Application of Rouché's Theorem. Show polynomial has exactly one root in each quadrant.
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Show that the complex polynomial given by $z^4+2z^2-z+1$ has exactly one root in each quadrant.
I know by the fundamental theorem of Algebra, that the polynomial has exactly four roots.
Now, to show it has exactly one root in each quadrant, it suffices to show that there are two non-real roots in the open right and left half planes, since zeros of polynomials come in conjugate pairs, and they're just the reflection about the real axis in the complex plane.
I think I have to apply Rouché's theorem and compare $z^4+2z^2-z+1$ to a nice function, but I am not immediately sure what that function may be.
Any help would be much appreciated. Thanks in advance!
complex-analysis
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
asked Dec 6 at 1:30
Math is Life
574
add a comment |
up vote
3
down vote
favorite
Show that the complex polynomial given by $z^4+2z^2-z+1$ has exactly one root in each quadrant.
I know by the fundamental theorem of Algebra, that the polynomial has exactly four roots.
Now, to show it has exactly one root in each quadrant, it suffices to show that there are two non-real roots in the open right and left half planes, since zeros of polynomials come in conjugate pairs, and they're just the reflection about the real axis in the complex plane.
I think I have to apply Rouché's theorem and compare $z^4+2z^2-z+1$ to a nice function, but I am not immediately sure what that function may be.
Any help would be much appreciated. Thanks in advance!
complex-analysis
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
asked Dec 6 at 1:30
Math is Life
574
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Show that the complex polynomial given by $z^4+2z^2-z+1$ has exactly one root in each quadrant.
I know by the fundamental theorem of Algebra, that the polynomial has exactly four roots.
Now, to show it has exactly one root in each quadrant, it suffices to show that there are two non-real roots in the open right and left half planes, since zeros of polynomials come in conjugate pairs, and they're just the reflection about the real axis in the complex plane.
I think I have to apply Rouché's theorem and compare $z^4+2z^2-z+1$ to a nice function, but I am not immediately sure what that function may be.
Any help would be much appreciated. Thanks in advance!
complex-analysis
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
asked Dec 6 at 1:30
Math is Life
574
Show that the complex polynomial given by $z^4+2z^2-z+1$ has exactly one root in each quadrant.
I know by the fundamental theorem of Algebra, that the polynomial has exactly four roots.
Now, to show it has exactly one root in each quadrant, it suffices to show that there are two non-real roots in the open right and left half planes, since zeros of polynomials come in conjugate pairs, and they're just the reflection about the real axis in the complex plane.
I think I have to apply Rouché's theorem and compare $z^4+2z^2-z+1$ to a nice function, but I am not immediately sure what that function may be.
Any help would be much appreciated. Thanks in advance!
complex-analysis
complex-analysis
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
asked Dec 6 at 1:30
Math is Life
574
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
asked Dec 6 at 1:30
Math is Life
574
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
edited Dec 6 at 1:39
Jean-Claude Arbaut
14.7k63363
14.7k63363
asked Dec 6 at 1:30
Math is Life
574
asked Dec 6 at 1:30
Math is Life
574
asked Dec 6 at 1:30
Math is Life
574
574
add a comment |
add a comment |
1 Answer
1
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oldest
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up vote
1
down vote
accepted
We usually apply Rouché's theorem inside the disk and here is not the case. Instead, we use the argument principle.
It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs.
Denote $P(z)=z^4+2z^2-z+1$.
$P(z)$ has no zeros on the real axis because $P(z)=z^4+z^2+(z-frac12)^2+frac34>0$.
$P(z)$ has no zeros on the imaginary axis because $P(iy)=y^4-y^2-iy+1ne0$.
Let $L_1=[0,R]$, $L_2=[0,Ri]$, $Gamma=Re^{itheta}, 0le thetalefracpi2$. Consider the quarter circle $C=L_1+ Gamma- L_2$. From the argument principle,
$$underbrace{text{number of zeros}}_{:=N} - underbrace{text{number of poles}}_{=0}=frac1{2pi i}oint_Cfrac{P'(z)}{P(z)}~mathrm dz=frac{Delta_Carg P(z)}{2pi},$$
or
begin{align*}
N&=frac{Delta_{L_1}arg P(z)}{2pi}+frac{Delta_Gammaarg P(z)}{2pi}-frac{Delta_{L_2}arg P(z)}{2pi}.
end{align*}
We have $Delta_{L_1}arg P(z)=0$ because $P(x)>0~forall~ xinBbb R$,
$$Delta_Gammaarg P(z)=Delta_Gammaarg z^4left(1+frac{2z^2-z+1}{z^4}right)to4cdotfrac{pi}{2}quadtext{as}~ Rtoinfty,$$
and
begin{align*}
Delta_{L_2}arg P(z)&=arg P(iR)=arg{(R^4-R^2+1-iR)}\
&=argleft(1-frac{iR}{R^4-R^2+1}right)to 0 quadtext{as}~ Rtoinfty.
end{align*}
Therefore $N=1$, there is exactly one root in the first quadrant.
answered Dec 6 at 4:01
Tianlalu
3,01021038
Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though?
– Math is Life
Dec 6 at 18:58
It is the variation of argument. The proof of Rouché's theorem is also based on this.
– Tianlalu
Dec 6 at 19:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We usually apply Rouché's theorem inside the disk and here is not the case. Instead, we use the argument principle.
It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs.
Denote $P(z)=z^4+2z^2-z+1$.
$P(z)$ has no zeros on the real axis because $P(z)=z^4+z^2+(z-frac12)^2+frac34>0$.
$P(z)$ has no zeros on the imaginary axis because $P(iy)=y^4-y^2-iy+1ne0$.
Let $L_1=[0,R]$, $L_2=[0,Ri]$, $Gamma=Re^{itheta}, 0le thetalefracpi2$. Consider the quarter circle $C=L_1+ Gamma- L_2$. From the argument principle,
$$underbrace{text{number of zeros}}_{:=N} - underbrace{text{number of poles}}_{=0}=frac1{2pi i}oint_Cfrac{P'(z)}{P(z)}~mathrm dz=frac{Delta_Carg P(z)}{2pi},$$
or
begin{align*}
N&=frac{Delta_{L_1}arg P(z)}{2pi}+frac{Delta_Gammaarg P(z)}{2pi}-frac{Delta_{L_2}arg P(z)}{2pi}.
end{align*}
We have $Delta_{L_1}arg P(z)=0$ because $P(x)>0~forall~ xinBbb R$,
$$Delta_Gammaarg P(z)=Delta_Gammaarg z^4left(1+frac{2z^2-z+1}{z^4}right)to4cdotfrac{pi}{2}quadtext{as}~ Rtoinfty,$$
and
begin{align*}
Delta_{L_2}arg P(z)&=arg P(iR)=arg{(R^4-R^2+1-iR)}\
&=argleft(1-frac{iR}{R^4-R^2+1}right)to 0 quadtext{as}~ Rtoinfty.
end{align*}
Therefore $N=1$, there is exactly one root in the first quadrant.
answered Dec 6 at 4:01
Tianlalu
3,01021038
Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though?
– Math is Life
Dec 6 at 18:58
It is the variation of argument. The proof of Rouché's theorem is also based on this.
– Tianlalu
Dec 6 at 19:28
add a comment |
up vote
1
down vote
accepted
We usually apply Rouché's theorem inside the disk and here is not the case. Instead, we use the argument principle.
It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs.
Denote $P(z)=z^4+2z^2-z+1$.
$P(z)$ has no zeros on the real axis because $P(z)=z^4+z^2+(z-frac12)^2+frac34>0$.
$P(z)$ has no zeros on the imaginary axis because $P(iy)=y^4-y^2-iy+1ne0$.
Let $L_1=[0,R]$, $L_2=[0,Ri]$, $Gamma=Re^{itheta}, 0le thetalefracpi2$. Consider the quarter circle $C=L_1+ Gamma- L_2$. From the argument principle,
$$underbrace{text{number of zeros}}_{:=N} - underbrace{text{number of poles}}_{=0}=frac1{2pi i}oint_Cfrac{P'(z)}{P(z)}~mathrm dz=frac{Delta_Carg P(z)}{2pi},$$
or
begin{align*}
N&=frac{Delta_{L_1}arg P(z)}{2pi}+frac{Delta_Gammaarg P(z)}{2pi}-frac{Delta_{L_2}arg P(z)}{2pi}.
end{align*}
We have $Delta_{L_1}arg P(z)=0$ because $P(x)>0~forall~ xinBbb R$,
$$Delta_Gammaarg P(z)=Delta_Gammaarg z^4left(1+frac{2z^2-z+1}{z^4}right)to4cdotfrac{pi}{2}quadtext{as}~ Rtoinfty,$$
and
begin{align*}
Delta_{L_2}arg P(z)&=arg P(iR)=arg{(R^4-R^2+1-iR)}\
&=argleft(1-frac{iR}{R^4-R^2+1}right)to 0 quadtext{as}~ Rtoinfty.
end{align*}
Therefore $N=1$, there is exactly one root in the first quadrant.
answered Dec 6 at 4:01
Tianlalu
3,01021038
Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though?
– Math is Life
Dec 6 at 18:58
It is the variation of argument. The proof of Rouché's theorem is also based on this.
– Tianlalu
Dec 6 at 19:28
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We usually apply Rouché's theorem inside the disk and here is not the case. Instead, we use the argument principle.
It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs.
Denote $P(z)=z^4+2z^2-z+1$.
$P(z)$ has no zeros on the real axis because $P(z)=z^4+z^2+(z-frac12)^2+frac34>0$.
$P(z)$ has no zeros on the imaginary axis because $P(iy)=y^4-y^2-iy+1ne0$.
Let $L_1=[0,R]$, $L_2=[0,Ri]$, $Gamma=Re^{itheta}, 0le thetalefracpi2$. Consider the quarter circle $C=L_1+ Gamma- L_2$. From the argument principle,
$$underbrace{text{number of zeros}}_{:=N} - underbrace{text{number of poles}}_{=0}=frac1{2pi i}oint_Cfrac{P'(z)}{P(z)}~mathrm dz=frac{Delta_Carg P(z)}{2pi},$$
or
begin{align*}
N&=frac{Delta_{L_1}arg P(z)}{2pi}+frac{Delta_Gammaarg P(z)}{2pi}-frac{Delta_{L_2}arg P(z)}{2pi}.
end{align*}
We have $Delta_{L_1}arg P(z)=0$ because $P(x)>0~forall~ xinBbb R$,
$$Delta_Gammaarg P(z)=Delta_Gammaarg z^4left(1+frac{2z^2-z+1}{z^4}right)to4cdotfrac{pi}{2}quadtext{as}~ Rtoinfty,$$
and
begin{align*}
Delta_{L_2}arg P(z)&=arg P(iR)=arg{(R^4-R^2+1-iR)}\
&=argleft(1-frac{iR}{R^4-R^2+1}right)to 0 quadtext{as}~ Rtoinfty.
end{align*}
Therefore $N=1$, there is exactly one root in the first quadrant.
answered Dec 6 at 4:01
Tianlalu
3,01021038
We usually apply Rouché's theorem inside the disk and here is not the case. Instead, we use the argument principle.
It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs.
Denote $P(z)=z^4+2z^2-z+1$.
$P(z)$ has no zeros on the real axis because $P(z)=z^4+z^2+(z-frac12)^2+frac34>0$.
$P(z)$ has no zeros on the imaginary axis because $P(iy)=y^4-y^2-iy+1ne0$.
Let $L_1=[0,R]$, $L_2=[0,Ri]$, $Gamma=Re^{itheta}, 0le thetalefracpi2$. Consider the quarter circle $C=L_1+ Gamma- L_2$. From the argument principle,
$$underbrace{text{number of zeros}}_{:=N} - underbrace{text{number of poles}}_{=0}=frac1{2pi i}oint_Cfrac{P'(z)}{P(z)}~mathrm dz=frac{Delta_Carg P(z)}{2pi},$$
or
begin{align*}
N&=frac{Delta_{L_1}arg P(z)}{2pi}+frac{Delta_Gammaarg P(z)}{2pi}-frac{Delta_{L_2}arg P(z)}{2pi}.
end{align*}
We have $Delta_{L_1}arg P(z)=0$ because $P(x)>0~forall~ xinBbb R$,
$$Delta_Gammaarg P(z)=Delta_Gammaarg z^4left(1+frac{2z^2-z+1}{z^4}right)to4cdotfrac{pi}{2}quadtext{as}~ Rtoinfty,$$
and
begin{align*}
Delta_{L_2}arg P(z)&=arg P(iR)=arg{(R^4-R^2+1-iR)}\
&=argleft(1-frac{iR}{R^4-R^2+1}right)to 0 quadtext{as}~ Rtoinfty.
end{align*}
Therefore $N=1$, there is exactly one root in the first quadrant.
answered Dec 6 at 4:01
Tianlalu
3,01021038
edited Dec 6 at 5:05
answered Dec 6 at 4:01
Tianlalu
3,01021038
answered Dec 6 at 4:01
Tianlalu
3,01021038
answered Dec 6 at 4:01
Tianlalu
3,01021038
3,01021038
Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though?
– Math is Life
Dec 6 at 18:58
It is the variation of argument. The proof of Rouché's theorem is also based on this.
– Tianlalu
Dec 6 at 19:28
add a comment |
Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though?
– Math is Life
Dec 6 at 18:58
It is the variation of argument. The proof of Rouché's theorem is also based on this.
– Tianlalu
Dec 6 at 19:28
Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though?
– Math is Life
Dec 6 at 18:58
Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though?
– Math is Life
Dec 6 at 18:58
It is the variation of argument. The proof of Rouché's theorem is also based on this.
– Tianlalu
Dec 6 at 19:28
It is the variation of argument. The proof of Rouché's theorem is also based on this.
– Tianlalu
Dec 6 at 19:28
add a comment |
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