Rouche's theorem in the right half plane
$begingroup$
I need to determine the number of zeros in the right half plane Re z>0 of the polynomial:
$$
f(z)=z^3-z+1
$$
My attempt to solve the problem:
I'm using Rouché's theorem and consider
$$
g(z)=z^3+1 ~~ and ~~
p(z)= -z
$$
and it can be seen that $|g(z)|>|p(z)|$ on the imaginary axis and for $|z|$ large.
My conclusion is that the function $f(z)$ has $3$ zeros in the right half plane. Is this correct?
(A solution from the book showed that it has $2$ zeros in $Re, z>0$ but I'm not sure if it's a typo)
complex-analysis rouches-theorem
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
asked Dec 14 '18 at 12:50
user608881user608881
283
$endgroup$
add a comment |
$begingroup$
I need to determine the number of zeros in the right half plane Re z>0 of the polynomial:
$$
f(z)=z^3-z+1
$$
My attempt to solve the problem:
I'm using Rouché's theorem and consider
$$
g(z)=z^3+1 ~~ and ~~
p(z)= -z
$$
and it can be seen that $|g(z)|>|p(z)|$ on the imaginary axis and for $|z|$ large.
My conclusion is that the function $f(z)$ has $3$ zeros in the right half plane. Is this correct?
(A solution from the book showed that it has $2$ zeros in $Re, z>0$ but I'm not sure if it's a typo)
complex-analysis rouches-theorem
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
asked Dec 14 '18 at 12:50
user608881user608881
283
$endgroup$
$begingroup$
Based on the partial fraction decomposition $$ frac{p(z)}{z^3+1}=1+frac{1}{3(z+1)}-frac{z+1}{3(z^2-z+1)} $$ one can improve the factorization $p(z)=(z+1)(z^2-z+1)-z$ to begin{align} p(z)&=(z+1)(z^2-z+1)+tfrac13(z^2-z+1)-tfrac13(z+1)^2\ &=left(z+tfrac43right)left(z^2-tfrac43z+tfrac23right)+tfrac19(z+1) end{align} giving the root locations close to $-frac43$ and $frac23pm ifrac{sqrt2}3$.
$endgroup$
– LutzL
Dec 14 '18 at 13:51
add a comment |
$begingroup$
I need to determine the number of zeros in the right half plane Re z>0 of the polynomial:
$$
f(z)=z^3-z+1
$$
My attempt to solve the problem:
I'm using Rouché's theorem and consider
$$
g(z)=z^3+1 ~~ and ~~
p(z)= -z
$$
and it can be seen that $|g(z)|>|p(z)|$ on the imaginary axis and for $|z|$ large.
My conclusion is that the function $f(z)$ has $3$ zeros in the right half plane. Is this correct?
(A solution from the book showed that it has $2$ zeros in $Re, z>0$ but I'm not sure if it's a typo)
complex-analysis rouches-theorem
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
asked Dec 14 '18 at 12:50
user608881user608881
283
$endgroup$
I need to determine the number of zeros in the right half plane Re z>0 of the polynomial:
$$
f(z)=z^3-z+1
$$
My attempt to solve the problem:
I'm using Rouché's theorem and consider
$$
g(z)=z^3+1 ~~ and ~~
p(z)= -z
$$
and it can be seen that $|g(z)|>|p(z)|$ on the imaginary axis and for $|z|$ large.
My conclusion is that the function $f(z)$ has $3$ zeros in the right half plane. Is this correct?
(A solution from the book showed that it has $2$ zeros in $Re, z>0$ but I'm not sure if it's a typo)
complex-analysis rouches-theorem
complex-analysis rouches-theorem
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
asked Dec 14 '18 at 12:50
user608881user608881
283
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
asked Dec 14 '18 at 12:50
user608881user608881
283
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
edited Dec 14 '18 at 13:16
LutzL
58.3k42054
58.3k42054
asked Dec 14 '18 at 12:50
user608881user608881
283
asked Dec 14 '18 at 12:50
user608881user608881
283
asked Dec 14 '18 at 12:50
user608881user608881
283
283
$begingroup$
Based on the partial fraction decomposition $$ frac{p(z)}{z^3+1}=1+frac{1}{3(z+1)}-frac{z+1}{3(z^2-z+1)} $$ one can improve the factorization $p(z)=(z+1)(z^2-z+1)-z$ to begin{align} p(z)&=(z+1)(z^2-z+1)+tfrac13(z^2-z+1)-tfrac13(z+1)^2\ &=left(z+tfrac43right)left(z^2-tfrac43z+tfrac23right)+tfrac19(z+1) end{align} giving the root locations close to $-frac43$ and $frac23pm ifrac{sqrt2}3$.
$endgroup$
– LutzL
Dec 14 '18 at 13:51
add a comment |
$begingroup$
Based on the partial fraction decomposition $$ frac{p(z)}{z^3+1}=1+frac{1}{3(z+1)}-frac{z+1}{3(z^2-z+1)} $$ one can improve the factorization $p(z)=(z+1)(z^2-z+1)-z$ to begin{align} p(z)&=(z+1)(z^2-z+1)+tfrac13(z^2-z+1)-tfrac13(z+1)^2\ &=left(z+tfrac43right)left(z^2-tfrac43z+tfrac23right)+tfrac19(z+1) end{align} giving the root locations close to $-frac43$ and $frac23pm ifrac{sqrt2}3$.
$endgroup$
– LutzL
Dec 14 '18 at 13:51
$begingroup$
Based on the partial fraction decomposition $$ frac{p(z)}{z^3+1}=1+frac{1}{3(z+1)}-frac{z+1}{3(z^2-z+1)} $$ one can improve the factorization $p(z)=(z+1)(z^2-z+1)-z$ to begin{align} p(z)&=(z+1)(z^2-z+1)+tfrac13(z^2-z+1)-tfrac13(z+1)^2\ &=left(z+tfrac43right)left(z^2-tfrac43z+tfrac23right)+tfrac19(z+1) end{align} giving the root locations close to $-frac43$ and $frac23pm ifrac{sqrt2}3$.
$endgroup$
– LutzL
Dec 14 '18 at 13:51
$begingroup$
Based on the partial fraction decomposition $$ frac{p(z)}{z^3+1}=1+frac{1}{3(z+1)}-frac{z+1}{3(z^2-z+1)} $$ one can improve the factorization $p(z)=(z+1)(z^2-z+1)-z$ to begin{align} p(z)&=(z+1)(z^2-z+1)+tfrac13(z^2-z+1)-tfrac13(z+1)^2\ &=left(z+tfrac43right)left(z^2-tfrac43z+tfrac23right)+tfrac19(z+1) end{align} giving the root locations close to $-frac43$ and $frac23pm ifrac{sqrt2}3$.
$endgroup$
– LutzL
Dec 14 '18 at 13:51
add a comment |
1 Answer
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$begingroup$
You correctly derived that $f$ and $g$ have the same number of zeros in the right half-plane.
Only the final conclusion is wrong: $g(z) = z^3+1$ has two zeros in the right half-plane (the third zero is $z=-1$).
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You correctly derived that $f$ and $g$ have the same number of zeros in the right half-plane.
Only the final conclusion is wrong: $g(z) = z^3+1$ has two zeros in the right half-plane (the third zero is $z=-1$).
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
$endgroup$
add a comment |
$begingroup$
You correctly derived that $f$ and $g$ have the same number of zeros in the right half-plane.
Only the final conclusion is wrong: $g(z) = z^3+1$ has two zeros in the right half-plane (the third zero is $z=-1$).
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
$endgroup$
add a comment |
$begingroup$
You correctly derived that $f$ and $g$ have the same number of zeros in the right half-plane.
Only the final conclusion is wrong: $g(z) = z^3+1$ has two zeros in the right half-plane (the third zero is $z=-1$).
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
$endgroup$
You correctly derived that $f$ and $g$ have the same number of zeros in the right half-plane.
Only the final conclusion is wrong: $g(z) = z^3+1$ has two zeros in the right half-plane (the third zero is $z=-1$).
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
edited Dec 14 '18 at 13:20
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
answered Dec 14 '18 at 13:12
Martin RMartin R
28.7k33356
28.7k33356
add a comment |
add a comment |
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$begingroup$
Based on the partial fraction decomposition $$ frac{p(z)}{z^3+1}=1+frac{1}{3(z+1)}-frac{z+1}{3(z^2-z+1)} $$ one can improve the factorization $p(z)=(z+1)(z^2-z+1)-z$ to begin{align} p(z)&=(z+1)(z^2-z+1)+tfrac13(z^2-z+1)-tfrac13(z+1)^2\ &=left(z+tfrac43right)left(z^2-tfrac43z+tfrac23right)+tfrac19(z+1) end{align} giving the root locations close to $-frac43$ and $frac23pm ifrac{sqrt2}3$.
$endgroup$
– LutzL
Dec 14 '18 at 13:51