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Complex analysis, P(z) [closed]

-1

$begingroup$

Show that for any polynomial P(z)
$max_=1|frac1z^2 - P(z)| geq 1$

complex-analysis polynomials

share|cite|improve this question

asked Mar 20 at 10:59

saharockksaharockk

81

$endgroup$

closed as off-topic by Martin R, John Omielan, Lee David Chung Lin, dantopa, Michael Rybkin Mar 21 at 2:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, John Omielan, Michael Rybkin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried so far?
  $endgroup$
  – Benedict W. J. Irwin
  Mar 20 at 11:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Assume that the maximum is $<1$. Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$. Apply the Argument principle together with this inequality to $1/z^2$ and $P$ to conclude that $1/z^2$ and $f$ have the same variation of the argument around $0$ along the loop $|z|=1$. But since this value is $-2$ for $1/z^2$, so should be for $P$. This implies that $P$ should have some poles inside $|z|<1$, which is a contradiction.
  $endgroup$
  – user647486
  Mar 20 at 11:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Essentially a duplicate of Show that for any polynomial $p(z)$ there is a $z$ with $|z|=1$ such that $|p(z)-1/z|geq 1$..
  $endgroup$
  – Martin R
  Mar 20 at 12:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Complex zeros of the polynomials $sum_k=0^n z^k/k!$, inside balls
  $endgroup$
  – Lee David Chung Lin
  Mar 21 at 0:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 21 at 0:57
  

  
  
  
  

add a comment | 

-1

$begingroup$

Show that for any polynomial P(z)
$max_=1|frac1z^2 - P(z)| geq 1$

complex-analysis polynomials

share|cite|improve this question

asked Mar 20 at 10:59

saharockksaharockk

81

$endgroup$

closed as off-topic by Martin R, John Omielan, Lee David Chung Lin, dantopa, Michael Rybkin Mar 21 at 2:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, John Omielan, Michael Rybkin

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried so far?
  $endgroup$
  – Benedict W. J. Irwin
  Mar 20 at 11:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Assume that the maximum is $<1$. Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$. Apply the Argument principle together with this inequality to $1/z^2$ and $P$ to conclude that $1/z^2$ and $f$ have the same variation of the argument around $0$ along the loop $|z|=1$. But since this value is $-2$ for $1/z^2$, so should be for $P$. This implies that $P$ should have some poles inside $|z|<1$, which is a contradiction.
  $endgroup$
  – user647486
  Mar 20 at 11:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Essentially a duplicate of Show that for any polynomial $p(z)$ there is a $z$ with $|z|=1$ such that $|p(z)-1/z|geq 1$..
  $endgroup$
  – Martin R
  Mar 20 at 12:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Complex zeros of the polynomials $sum_k=0^n z^k/k!$, inside balls
  $endgroup$
  – Lee David Chung Lin
  Mar 21 at 0:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 21 at 0:57
  

  
  
  
  

add a comment | 

$begingroup$

Show that for any polynomial P(z)
$max_=1|frac1z^2 - P(z)| geq 1$

complex-analysis polynomials

share|cite|improve this question

asked Mar 20 at 10:59

saharockksaharockk

81

$endgroup$

share|cite|improve this question

asked Mar 20 at 10:59

saharockksaharockk

81

share|cite|improve this question

asked Mar 20 at 10:59

saharockksaharockk

81

asked Mar 20 at 10:59

saharockksaharockk

81

asked Mar 20 at 10:59

saharockksaharockk

81

2 Answers
2

active

oldest

votes

0

$begingroup$

Assume that the maximum is $<1$.

Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$.

Neither $1/z^2$ nor $P$ has poles on $|z|=1$ and by the inequality $P$ doesn't vanish on $|z|=1$, nor does $1/z^2$. This allows us to use the Argument principle to compute that $-2$ is the winding number of the image of the unit circle $|z|=1$ by $1/z^2$ around $0$.

Likewise, $0$ is the winding number of the image of $C$ by $P(z)$ around $0$.

Since $|1/z^2-P(z)|<1=|1/z^2|$, we have that $1/z^2$ is not a negative multiple of $P(z)$, for $|z|=1$. Therefore,

$$H(t,z)=t/z^2+(1-t)P(z)$$

provides a homotopy between those two curves that never touches $0$. Therefore, those two curves ought to have the same winding number with respect to $0$.

Contradiction.

---

Or you could apply Rouche's theorem to $1$ and $z^2P(z)$, since $$|z^2P(z)-1|<|z^2|=1$$
for $|z|=1$. The theorem implies that $1$ and $z^2P(z)$ have the same number of zeros inside $|z|<1$. But $1$ never vanishes and $z^2P(z)$ has two zeros in $|z|<1$.

share|cite|improve this answer

edited Mar 20 at 12:11

answered Mar 20 at 11:28

user647486user647486

832110

$endgroup$

add a comment | 

0

$begingroup$

Consider the entire function $f(z)=1-z^2P(z)$. For $|z|=1$ we have

$$|frac1z^2 - P(z)| =|f(z)|.$$

Hence we have to show that $m:=max =1 ge 1.$

If $M:=max f(z)$, then the maximum - principle says:

$$M=m.$$

From $|f(0)|=1$, we get $M ge1.$

share|cite|improve this answer

answered Mar 20 at 12:15

FredFred

48.7k11849

$endgroup$

add a comment | 

0

$begingroup$

Assume that the maximum is $<1$.

Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$.

Neither $1/z^2$ nor $P$ has poles on $|z|=1$ and by the inequality $P$ doesn't vanish on $|z|=1$, nor does $1/z^2$. This allows us to use the Argument principle to compute that $-2$ is the winding number of the image of the unit circle $|z|=1$ by $1/z^2$ around $0$.

Likewise, $0$ is the winding number of the image of $C$ by $P(z)$ around $0$.

Since $|1/z^2-P(z)|<1=|1/z^2|$, we have that $1/z^2$ is not a negative multiple of $P(z)$, for $|z|=1$. Therefore,

$$H(t,z)=t/z^2+(1-t)P(z)$$

provides a homotopy between those two curves that never touches $0$. Therefore, those two curves ought to have the same winding number with respect to $0$.

Contradiction.

---

Or you could apply Rouche's theorem to $1$ and $z^2P(z)$, since $$|z^2P(z)-1|<|z^2|=1$$
for $|z|=1$. The theorem implies that $1$ and $z^2P(z)$ have the same number of zeros inside $|z|<1$. But $1$ never vanishes and $z^2P(z)$ has two zeros in $|z|<1$.

share|cite|improve this answer

edited Mar 20 at 12:11

answered Mar 20 at 11:28

user647486user647486

832110

$endgroup$

add a comment | 

0

$begingroup$

Assume that the maximum is $<1$.

Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$.

Neither $1/z^2$ nor $P$ has poles on $|z|=1$ and by the inequality $P$ doesn't vanish on $|z|=1$, nor does $1/z^2$. This allows us to use the Argument principle to compute that $-2$ is the winding number of the image of the unit circle $|z|=1$ by $1/z^2$ around $0$.

Likewise, $0$ is the winding number of the image of $C$ by $P(z)$ around $0$.

Since $|1/z^2-P(z)|<1=|1/z^2|$, we have that $1/z^2$ is not a negative multiple of $P(z)$, for $|z|=1$. Therefore,

$$H(t,z)=t/z^2+(1-t)P(z)$$

provides a homotopy between those two curves that never touches $0$. Therefore, those two curves ought to have the same winding number with respect to $0$.

Contradiction.

---

Or you could apply Rouche's theorem to $1$ and $z^2P(z)$, since $$|z^2P(z)-1|<|z^2|=1$$
for $|z|=1$. The theorem implies that $1$ and $z^2P(z)$ have the same number of zeros inside $|z|<1$. But $1$ never vanishes and $z^2P(z)$ has two zeros in $|z|<1$.

share|cite|improve this answer

edited Mar 20 at 12:11

answered Mar 20 at 11:28

user647486user647486

832110

$endgroup$

add a comment | 

$begingroup$

Assume that the maximum is $<1$.

Then, on $|z|=1$ $|1/z^2-P(z)|<1=|1/z^2|$.

Neither $1/z^2$ nor $P$ has poles on $|z|=1$ and by the inequality $P$ doesn't vanish on $|z|=1$, nor does $1/z^2$. This allows us to use the Argument principle to compute that $-2$ is the winding number of the image of the unit circle $|z|=1$ by $1/z^2$ around $0$.

Likewise, $0$ is the winding number of the image of $C$ by $P(z)$ around $0$.

Since $|1/z^2-P(z)|<1=|1/z^2|$, we have that $1/z^2$ is not a negative multiple of $P(z)$, for $|z|=1$. Therefore,

$$H(t,z)=t/z^2+(1-t)P(z)$$

provides a homotopy between those two curves that never touches $0$. Therefore, those two curves ought to have the same winding number with respect to $0$.

Contradiction.

---

Or you could apply Rouche's theorem to $1$ and $z^2P(z)$, since $$|z^2P(z)-1|<|z^2|=1$$
for $|z|=1$. The theorem implies that $1$ and $z^2P(z)$ have the same number of zeros inside $|z|<1$. But $1$ never vanishes and $z^2P(z)$ has two zeros in $|z|<1$.

share|cite|improve this answer

edited Mar 20 at 12:11

answered Mar 20 at 11:28

user647486user647486

832110

$endgroup$

share|cite|improve this answer

edited Mar 20 at 12:11

answered Mar 20 at 11:28

user647486user647486

832110

answered Mar 20 at 11:28

user647486user647486

832110

answered Mar 20 at 11:28

user647486user647486

832110

0

$begingroup$

Consider the entire function $f(z)=1-z^2P(z)$. For $|z|=1$ we have

$$|frac1z^2 - P(z)| =|f(z)|.$$

Hence we have to show that $m:=max =1 ge 1.$

If $M:=max f(z)$, then the maximum - principle says:

$$M=m.$$

From $|f(0)|=1$, we get $M ge1.$

share|cite|improve this answer

answered Mar 20 at 12:15

FredFred

48.7k11849

$endgroup$

add a comment | 

0

$begingroup$

Consider the entire function $f(z)=1-z^2P(z)$. For $|z|=1$ we have

$$|frac1z^2 - P(z)| =|f(z)|.$$

Hence we have to show that $m:=max =1 ge 1.$

If $M:=max f(z)$, then the maximum - principle says:

$$M=m.$$

From $|f(0)|=1$, we get $M ge1.$

share|cite|improve this answer

answered Mar 20 at 12:15

FredFred

48.7k11849

$endgroup$

add a comment | 

$begingroup$

Consider the entire function $f(z)=1-z^2P(z)$. For $|z|=1$ we have

$$|frac1z^2 - P(z)| =|f(z)|.$$

Hence we have to show that $m:=max =1 ge 1.$

If $M:=max f(z)$, then the maximum - principle says:

$$M=m.$$

From $|f(0)|=1$, we get $M ge1.$

share|cite|improve this answer

answered Mar 20 at 12:15

FredFred

48.7k11849

$endgroup$

share|cite|improve this answer

answered Mar 20 at 12:15

FredFred

48.7k11849

answered Mar 20 at 12:15

FredFred

48.7k11849

answered Mar 20 at 12:15

FredFred

48.7k11849