How many roots does $g(z)=z^7-2z^5+6z^3-z+1$ have inside the unit disk - Rouche's Theorem Application...
$begingroup$
$g(z)=z^7-2z^5+6z^3-z+1$
and choose $f(z)=2z^5-6z^3$.
On $mid zmid =1$, we have
$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$
so $$mid f(z)-g(z)mid leq mid f(z)mid$$
and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.
Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.
Is this argument correct?
complex-analysis proof-verification complex-numbers roots rouches-theorem
$endgroup$
add a comment |
$begingroup$
$g(z)=z^7-2z^5+6z^3-z+1$
and choose $f(z)=2z^5-6z^3$.
On $mid zmid =1$, we have
$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$
so $$mid f(z)-g(z)mid leq mid f(z)mid$$
and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.
Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.
Is this argument correct?
complex-analysis proof-verification complex-numbers roots rouches-theorem
$endgroup$
$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07
add a comment |
$begingroup$
$g(z)=z^7-2z^5+6z^3-z+1$
and choose $f(z)=2z^5-6z^3$.
On $mid zmid =1$, we have
$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$
so $$mid f(z)-g(z)mid leq mid f(z)mid$$
and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.
Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.
Is this argument correct?
complex-analysis proof-verification complex-numbers roots rouches-theorem
$endgroup$
$g(z)=z^7-2z^5+6z^3-z+1$
and choose $f(z)=2z^5-6z^3$.
On $mid zmid =1$, we have
$mid f(z)-g(z) mid=1$ and $mid f(z) mid=4$
so $$mid f(z)-g(z)mid leq mid f(z)mid$$
and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.
Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.
Is this argument correct?
complex-analysis proof-verification complex-numbers roots rouches-theorem
complex-analysis proof-verification complex-numbers roots rouches-theorem
asked Dec 18 '18 at 21:49
MikeMike
738415
738415
$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07
add a comment |
$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07
$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07
$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07
add a comment |
1 Answer
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$begingroup$
I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.
Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
$$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
$$lvert f(z)rvert= lvert 6z^3rvert=6.$$
So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.
Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
$$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
$$lvert f(z)rvert= lvert 6z^3rvert=6.$$
So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.
$endgroup$
add a comment |
$begingroup$
I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.
Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
$$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
$$lvert f(z)rvert= lvert 6z^3rvert=6.$$
So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.
$endgroup$
add a comment |
$begingroup$
I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.
Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
$$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
$$lvert f(z)rvert= lvert 6z^3rvert=6.$$
So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.
$endgroup$
I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $lvert f(z)-g(z)rvertne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.
Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
$$ lvert h(z)rvertle lvert zrvert^7+2lvert zrvert^5+lvert zrvert +1le 5,$$
$$lvert f(z)rvert= lvert 6z^3rvert=6.$$
So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.
answered Dec 18 '18 at 22:11
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41641
10.5k41641
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$begingroup$
why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$
$endgroup$
– MoonKnight
Dec 18 '18 at 22:07