Mixing
How to do partial fraction decomposition with complex roots?
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I want to determine $intfrac{1}{x^2+x+1}, dx$ which I approached by partial fraction decomposition.
The complex roots of the denominator are $z_{1}=-0.5+i frac{sqrt 3}{2}$ and $z_2=-0.5-ifrac{sqrt 3}{2}$.
So $$frac{1}{x^2+x+1}=frac{Bx+C}{(x-z_1)(x-z_2)}$$ which yields $B=0$ and $C=1$. However, this would just take me back to me where I started - how do I proceed here?
integration complex-numbers partial-fractions
asked Jan 22 at 20:07
TeslaTesla
885426
$endgroup$
add a comment |
$begingroup$
I want to determine $intfrac{1}{x^2+x+1}, dx$ which I approached by partial fraction decomposition.
The complex roots of the denominator are $z_{1}=-0.5+i frac{sqrt 3}{2}$ and $z_2=-0.5-ifrac{sqrt 3}{2}$.
So $$frac{1}{x^2+x+1}=frac{Bx+C}{(x-z_1)(x-z_2)}$$ which yields $B=0$ and $C=1$. However, this would just take me back to me where I started - how do I proceed here?
integration complex-numbers partial-fractions
asked Jan 22 at 20:07
TeslaTesla
885426
$endgroup$
$begingroup$
Why not $$frac{1}{x^2+x+1}=frac{B}{x-z_1}+frac{C}{x-z_2}?$$
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:10
$begingroup$
Thanks. that's what I did for the exercises where the roots were real. For some reason, I stumbled upon this site that instructed something with $Bx+C$..
$endgroup$
– Tesla
Jan 22 at 20:13
add a comment |
$begingroup$
I want to determine $intfrac{1}{x^2+x+1}, dx$ which I approached by partial fraction decomposition.
The complex roots of the denominator are $z_{1}=-0.5+i frac{sqrt 3}{2}$ and $z_2=-0.5-ifrac{sqrt 3}{2}$.
So $$frac{1}{x^2+x+1}=frac{Bx+C}{(x-z_1)(x-z_2)}$$ which yields $B=0$ and $C=1$. However, this would just take me back to me where I started - how do I proceed here?
integration complex-numbers partial-fractions
asked Jan 22 at 20:07
TeslaTesla
885426
$endgroup$
I want to determine $intfrac{1}{x^2+x+1}, dx$ which I approached by partial fraction decomposition.
The complex roots of the denominator are $z_{1}=-0.5+i frac{sqrt 3}{2}$ and $z_2=-0.5-ifrac{sqrt 3}{2}$.
So $$frac{1}{x^2+x+1}=frac{Bx+C}{(x-z_1)(x-z_2)}$$ which yields $B=0$ and $C=1$. However, this would just take me back to me where I started - how do I proceed here?
integration complex-numbers partial-fractions
integration complex-numbers partial-fractions
asked Jan 22 at 20:07
TeslaTesla
885426
asked Jan 22 at 20:07
TeslaTesla
885426
asked Jan 22 at 20:07
TeslaTesla
885426
asked Jan 22 at 20:07
TeslaTesla
885426
asked Jan 22 at 20:07
TeslaTesla
885426
885426
$begingroup$
Why not $$frac{1}{x^2+x+1}=frac{B}{x-z_1}+frac{C}{x-z_2}?$$
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:10
$begingroup$
Thanks. that's what I did for the exercises where the roots were real. For some reason, I stumbled upon this site that instructed something with $Bx+C$..
$endgroup$
– Tesla
Jan 22 at 20:13
add a comment |
$begingroup$
Why not $$frac{1}{x^2+x+1}=frac{B}{x-z_1}+frac{C}{x-z_2}?$$
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:10
$begingroup$
Thanks. that's what I did for the exercises where the roots were real. For some reason, I stumbled upon this site that instructed something with $Bx+C$..
$endgroup$
– Tesla
Jan 22 at 20:13
$begingroup$
Why not $$frac{1}{x^2+x+1}=frac{B}{x-z_1}+frac{C}{x-z_2}?$$
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:10
$begingroup$
Why not $$frac{1}{x^2+x+1}=frac{B}{x-z_1}+frac{C}{x-z_2}?$$
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:10
$begingroup$
Thanks. that's what I did for the exercises where the roots were real. For some reason, I stumbled upon this site that instructed something with $Bx+C$..
$endgroup$
– Tesla
Jan 22 at 20:13
$begingroup$
Thanks. that's what I did for the exercises where the roots were real. For some reason, I stumbled upon this site that instructed something with $Bx+C$..
$endgroup$
– Tesla
Jan 22 at 20:13
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
These numbers (the complex cubic roots of unity) are usually denoted $j$ and $bar j$ ($=j^2$). You decompose into partial fractions as with real roots:
$$frac{1}{x^2+x+1}=frac{A}{x-j}+frac{B}{x-bar j}.$$
To determine $A$ and $B$, multiply both sides of this equality by $x^2+x+1$ and simplify. You'll obtain
$$1=A(x-bar j)+B(x-j).$$
Setting $x=j$, this equality becomes $;1=A(j-bar j)=2operatorname{Im}(j)=isqrt 3$, whence $A=-dfrac i{sqrt 3}$.
Can you continue?
Effective computation of the integral:
You don't have to decompose this fraction over the complex numbers. Completing the square and a simple substitution reduces it to the basic formula:
$$intfrac{mathrm dx}{x^2+a^2}=frac1a,arctanfrac xa.$$
Now rewrite the quadratic polynomial as
$$x^2+x+1=Bigl(x+frac12Bigr)^2+frac 34$$
and integrate after you've set $t=x+frac12$.
answered Jan 22 at 20:20
BernardBernard
122k741116
$endgroup$
$begingroup$
Thanks. I think you forgot to put $A$ twice in your last equations? Yes I think I can do the rest. Now I just set $x=bar j$ to solve for B?
$endgroup$
– Tesla
Jan 22 at 20:29
$begingroup$
So is it correct that I should get $$F(x)=frac{i}{sqrt 3} ln(vert x-j vert)-frac{i}{sqrt 3} ln(vert x-bar j vert) +c$$?
$endgroup$
– Tesla
Jan 22 at 20:42
$begingroup$
It is not that simple, because the complex logarithm requires branch cut. I'll add the way to calculate this integral in a moment.
$endgroup$
– Bernard
Jan 22 at 20:53
$begingroup$
@Tesla: I've added hints on how to calculate the integral.
$endgroup$
– Bernard
Jan 22 at 21:03
add a comment |
$begingroup$
What you should do is$$frac1{x^2+x+1}=frac A{x-z_1}+frac B{x-z_2}.$$
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
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add a comment |
$begingroup$
$frac {1}{x^2 + x +1} = frac {A}{x-z_1} + frac {B}{x-z_2}$
Now there are two ways you can go by.
$A(x-z_2) + B(x-z_1) = 1$
Multiply it out and solve the system of equations.
$Ax+Bx = 0x\
A = -B\
-Az_2-Bz_1 = 1\
A = frac {1}{z_1-z_2}$
But, this is a nice trick.
$(x-z_1)frac {1}{x^2 + x +1} = A + (x-z_1)frac {B}{x-z_2}\
lim_limits{xto z_1} frac {1}{x-z_2} = A$
Similarly, $lim_limits{xto z_2} frac {1}{x-z_1} = B$
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
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add a comment |
$begingroup$
You'd have $$frac{1}{x^2+x+1}=frac{A}{(x-z_1)} + frac{B}{(x-z_2)} = frac{A}{x-left(-0.5+i frac{sqrt 3}{2}right)} + frac{B}{x-left(-0.5-ifrac{sqrt 3}{2}right)}$$
Multiply both sides by $x^2 +x+1$ to get:
$$1= Aleft(x-left(-frac 12-ifrac{sqrt 3} 2right)right) + Bleft(x-left(-frac 12+ifrac{sqrt 3}2right)right)$$
So, $A+B=0$, and
Take it from here.
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
These numbers (the complex cubic roots of unity) are usually denoted $j$ and $bar j$ ($=j^2$). You decompose into partial fractions as with real roots:
$$frac{1}{x^2+x+1}=frac{A}{x-j}+frac{B}{x-bar j}.$$
To determine $A$ and $B$, multiply both sides of this equality by $x^2+x+1$ and simplify. You'll obtain
$$1=A(x-bar j)+B(x-j).$$
Setting $x=j$, this equality becomes $;1=A(j-bar j)=2operatorname{Im}(j)=isqrt 3$, whence $A=-dfrac i{sqrt 3}$.
Can you continue?
Effective computation of the integral:
You don't have to decompose this fraction over the complex numbers. Completing the square and a simple substitution reduces it to the basic formula:
$$intfrac{mathrm dx}{x^2+a^2}=frac1a,arctanfrac xa.$$
Now rewrite the quadratic polynomial as
$$x^2+x+1=Bigl(x+frac12Bigr)^2+frac 34$$
and integrate after you've set $t=x+frac12$.
answered Jan 22 at 20:20
BernardBernard
122k741116
$endgroup$
$begingroup$
Thanks. I think you forgot to put $A$ twice in your last equations? Yes I think I can do the rest. Now I just set $x=bar j$ to solve for B?
$endgroup$
– Tesla
Jan 22 at 20:29
$begingroup$
So is it correct that I should get $$F(x)=frac{i}{sqrt 3} ln(vert x-j vert)-frac{i}{sqrt 3} ln(vert x-bar j vert) +c$$?
$endgroup$
– Tesla
Jan 22 at 20:42
$begingroup$
It is not that simple, because the complex logarithm requires branch cut. I'll add the way to calculate this integral in a moment.
$endgroup$
– Bernard
Jan 22 at 20:53
$begingroup$
@Tesla: I've added hints on how to calculate the integral.
$endgroup$
– Bernard
Jan 22 at 21:03
add a comment |
$begingroup$
These numbers (the complex cubic roots of unity) are usually denoted $j$ and $bar j$ ($=j^2$). You decompose into partial fractions as with real roots:
$$frac{1}{x^2+x+1}=frac{A}{x-j}+frac{B}{x-bar j}.$$
To determine $A$ and $B$, multiply both sides of this equality by $x^2+x+1$ and simplify. You'll obtain
$$1=A(x-bar j)+B(x-j).$$
Setting $x=j$, this equality becomes $;1=A(j-bar j)=2operatorname{Im}(j)=isqrt 3$, whence $A=-dfrac i{sqrt 3}$.
Can you continue?
Effective computation of the integral:
You don't have to decompose this fraction over the complex numbers. Completing the square and a simple substitution reduces it to the basic formula:
$$intfrac{mathrm dx}{x^2+a^2}=frac1a,arctanfrac xa.$$
Now rewrite the quadratic polynomial as
$$x^2+x+1=Bigl(x+frac12Bigr)^2+frac 34$$
and integrate after you've set $t=x+frac12$.
answered Jan 22 at 20:20
BernardBernard
122k741116
$endgroup$
$begingroup$
Thanks. I think you forgot to put $A$ twice in your last equations? Yes I think I can do the rest. Now I just set $x=bar j$ to solve for B?
$endgroup$
– Tesla
Jan 22 at 20:29
$begingroup$
So is it correct that I should get $$F(x)=frac{i}{sqrt 3} ln(vert x-j vert)-frac{i}{sqrt 3} ln(vert x-bar j vert) +c$$?
$endgroup$
– Tesla
Jan 22 at 20:42
$begingroup$
It is not that simple, because the complex logarithm requires branch cut. I'll add the way to calculate this integral in a moment.
$endgroup$
– Bernard
Jan 22 at 20:53
$begingroup$
@Tesla: I've added hints on how to calculate the integral.
$endgroup$
– Bernard
Jan 22 at 21:03
add a comment |
$begingroup$
These numbers (the complex cubic roots of unity) are usually denoted $j$ and $bar j$ ($=j^2$). You decompose into partial fractions as with real roots:
$$frac{1}{x^2+x+1}=frac{A}{x-j}+frac{B}{x-bar j}.$$
To determine $A$ and $B$, multiply both sides of this equality by $x^2+x+1$ and simplify. You'll obtain
$$1=A(x-bar j)+B(x-j).$$
Setting $x=j$, this equality becomes $;1=A(j-bar j)=2operatorname{Im}(j)=isqrt 3$, whence $A=-dfrac i{sqrt 3}$.
Can you continue?
Effective computation of the integral:
You don't have to decompose this fraction over the complex numbers. Completing the square and a simple substitution reduces it to the basic formula:
$$intfrac{mathrm dx}{x^2+a^2}=frac1a,arctanfrac xa.$$
Now rewrite the quadratic polynomial as
$$x^2+x+1=Bigl(x+frac12Bigr)^2+frac 34$$
and integrate after you've set $t=x+frac12$.
answered Jan 22 at 20:20
BernardBernard
122k741116
$endgroup$
These numbers (the complex cubic roots of unity) are usually denoted $j$ and $bar j$ ($=j^2$). You decompose into partial fractions as with real roots:
$$frac{1}{x^2+x+1}=frac{A}{x-j}+frac{B}{x-bar j}.$$
To determine $A$ and $B$, multiply both sides of this equality by $x^2+x+1$ and simplify. You'll obtain
$$1=A(x-bar j)+B(x-j).$$
Setting $x=j$, this equality becomes $;1=A(j-bar j)=2operatorname{Im}(j)=isqrt 3$, whence $A=-dfrac i{sqrt 3}$.
Can you continue?
Effective computation of the integral:
You don't have to decompose this fraction over the complex numbers. Completing the square and a simple substitution reduces it to the basic formula:
$$intfrac{mathrm dx}{x^2+a^2}=frac1a,arctanfrac xa.$$
Now rewrite the quadratic polynomial as
$$x^2+x+1=Bigl(x+frac12Bigr)^2+frac 34$$
and integrate after you've set $t=x+frac12$.
answered Jan 22 at 20:20
BernardBernard
122k741116
edited Jan 22 at 21:02
answered Jan 22 at 20:20
BernardBernard
122k741116
answered Jan 22 at 20:20
BernardBernard
122k741116
answered Jan 22 at 20:20
BernardBernard
122k741116
122k741116
$begingroup$
Thanks. I think you forgot to put $A$ twice in your last equations? Yes I think I can do the rest. Now I just set $x=bar j$ to solve for B?
$endgroup$
– Tesla
Jan 22 at 20:29
$begingroup$
So is it correct that I should get $$F(x)=frac{i}{sqrt 3} ln(vert x-j vert)-frac{i}{sqrt 3} ln(vert x-bar j vert) +c$$?
$endgroup$
– Tesla
Jan 22 at 20:42
$begingroup$
It is not that simple, because the complex logarithm requires branch cut. I'll add the way to calculate this integral in a moment.
$endgroup$
– Bernard
Jan 22 at 20:53
$begingroup$
@Tesla: I've added hints on how to calculate the integral.
$endgroup$
– Bernard
Jan 22 at 21:03
add a comment |
$begingroup$
Thanks. I think you forgot to put $A$ twice in your last equations? Yes I think I can do the rest. Now I just set $x=bar j$ to solve for B?
$endgroup$
– Tesla
Jan 22 at 20:29
$begingroup$
So is it correct that I should get $$F(x)=frac{i}{sqrt 3} ln(vert x-j vert)-frac{i}{sqrt 3} ln(vert x-bar j vert) +c$$?
$endgroup$
– Tesla
Jan 22 at 20:42
$begingroup$
It is not that simple, because the complex logarithm requires branch cut. I'll add the way to calculate this integral in a moment.
$endgroup$
– Bernard
Jan 22 at 20:53
$begingroup$
@Tesla: I've added hints on how to calculate the integral.
$endgroup$
– Bernard
Jan 22 at 21:03
$begingroup$
Thanks. I think you forgot to put $A$ twice in your last equations? Yes I think I can do the rest. Now I just set $x=bar j$ to solve for B?
$endgroup$
– Tesla
Jan 22 at 20:29
$begingroup$
Thanks. I think you forgot to put $A$ twice in your last equations? Yes I think I can do the rest. Now I just set $x=bar j$ to solve for B?
$endgroup$
– Tesla
Jan 22 at 20:29
$begingroup$
So is it correct that I should get $$F(x)=frac{i}{sqrt 3} ln(vert x-j vert)-frac{i}{sqrt 3} ln(vert x-bar j vert) +c$$?
$endgroup$
– Tesla
Jan 22 at 20:42
$begingroup$
So is it correct that I should get $$F(x)=frac{i}{sqrt 3} ln(vert x-j vert)-frac{i}{sqrt 3} ln(vert x-bar j vert) +c$$?
$endgroup$
– Tesla
Jan 22 at 20:42
$begingroup$
It is not that simple, because the complex logarithm requires branch cut. I'll add the way to calculate this integral in a moment.
$endgroup$
– Bernard
Jan 22 at 20:53
$begingroup$
It is not that simple, because the complex logarithm requires branch cut. I'll add the way to calculate this integral in a moment.
$endgroup$
– Bernard
Jan 22 at 20:53
$begingroup$
@Tesla: I've added hints on how to calculate the integral.
$endgroup$
– Bernard
Jan 22 at 21:03
$begingroup$
@Tesla: I've added hints on how to calculate the integral.
$endgroup$
– Bernard
Jan 22 at 21:03
add a comment |
$begingroup$
What you should do is$$frac1{x^2+x+1}=frac A{x-z_1}+frac B{x-z_2}.$$
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
What you should do is$$frac1{x^2+x+1}=frac A{x-z_1}+frac B{x-z_2}.$$
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
What you should do is$$frac1{x^2+x+1}=frac A{x-z_1}+frac B{x-z_2}.$$
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
What you should do is$$frac1{x^2+x+1}=frac A{x-z_1}+frac B{x-z_2}.$$
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 20:11
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
add a comment |
add a comment |
$begingroup$
$frac {1}{x^2 + x +1} = frac {A}{x-z_1} + frac {B}{x-z_2}$
Now there are two ways you can go by.
$A(x-z_2) + B(x-z_1) = 1$
Multiply it out and solve the system of equations.
$Ax+Bx = 0x\
A = -B\
-Az_2-Bz_1 = 1\
A = frac {1}{z_1-z_2}$
But, this is a nice trick.
$(x-z_1)frac {1}{x^2 + x +1} = A + (x-z_1)frac {B}{x-z_2}\
lim_limits{xto z_1} frac {1}{x-z_2} = A$
Similarly, $lim_limits{xto z_2} frac {1}{x-z_1} = B$
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$frac {1}{x^2 + x +1} = frac {A}{x-z_1} + frac {B}{x-z_2}$
Now there are two ways you can go by.
$A(x-z_2) + B(x-z_1) = 1$
Multiply it out and solve the system of equations.
$Ax+Bx = 0x\
A = -B\
-Az_2-Bz_1 = 1\
A = frac {1}{z_1-z_2}$
But, this is a nice trick.
$(x-z_1)frac {1}{x^2 + x +1} = A + (x-z_1)frac {B}{x-z_2}\
lim_limits{xto z_1} frac {1}{x-z_2} = A$
Similarly, $lim_limits{xto z_2} frac {1}{x-z_1} = B$
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$frac {1}{x^2 + x +1} = frac {A}{x-z_1} + frac {B}{x-z_2}$
Now there are two ways you can go by.
$A(x-z_2) + B(x-z_1) = 1$
Multiply it out and solve the system of equations.
$Ax+Bx = 0x\
A = -B\
-Az_2-Bz_1 = 1\
A = frac {1}{z_1-z_2}$
But, this is a nice trick.
$(x-z_1)frac {1}{x^2 + x +1} = A + (x-z_1)frac {B}{x-z_2}\
lim_limits{xto z_1} frac {1}{x-z_2} = A$
Similarly, $lim_limits{xto z_2} frac {1}{x-z_1} = B$
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
$endgroup$
$frac {1}{x^2 + x +1} = frac {A}{x-z_1} + frac {B}{x-z_2}$
Now there are two ways you can go by.
$A(x-z_2) + B(x-z_1) = 1$
Multiply it out and solve the system of equations.
$Ax+Bx = 0x\
A = -B\
-Az_2-Bz_1 = 1\
A = frac {1}{z_1-z_2}$
But, this is a nice trick.
$(x-z_1)frac {1}{x^2 + x +1} = A + (x-z_1)frac {B}{x-z_2}\
lim_limits{xto z_1} frac {1}{x-z_2} = A$
Similarly, $lim_limits{xto z_2} frac {1}{x-z_1} = B$
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
answered Jan 22 at 21:32
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
$begingroup$
You'd have $$frac{1}{x^2+x+1}=frac{A}{(x-z_1)} + frac{B}{(x-z_2)} = frac{A}{x-left(-0.5+i frac{sqrt 3}{2}right)} + frac{B}{x-left(-0.5-ifrac{sqrt 3}{2}right)}$$
Multiply both sides by $x^2 +x+1$ to get:
$$1= Aleft(x-left(-frac 12-ifrac{sqrt 3} 2right)right) + Bleft(x-left(-frac 12+ifrac{sqrt 3}2right)right)$$
So, $A+B=0$, and
Take it from here.
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
You'd have $$frac{1}{x^2+x+1}=frac{A}{(x-z_1)} + frac{B}{(x-z_2)} = frac{A}{x-left(-0.5+i frac{sqrt 3}{2}right)} + frac{B}{x-left(-0.5-ifrac{sqrt 3}{2}right)}$$
Multiply both sides by $x^2 +x+1$ to get:
$$1= Aleft(x-left(-frac 12-ifrac{sqrt 3} 2right)right) + Bleft(x-left(-frac 12+ifrac{sqrt 3}2right)right)$$
So, $A+B=0$, and
Take it from here.
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
You'd have $$frac{1}{x^2+x+1}=frac{A}{(x-z_1)} + frac{B}{(x-z_2)} = frac{A}{x-left(-0.5+i frac{sqrt 3}{2}right)} + frac{B}{x-left(-0.5-ifrac{sqrt 3}{2}right)}$$
Multiply both sides by $x^2 +x+1$ to get:
$$1= Aleft(x-left(-frac 12-ifrac{sqrt 3} 2right)right) + Bleft(x-left(-frac 12+ifrac{sqrt 3}2right)right)$$
So, $A+B=0$, and
Take it from here.
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
$endgroup$
You'd have $$frac{1}{x^2+x+1}=frac{A}{(x-z_1)} + frac{B}{(x-z_2)} = frac{A}{x-left(-0.5+i frac{sqrt 3}{2}right)} + frac{B}{x-left(-0.5-ifrac{sqrt 3}{2}right)}$$
Multiply both sides by $x^2 +x+1$ to get:
$$1= Aleft(x-left(-frac 12-ifrac{sqrt 3} 2right)right) + Bleft(x-left(-frac 12+ifrac{sqrt 3}2right)right)$$
So, $A+B=0$, and
Take it from here.
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
edited Jan 22 at 21:22
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
answered Jan 22 at 20:12
jordan_glenjordan_glen
1
1
add a comment |
add a comment |
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$begingroup$
Why not $$frac{1}{x^2+x+1}=frac{B}{x-z_1}+frac{C}{x-z_2}?$$
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:10
$begingroup$
Thanks. that's what I did for the exercises where the roots were real. For some reason, I stumbled upon this site that instructed something with $Bx+C$..
$endgroup$
– Tesla
Jan 22 at 20:13