Theorem for $p(z)$ and $g(z)=-z(z-3)^3$, but I don't know how to confine on $|z-3|=1$. The problem should be solved without using a calculator.

Advice appreciated.
Thanks

complex-analysis roots rouches-theorem

share|cite|improve this question

edited Jul 19 at 3:15

Nosrati

19.6k41544

asked Jul 18 at 20:05

Clip

64

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 3:15

Nosrati

19.6k41544

edited Jul 19 at 3:15

Nosrati

19.6k41544

edited Jul 19 at 3:15

Nosrati

19.6k41544

19.6k41544

asked Jul 18 at 20:05

Clip

64

asked Jul 18 at 20:05

Clip

64

asked Jul 18 at 20:05

Clip

64

64

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
0
down vote

You get
$$
|z^7-1|ge 2^7-1=127
$$
and
$$
|z(z-3)^3|le 4
$$
on the circle $|z-3|=1$. Thus $p$ has the same number of roots inside the circle as $z^7+1$.

---

Roots of $z^7+alpha z(z-3)^3+1$ for $αin[0,1]$, red for $α=1$.

share|cite|improve this answer

edited Jul 18 at 22:16

answered Jul 18 at 21:52

LutzL

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought of this, too. However, I could not come up with a way to account for the number of zeros that $z^7+1$ actually has within $|z-3| leq 1$ without using a calculator or graphing tool.
  – Clip
  Jul 20 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As $0=z^7+1$ implies $|z|=1$, all roots are on the unit circle around $0$ and thus well outside the unit circle around $3$.
  – LutzL
  Jul 20 at 13:33
  

  
  
  
  

add a comment | 

up vote
0
down vote

It's better to work with unit circle $|z|leqslant1$, For this purpose let $w=z-3$ then $p(z)=(w+3)^7+w^4+3w^3+1$ and for $|w|=1$ we have
$$|f(w)|=|w^4+3w^3+1|leqslant|w|^4+3|w|^3+1=5<|w+3|^7=|g(w)|$$
therefore the number of zeros $p(w)$ and $g(w)$ in $|w|leqslant1$ are the same, but $g(w)$ has no zeros there.

share|cite|improve this answer

edited Jul 18 at 22:19

answered Jul 18 at 22:05

Nosrati

19.6k41544

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up vote
0
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Let $q(z) = z^7.$ Then on the circle $z= 3 +e^it, 0le t le 2pi,$

$$|p(z)-q(z)| = |z(z-3)^3 +1| le |z(z-3)^3| +1 le 4cdot 1^3 + 1 = 5 .$$

But note $|q|ge 2^7$ on this circle. Since $5< 2^7,$ $p,q,$ have the same number of zeros in $D(3,1).$ Since $q=0$ has no zeroes in this disc, $p$ has no zeros in this disc by Rouche.

share|cite|improve this answer

answered Jul 18 at 22:29

zhw.

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

You get
$$
|z^7-1|ge 2^7-1=127
$$
and
$$
|z(z-3)^3|le 4
$$
on the circle $|z-3|=1$. Thus $p$ has the same number of roots inside the circle as $z^7+1$.

---

Roots of $z^7+alpha z(z-3)^3+1$ for $αin[0,1]$, red for $α=1$.

share|cite|improve this answer

edited Jul 18 at 22:16

answered Jul 18 at 21:52

LutzL

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought of this, too. However, I could not come up with a way to account for the number of zeros that $z^7+1$ actually has within $|z-3| leq 1$ without using a calculator or graphing tool.
  – Clip
  Jul 20 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As $0=z^7+1$ implies $|z|=1$, all roots are on the unit circle around $0$ and thus well outside the unit circle around $3$.
  – LutzL
  Jul 20 at 13:33
  

  
  
  
  

add a comment | 

up vote
0
down vote

You get
$$
|z^7-1|ge 2^7-1=127
$$
and
$$
|z(z-3)^3|le 4
$$
on the circle $|z-3|=1$. Thus $p$ has the same number of roots inside the circle as $z^7+1$.

---

Roots of $z^7+alpha z(z-3)^3+1$ for $αin[0,1]$, red for $α=1$.

share|cite|improve this answer

edited Jul 18 at 22:16

answered Jul 18 at 21:52

LutzL

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought of this, too. However, I could not come up with a way to account for the number of zeros that $z^7+1$ actually has within $|z-3| leq 1$ without using a calculator or graphing tool.
  – Clip
  Jul 20 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As $0=z^7+1$ implies $|z|=1$, all roots are on the unit circle around $0$ and thus well outside the unit circle around $3$.
  – LutzL
  Jul 20 at 13:33
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

You get
$$
|z^7-1|ge 2^7-1=127
$$
and
$$
|z(z-3)^3|le 4
$$
on the circle $|z-3|=1$. Thus $p$ has the same number of roots inside the circle as $z^7+1$.

---

Roots of $z^7+alpha z(z-3)^3+1$ for $αin[0,1]$, red for $α=1$.

share|cite|improve this answer

edited Jul 18 at 22:16

answered Jul 18 at 21:52

LutzL

49.8k31849

You get
$$
|z^7-1|ge 2^7-1=127
$$
and
$$
|z(z-3)^3|le 4
$$
on the circle $|z-3|=1$. Thus $p$ has the same number of roots inside the circle as $z^7+1$.

---

Roots of $z^7+alpha z(z-3)^3+1$ for $αin[0,1]$, red for $α=1$.

share|cite|improve this answer

edited Jul 18 at 22:16

answered Jul 18 at 21:52

LutzL

49.8k31849

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 22:16

edited Jul 18 at 22:16

edited Jul 18 at 22:16

answered Jul 18 at 21:52

LutzL

49.8k31849

answered Jul 18 at 21:52

LutzL

49.8k31849

answered Jul 18 at 21:52

LutzL

49.8k31849

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought of this, too. However, I could not come up with a way to account for the number of zeros that $z^7+1$ actually has within $|z-3| leq 1$ without using a calculator or graphing tool.
  – Clip
  Jul 20 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As $0=z^7+1$ implies $|z|=1$, all roots are on the unit circle around $0$ and thus well outside the unit circle around $3$.
  – LutzL
  Jul 20 at 13:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought of this, too. However, I could not come up with a way to account for the number of zeros that $z^7+1$ actually has within $|z-3| leq 1$ without using a calculator or graphing tool.
  – Clip
  Jul 20 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As $0=z^7+1$ implies $|z|=1$, all roots are on the unit circle around $0$ and thus well outside the unit circle around $3$.
  – LutzL
  Jul 20 at 13:33
  

  
  
  
  

I thought of this, too. However, I could not come up with a way to account for the number of zeros that $z^7+1$ actually has within $|z-3| leq 1$ without using a calculator or graphing tool.
– Clip
Jul 20 at 13:19

I thought of this, too. However, I could not come up with a way to account for the number of zeros that $z^7+1$ actually has within $|z-3| leq 1$ without using a calculator or graphing tool.
– Clip
Jul 20 at 13:19

As $0=z^7+1$ implies $|z|=1$, all roots are on the unit circle around $0$ and thus well outside the unit circle around $3$.
– LutzL
Jul 20 at 13:33

As $0=z^7+1$ implies $|z|=1$, all roots are on the unit circle around $0$ and thus well outside the unit circle around $3$.
– LutzL
Jul 20 at 13:33

add a comment | 

up vote
0
down vote

It's better to work with unit circle $|z|leqslant1$, For this purpose let $w=z-3$ then $p(z)=(w+3)^7+w^4+3w^3+1$ and for $|w|=1$ we have
$$|f(w)|=|w^4+3w^3+1|leqslant|w|^4+3|w|^3+1=5<|w+3|^7=|g(w)|$$
therefore the number of zeros $p(w)$ and $g(w)$ in $|w|leqslant1$ are the same, but $g(w)$ has no zeros there.

share|cite|improve this answer

edited Jul 18 at 22:19

answered Jul 18 at 22:05

Nosrati

19.6k41544

add a comment | 

up vote
0
down vote

It's better to work with unit circle $|z|leqslant1$, For this purpose let $w=z-3$ then $p(z)=(w+3)^7+w^4+3w^3+1$ and for $|w|=1$ we have
$$|f(w)|=|w^4+3w^3+1|leqslant|w|^4+3|w|^3+1=5<|w+3|^7=|g(w)|$$
therefore the number of zeros $p(w)$ and $g(w)$ in $|w|leqslant1$ are the same, but $g(w)$ has no zeros there.

share|cite|improve this answer

edited Jul 18 at 22:19

answered Jul 18 at 22:05

Nosrati

19.6k41544

add a comment | 

up vote
0
down vote

up vote
0
down vote

It's better to work with unit circle $|z|leqslant1$, For this purpose let $w=z-3$ then $p(z)=(w+3)^7+w^4+3w^3+1$ and for $|w|=1$ we have
$$|f(w)|=|w^4+3w^3+1|leqslant|w|^4+3|w|^3+1=5<|w+3|^7=|g(w)|$$
therefore the number of zeros $p(w)$ and $g(w)$ in $|w|leqslant1$ are the same, but $g(w)$ has no zeros there.

share|cite|improve this answer

edited Jul 18 at 22:19

answered Jul 18 at 22:05

Nosrati

19.6k41544

It's better to work with unit circle $|z|leqslant1$, For this purpose let $w=z-3$ then $p(z)=(w+3)^7+w^4+3w^3+1$ and for $|w|=1$ we have
$$|f(w)|=|w^4+3w^3+1|leqslant|w|^4+3|w|^3+1=5<|w+3|^7=|g(w)|$$
therefore the number of zeros $p(w)$ and $g(w)$ in $|w|leqslant1$ are the same, but $g(w)$ has no zeros there.

share|cite|improve this answer

edited Jul 18 at 22:19

answered Jul 18 at 22:05

Nosrati

19.6k41544

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 22:19

edited Jul 18 at 22:19

edited Jul 18 at 22:19

answered Jul 18 at 22:05

Nosrati

19.6k41544

answered Jul 18 at 22:05

Nosrati

19.6k41544

answered Jul 18 at 22:05

Nosrati

19.6k41544

19.6k41544

add a comment | 

add a comment | 

up vote
0
down vote

Let $q(z) = z^7.$ Then on the circle $z= 3 +e^it, 0le t le 2pi,$

$$|p(z)-q(z)| = |z(z-3)^3 +1| le |z(z-3)^3| +1 le 4cdot 1^3 + 1 = 5 .$$

But note $|q|ge 2^7$ on this circle. Since $5< 2^7,$ $p,q,$ have the same number of zeros in $D(3,1).$ Since $q=0$ has no zeroes in this disc, $p$ has no zeros in this disc by Rouche.

share|cite|improve this answer

answered Jul 18 at 22:29

zhw.

65.8k42870

add a comment | 

up vote
0
down vote

Let $q(z) = z^7.$ Then on the circle $z= 3 +e^it, 0le t le 2pi,$

$$|p(z)-q(z)| = |z(z-3)^3 +1| le |z(z-3)^3| +1 le 4cdot 1^3 + 1 = 5 .$$

But note $|q|ge 2^7$ on this circle. Since $5< 2^7,$ $p,q,$ have the same number of zeros in $D(3,1).$ Since $q=0$ has no zeroes in this disc, $p$ has no zeros in this disc by Rouche.

share|cite|improve this answer

answered Jul 18 at 22:29

zhw.

65.8k42870

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $q(z) = z^7.$ Then on the circle $z= 3 +e^it, 0le t le 2pi,$

$$|p(z)-q(z)| = |z(z-3)^3 +1| le |z(z-3)^3| +1 le 4cdot 1^3 + 1 = 5 .$$

But note $|q|ge 2^7$ on this circle. Since $5< 2^7,$ $p,q,$ have the same number of zeros in $D(3,1).$ Since $q=0$ has no zeroes in this disc, $p$ has no zeros in this disc by Rouche.

share|cite|improve this answer

answered Jul 18 at 22:29

zhw.

65.8k42870

Let $q(z) = z^7.$ Then on the circle $z= 3 +e^it, 0le t le 2pi,$

$$|p(z)-q(z)| = |z(z-3)^3 +1| le |z(z-3)^3| +1 le 4cdot 1^3 + 1 = 5 .$$

But note $|q|ge 2^7$ on this circle. Since $5< 2^7,$ $p,q,$ have the same number of zeros in $D(3,1).$ Since $q=0$ has no zeroes in this disc, $p$ has no zeros in this disc by Rouche.

share|cite|improve this answer

answered Jul 18 at 22:29

zhw.

65.8k42870

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 22:29

zhw.

65.8k42870

answered Jul 18 at 22:29

zhw.

65.8k42870

answered Jul 18 at 22:29

zhw.

65.8k42870

65.8k42870

add a comment | 

add a comment | 

 

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