Find number of roots of a polynomial inside $ |z| < 2 $
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The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
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add a comment |
$begingroup$
The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
$endgroup$
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
add a comment |
$begingroup$
The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
$endgroup$
The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
asked Jan 6 at 9:29
calmcalm
1387
edited Jan 6 at 12:01
calm
asked Jan 6 at 9:29
calmcalm
1387
asked Jan 6 at 9:29
calmcalm
1387
asked Jan 6 at 9:29
calmcalm
1387
1387
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
add a comment |
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
add a comment |
1 Answer
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$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
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add a comment |
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1 Answer
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$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
$endgroup$
add a comment |
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
$endgroup$
add a comment |
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
$endgroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
answered Jan 6 at 13:38
LutzLLutzL
59.8k42057
59.8k42057
add a comment |
add a comment |
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$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02