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Laurent series expansion of $f(z)=frac 1z^2+z+1$ for $|z| > 1$ centered at $z=0$
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Since $|z| > 1$, with finding a convergent series in mind I wanted to write an equivalent closed-form expression of $f$ in terms of $frac 1z$ because $|frac 1z| < 1$. So I tried to factor out $frac 1z^2$ and writing $f$ as follows:
$$
f(z) = frac 1z^2frac 11 - (-frac 1z - frac 1z^2).
$$
However, $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < 1+1=2$, so I was unable to bound $|-frac 1z - frac 1z^2|$ by $1$.
What is a suitable expression of $f$ that I would need so that I can finally rewrite it in terms of a convergent series?
Should I consider splitting into two cases: $1 < |z| le 2$ and $|z| > 2$? It turns out that if $|z| > 2$, then $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < frac 12 + frac 12 = 1$.
complex-analysis laurent-series
asked Jul 18 at 19:02
Cookie
8,626123376
add a comment |Â
up vote
1
down vote
favorite
Since $|z| > 1$, with finding a convergent series in mind I wanted to write an equivalent closed-form expression of $f$ in terms of $frac 1z$ because $|frac 1z| < 1$. So I tried to factor out $frac 1z^2$ and writing $f$ as follows:
$$
f(z) = frac 1z^2frac 11 - (-frac 1z - frac 1z^2).
$$
However, $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < 1+1=2$, so I was unable to bound $|-frac 1z - frac 1z^2|$ by $1$.
What is a suitable expression of $f$ that I would need so that I can finally rewrite it in terms of a convergent series?
Should I consider splitting into two cases: $1 < |z| le 2$ and $|z| > 2$? It turns out that if $|z| > 2$, then $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < frac 12 + frac 12 = 1$.
complex-analysis laurent-series
asked Jul 18 at 19:02
Cookie
8,626123376
1
$frac1z^2+z+1=frac1/(w-w^2)z-w+frac1/(w^2-w)z-w^2=frac1zfrac1/(w-w^2)1-w/z+frac1zfrac1/(w^2-w)1-w^2/z$, where $w=e^2pi i/3$. Now expand $frac11-w/z=sum_n=0^inftyleft(fracwzright)^n$ and likewise the other term.
– user577471
Jul 18 at 19:13
How did you get the first equality? Did you factor $z^2+z+1$ into imaginary factors and then used partial fractions?
– Cookie
Jul 18 at 20:35
Yes, that is how you do many computations with rational functions, in particular Laurent expansions.
– user577471
Jul 18 at 20:46
I'm still not sure how you got the RHS of your first equality; doing the distinct factor case of the method of partial fractions, I got $$frac 1(z+w)(z-w) = frac-frac 12wz+w + frac frac 12wz-w$$ where $w=e^frac2pi i3$.
– Cookie
Jul 18 at 22:30
The roots of $z^2+z+1$ are not $w$ and $-w$, are $w$ and $overlinew=w^2=1/w$.
– user577471
Jul 19 at 14:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Since $|z| > 1$, with finding a convergent series in mind I wanted to write an equivalent closed-form expression of $f$ in terms of $frac 1z$ because $|frac 1z| < 1$. So I tried to factor out $frac 1z^2$ and writing $f$ as follows:
$$
f(z) = frac 1z^2frac 11 - (-frac 1z - frac 1z^2).
$$
However, $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < 1+1=2$, so I was unable to bound $|-frac 1z - frac 1z^2|$ by $1$.
What is a suitable expression of $f$ that I would need so that I can finally rewrite it in terms of a convergent series?
Should I consider splitting into two cases: $1 < |z| le 2$ and $|z| > 2$? It turns out that if $|z| > 2$, then $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < frac 12 + frac 12 = 1$.
complex-analysis laurent-series
asked Jul 18 at 19:02
Cookie
8,626123376
Since $|z| > 1$, with finding a convergent series in mind I wanted to write an equivalent closed-form expression of $f$ in terms of $frac 1z$ because $|frac 1z| < 1$. So I tried to factor out $frac 1z^2$ and writing $f$ as follows:
$$
f(z) = frac 1z^2frac 11 - (-frac 1z - frac 1z^2).
$$
However, $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < 1+1=2$, so I was unable to bound $|-frac 1z - frac 1z^2|$ by $1$.
What is a suitable expression of $f$ that I would need so that I can finally rewrite it in terms of a convergent series?
Should I consider splitting into two cases: $1 < |z| le 2$ and $|z| > 2$? It turns out that if $|z| > 2$, then $|-frac 1z - frac 1z^2| le frac 1+frac 1^2 < frac 12 + frac 12 = 1$.
complex-analysis laurent-series
asked Jul 18 at 19:02
Cookie
8,626123376
edited Jul 18 at 19:44
asked Jul 18 at 19:02
Cookie
8,626123376
asked Jul 18 at 19:02
Cookie
8,626123376
asked Jul 18 at 19:02
Cookie
8,626123376
8,626123376
1
$frac1z^2+z+1=frac1/(w-w^2)z-w+frac1/(w^2-w)z-w^2=frac1zfrac1/(w-w^2)1-w/z+frac1zfrac1/(w^2-w)1-w^2/z$, where $w=e^2pi i/3$. Now expand $frac11-w/z=sum_n=0^inftyleft(fracwzright)^n$ and likewise the other term.
– user577471
Jul 18 at 19:13
How did you get the first equality? Did you factor $z^2+z+1$ into imaginary factors and then used partial fractions?
– Cookie
Jul 18 at 20:35
Yes, that is how you do many computations with rational functions, in particular Laurent expansions.
– user577471
Jul 18 at 20:46
I'm still not sure how you got the RHS of your first equality; doing the distinct factor case of the method of partial fractions, I got $$frac 1(z+w)(z-w) = frac-frac 12wz+w + frac frac 12wz-w$$ where $w=e^frac2pi i3$.
– Cookie
Jul 18 at 22:30
The roots of $z^2+z+1$ are not $w$ and $-w$, are $w$ and $overlinew=w^2=1/w$.
– user577471
Jul 19 at 14:33
add a comment |Â
1
$frac1z^2+z+1=frac1/(w-w^2)z-w+frac1/(w^2-w)z-w^2=frac1zfrac1/(w-w^2)1-w/z+frac1zfrac1/(w^2-w)1-w^2/z$, where $w=e^2pi i/3$. Now expand $frac11-w/z=sum_n=0^inftyleft(fracwzright)^n$ and likewise the other term.
– user577471
Jul 18 at 19:13
How did you get the first equality? Did you factor $z^2+z+1$ into imaginary factors and then used partial fractions?
– Cookie
Jul 18 at 20:35
Yes, that is how you do many computations with rational functions, in particular Laurent expansions.
– user577471
Jul 18 at 20:46
I'm still not sure how you got the RHS of your first equality; doing the distinct factor case of the method of partial fractions, I got $$frac 1(z+w)(z-w) = frac-frac 12wz+w + frac frac 12wz-w$$ where $w=e^frac2pi i3$.
– Cookie
Jul 18 at 22:30
The roots of $z^2+z+1$ are not $w$ and $-w$, are $w$ and $overlinew=w^2=1/w$.
– user577471
Jul 19 at 14:33
1
1
$frac1z^2+z+1=frac1/(w-w^2)z-w+frac1/(w^2-w)z-w^2=frac1zfrac1/(w-w^2)1-w/z+frac1zfrac1/(w^2-w)1-w^2/z$, where $w=e^2pi i/3$. Now expand $frac11-w/z=sum_n=0^inftyleft(fracwzright)^n$ and likewise the other term.
– user577471
Jul 18 at 19:13
$frac1z^2+z+1=frac1/(w-w^2)z-w+frac1/(w^2-w)z-w^2=frac1zfrac1/(w-w^2)1-w/z+frac1zfrac1/(w^2-w)1-w^2/z$, where $w=e^2pi i/3$. Now expand $frac11-w/z=sum_n=0^inftyleft(fracwzright)^n$ and likewise the other term.
– user577471
Jul 18 at 19:13
How did you get the first equality? Did you factor $z^2+z+1$ into imaginary factors and then used partial fractions?
– Cookie
Jul 18 at 20:35
How did you get the first equality? Did you factor $z^2+z+1$ into imaginary factors and then used partial fractions?
– Cookie
Jul 18 at 20:35
Yes, that is how you do many computations with rational functions, in particular Laurent expansions.
– user577471
Jul 18 at 20:46
Yes, that is how you do many computations with rational functions, in particular Laurent expansions.
– user577471
Jul 18 at 20:46
I'm still not sure how you got the RHS of your first equality; doing the distinct factor case of the method of partial fractions, I got $$frac 1(z+w)(z-w) = frac-frac 12wz+w + frac frac 12wz-w$$ where $w=e^frac2pi i3$.
– Cookie
Jul 18 at 22:30
I'm still not sure how you got the RHS of your first equality; doing the distinct factor case of the method of partial fractions, I got $$frac 1(z+w)(z-w) = frac-frac 12wz+w + frac frac 12wz-w$$ where $w=e^frac2pi i3$.
– Cookie
Jul 18 at 22:30
The roots of $z^2+z+1$ are not $w$ and $-w$, are $w$ and $overlinew=w^2=1/w$.
– user577471
Jul 19 at 14:33
The roots of $z^2+z+1$ are not $w$ and $-w$, are $w$ and $overlinew=w^2=1/w$.
– user577471
Jul 19 at 14:33
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Since $|z|>1$ we can consider $|w|=left|dfrac1zright|<1$, then
beginalign
dfrac1z^2+z+1
&= dfracw^21+w+w^2 \
&= dfracw^2(1-w)1-w^3 \
&= (w^2-w^3)(1+w^3+w^6+cdots) \
&= dfrac1z^2-dfrac1z^3+dfrac1z^5-dfrac1z^6+dfrac1z^8-dfrac1z^9+cdots
endalign
edited Jul 19 at 1:50
Cookie
8,626123376
answered Jul 18 at 22:35
Nosrati
19.6k41544
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Since $|z|>1$ we can consider $|w|=left|dfrac1zright|<1$, then
beginalign
dfrac1z^2+z+1
&= dfracw^21+w+w^2 \
&= dfracw^2(1-w)1-w^3 \
&= (w^2-w^3)(1+w^3+w^6+cdots) \
&= dfrac1z^2-dfrac1z^3+dfrac1z^5-dfrac1z^6+dfrac1z^8-dfrac1z^9+cdots
endalign
edited Jul 19 at 1:50
Cookie
8,626123376
answered Jul 18 at 22:35
Nosrati
19.6k41544
add a comment |Â
up vote
3
down vote
accepted
Since $|z|>1$ we can consider $|w|=left|dfrac1zright|<1$, then
beginalign
dfrac1z^2+z+1
&= dfracw^21+w+w^2 \
&= dfracw^2(1-w)1-w^3 \
&= (w^2-w^3)(1+w^3+w^6+cdots) \
&= dfrac1z^2-dfrac1z^3+dfrac1z^5-dfrac1z^6+dfrac1z^8-dfrac1z^9+cdots
endalign
edited Jul 19 at 1:50
Cookie
8,626123376
answered Jul 18 at 22:35
Nosrati
19.6k41544
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Since $|z|>1$ we can consider $|w|=left|dfrac1zright|<1$, then
beginalign
dfrac1z^2+z+1
&= dfracw^21+w+w^2 \
&= dfracw^2(1-w)1-w^3 \
&= (w^2-w^3)(1+w^3+w^6+cdots) \
&= dfrac1z^2-dfrac1z^3+dfrac1z^5-dfrac1z^6+dfrac1z^8-dfrac1z^9+cdots
endalign
edited Jul 19 at 1:50
Cookie
8,626123376
answered Jul 18 at 22:35
Nosrati
19.6k41544
Since $|z|>1$ we can consider $|w|=left|dfrac1zright|<1$, then
beginalign
dfrac1z^2+z+1
&= dfracw^21+w+w^2 \
&= dfracw^2(1-w)1-w^3 \
&= (w^2-w^3)(1+w^3+w^6+cdots) \
&= dfrac1z^2-dfrac1z^3+dfrac1z^5-dfrac1z^6+dfrac1z^8-dfrac1z^9+cdots
endalign
edited Jul 19 at 1:50
Cookie
8,626123376
answered Jul 18 at 22:35
Nosrati
19.6k41544
edited Jul 19 at 1:50
Cookie
8,626123376
edited Jul 19 at 1:50
Cookie
8,626123376
edited Jul 19 at 1:50
Cookie
8,626123376
8,626123376
answered Jul 18 at 22:35
Nosrati
19.6k41544
answered Jul 18 at 22:35
Nosrati
19.6k41544
answered Jul 18 at 22:35
Nosrati
19.6k41544
19.6k41544
add a comment |Â
add a comment |Â
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1
$frac1z^2+z+1=frac1/(w-w^2)z-w+frac1/(w^2-w)z-w^2=frac1zfrac1/(w-w^2)1-w/z+frac1zfrac1/(w^2-w)1-w^2/z$, where $w=e^2pi i/3$. Now expand $frac11-w/z=sum_n=0^inftyleft(fracwzright)^n$ and likewise the other term.
– user577471
Jul 18 at 19:13
How did you get the first equality? Did you factor $z^2+z+1$ into imaginary factors and then used partial fractions?
– Cookie
Jul 18 at 20:35
Yes, that is how you do many computations with rational functions, in particular Laurent expansions.
– user577471
Jul 18 at 20:46
I'm still not sure how you got the RHS of your first equality; doing the distinct factor case of the method of partial fractions, I got $$frac 1(z+w)(z-w) = frac-frac 12wz+w + frac frac 12wz-w$$ where $w=e^frac2pi i3$.
– Cookie
Jul 18 at 22:30
The roots of $z^2+z+1$ are not $w$ and $-w$, are $w$ and $overlinew=w^2=1/w$.
– user577471
Jul 19 at 14:33