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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.
-
(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrite $frac1z+1 = frac12+z-1$?
I did similarly for Exer 8.18, and it seems that that's the point based on Exer 8.32 where I obtained 4 Laurent series (the 4th being convergent on the $emptyset$!)
For (Q1)
Take out $z-1$
$$frac1z+1 = frac12+z-1 = frac1(z-1)(frac2z-1+1) to texta Laurent series for |z-1| > 2$$Take out $2$
$$frac1z+1 = frac12+z-1 = frac1(2)(1+fracz-12) to texta Laurent series for |z-1| < 2$$
-
(Q2) For Exer 8.19, is there only one Laurent series unlike in Exer 8.17,18?
For (Q2)
$$fracz-2z+1 = 1 + frac-3z+1 textonly?$$
(Q3) For Exer 8.36, I came up with 2 Laurent series. Are they both valid?
Rewrite $frac1(z^2-4)(z-2) = frac1(z-2)^2(z+2)$. Rewrite $frac1z+2 = frac1z-2+4$:
- Take out $z-2$
$$frac1z+2 = frac1z-2+4 = frac1(z-2)(frac4z-2+1) to texta Laurent series for |z-2| > 4$$
--> This is not the book's answer, and it doesn't seem to have a $c_-1$. It looks like the integral will be 0.
(Q3.1) What's wrong with this Laurent series? I guess something like $C[2,1] subsetneq $, so it doesn't apply or something.
- Take out $4$
$$frac1z+2 = frac1z-2+4 = frac1(4)(1+fracz-24) to texta Laurent series for |z-2| < 4$$
--> This is the book's answer (apart from the region), and it gives the same answer as with Cauchy Integral Formula 5.1 (and later Residue Theorem 9.10) namely $frac- pi i8$
(Q3.2) Book says that for $frac1(4)(1+fracz-24)$, the region of convergence is $colorred0 < |z-2| < 4$. Why $0 <$?
calculus sequences-and-series complex-analysis convergence laurent-series
asked 9 hours ago
BCLC
6,96421973
add a comment |Â
up vote
1
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.
-
(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrite $frac1z+1 = frac12+z-1$?
I did similarly for Exer 8.18, and it seems that that's the point based on Exer 8.32 where I obtained 4 Laurent series (the 4th being convergent on the $emptyset$!)
For (Q1)
Take out $z-1$
$$frac1z+1 = frac12+z-1 = frac1(z-1)(frac2z-1+1) to texta Laurent series for |z-1| > 2$$Take out $2$
$$frac1z+1 = frac12+z-1 = frac1(2)(1+fracz-12) to texta Laurent series for |z-1| < 2$$
-
(Q2) For Exer 8.19, is there only one Laurent series unlike in Exer 8.17,18?
For (Q2)
$$fracz-2z+1 = 1 + frac-3z+1 textonly?$$
(Q3) For Exer 8.36, I came up with 2 Laurent series. Are they both valid?
Rewrite $frac1(z^2-4)(z-2) = frac1(z-2)^2(z+2)$. Rewrite $frac1z+2 = frac1z-2+4$:
- Take out $z-2$
$$frac1z+2 = frac1z-2+4 = frac1(z-2)(frac4z-2+1) to texta Laurent series for |z-2| > 4$$
--> This is not the book's answer, and it doesn't seem to have a $c_-1$. It looks like the integral will be 0.
(Q3.1) What's wrong with this Laurent series? I guess something like $C[2,1] subsetneq $, so it doesn't apply or something.
- Take out $4$
$$frac1z+2 = frac1z-2+4 = frac1(4)(1+fracz-24) to texta Laurent series for |z-2| < 4$$
--> This is the book's answer (apart from the region), and it gives the same answer as with Cauchy Integral Formula 5.1 (and later Residue Theorem 9.10) namely $frac- pi i8$
(Q3.2) Book says that for $frac1(4)(1+fracz-24)$, the region of convergence is $colorred0 < |z-2| < 4$. Why $0 <$?
calculus sequences-and-series complex-analysis convergence laurent-series
asked 9 hours ago
BCLC
6,96421973
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.
-
(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrite $frac1z+1 = frac12+z-1$?
I did similarly for Exer 8.18, and it seems that that's the point based on Exer 8.32 where I obtained 4 Laurent series (the 4th being convergent on the $emptyset$!)
For (Q1)
Take out $z-1$
$$frac1z+1 = frac12+z-1 = frac1(z-1)(frac2z-1+1) to texta Laurent series for |z-1| > 2$$Take out $2$
$$frac1z+1 = frac12+z-1 = frac1(2)(1+fracz-12) to texta Laurent series for |z-1| < 2$$
-
(Q2) For Exer 8.19, is there only one Laurent series unlike in Exer 8.17,18?
For (Q2)
$$fracz-2z+1 = 1 + frac-3z+1 textonly?$$
(Q3) For Exer 8.36, I came up with 2 Laurent series. Are they both valid?
Rewrite $frac1(z^2-4)(z-2) = frac1(z-2)^2(z+2)$. Rewrite $frac1z+2 = frac1z-2+4$:
- Take out $z-2$
$$frac1z+2 = frac1z-2+4 = frac1(z-2)(frac4z-2+1) to texta Laurent series for |z-2| > 4$$
--> This is not the book's answer, and it doesn't seem to have a $c_-1$. It looks like the integral will be 0.
(Q3.1) What's wrong with this Laurent series? I guess something like $C[2,1] subsetneq $, so it doesn't apply or something.
- Take out $4$
$$frac1z+2 = frac1z-2+4 = frac1(4)(1+fracz-24) to texta Laurent series for |z-2| < 4$$
--> This is the book's answer (apart from the region), and it gives the same answer as with Cauchy Integral Formula 5.1 (and later Residue Theorem 9.10) namely $frac- pi i8$
(Q3.2) Book says that for $frac1(4)(1+fracz-24)$, the region of convergence is $colorred0 < |z-2| < 4$. Why $0 <$?
calculus sequences-and-series complex-analysis convergence laurent-series
asked 9 hours ago
BCLC
6,96421973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.
-
(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrite $frac1z+1 = frac12+z-1$?
I did similarly for Exer 8.18, and it seems that that's the point based on Exer 8.32 where I obtained 4 Laurent series (the 4th being convergent on the $emptyset$!)
For (Q1)
Take out $z-1$
$$frac1z+1 = frac12+z-1 = frac1(z-1)(frac2z-1+1) to texta Laurent series for |z-1| > 2$$Take out $2$
$$frac1z+1 = frac12+z-1 = frac1(2)(1+fracz-12) to texta Laurent series for |z-1| < 2$$
-
(Q2) For Exer 8.19, is there only one Laurent series unlike in Exer 8.17,18?
For (Q2)
$$fracz-2z+1 = 1 + frac-3z+1 textonly?$$
(Q3) For Exer 8.36, I came up with 2 Laurent series. Are they both valid?
Rewrite $frac1(z^2-4)(z-2) = frac1(z-2)^2(z+2)$. Rewrite $frac1z+2 = frac1z-2+4$:
- Take out $z-2$
$$frac1z+2 = frac1z-2+4 = frac1(z-2)(frac4z-2+1) to texta Laurent series for |z-2| > 4$$
--> This is not the book's answer, and it doesn't seem to have a $c_-1$. It looks like the integral will be 0.
(Q3.1) What's wrong with this Laurent series? I guess something like $C[2,1] subsetneq $, so it doesn't apply or something.
- Take out $4$
$$frac1z+2 = frac1z-2+4 = frac1(4)(1+fracz-24) to texta Laurent series for |z-2| < 4$$
--> This is the book's answer (apart from the region), and it gives the same answer as with Cauchy Integral Formula 5.1 (and later Residue Theorem 9.10) namely $frac- pi i8$
(Q3.2) Book says that for $frac1(4)(1+fracz-24)$, the region of convergence is $colorred0 < |z-2| < 4$. Why $0 <$?
calculus sequences-and-series complex-analysis convergence laurent-series
asked 9 hours ago
BCLC
6,96421973
asked 9 hours ago
BCLC
6,96421973
asked 9 hours ago
BCLC
6,96421973
asked 9 hours ago
BCLC
6,96421973
6,96421973
add a comment |Â
add a comment |Â
1 Answer
1
active
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votes
up vote
1
down vote
Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.
Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$.
beginalign*
D_1:&quad 0< |z-1|<2\
D_2:&quad |z-1|>2
endalign*
The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.
The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.
We have in $D_1$ the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1z-1cdotfrac12left(1+fracz-12right)\
&=frac12cdotfrac1z-1sum_j=0^infty(-1)^jleft(fracz-12right)^j\
&=-sum_j=-1^inftyleft(-frac12right)^j(z-1)^j
endalign*
and in $D_2$ we have the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1(z-1)^2cdotfrac11+frac2z-1\
&=frac1(z-1)^2sum_j=0^infty(-1)^jleft(frac2z-1right)^j\
&=frac14sum_j=-infty^-2left(-frac12right)^j(z-1)^j
endalign*
Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.
Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion
beginalign*
frac1(z^2-4)(z-2)&=frac1(z-2)^2cdotfrac1z+2\
&=frac1(z-2)^3cdotfrac11+frac4z-2\
&=frac1(z-2)^3sum_j=0^infty(-4)^jfrac1(z-2)^jtag1\
&=cdots
endalign*
Ad Q.3.2: The function under consideration
beginalign*
frac1(z^2-4)(z-2)
endalign*
has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.
answered 2 hours ago
Markus Scheuer
55.7k449135
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.
Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$.
beginalign*
D_1:&quad 0< |z-1|<2\
D_2:&quad |z-1|>2
endalign*
The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.
The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.
We have in $D_1$ the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1z-1cdotfrac12left(1+fracz-12right)\
&=frac12cdotfrac1z-1sum_j=0^infty(-1)^jleft(fracz-12right)^j\
&=-sum_j=-1^inftyleft(-frac12right)^j(z-1)^j
endalign*
and in $D_2$ we have the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1(z-1)^2cdotfrac11+frac2z-1\
&=frac1(z-1)^2sum_j=0^infty(-1)^jleft(frac2z-1right)^j\
&=frac14sum_j=-infty^-2left(-frac12right)^j(z-1)^j
endalign*
Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.
Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion
beginalign*
frac1(z^2-4)(z-2)&=frac1(z-2)^2cdotfrac1z+2\
&=frac1(z-2)^3cdotfrac11+frac4z-2\
&=frac1(z-2)^3sum_j=0^infty(-4)^jfrac1(z-2)^jtag1\
&=cdots
endalign*
Ad Q.3.2: The function under consideration
beginalign*
frac1(z^2-4)(z-2)
endalign*
has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.
answered 2 hours ago
Markus Scheuer
55.7k449135
add a comment |Â
up vote
1
down vote
Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.
Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$.
beginalign*
D_1:&quad 0< |z-1|<2\
D_2:&quad |z-1|>2
endalign*
The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.
The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.
We have in $D_1$ the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1z-1cdotfrac12left(1+fracz-12right)\
&=frac12cdotfrac1z-1sum_j=0^infty(-1)^jleft(fracz-12right)^j\
&=-sum_j=-1^inftyleft(-frac12right)^j(z-1)^j
endalign*
and in $D_2$ we have the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1(z-1)^2cdotfrac11+frac2z-1\
&=frac1(z-1)^2sum_j=0^infty(-1)^jleft(frac2z-1right)^j\
&=frac14sum_j=-infty^-2left(-frac12right)^j(z-1)^j
endalign*
Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.
Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion
beginalign*
frac1(z^2-4)(z-2)&=frac1(z-2)^2cdotfrac1z+2\
&=frac1(z-2)^3cdotfrac11+frac4z-2\
&=frac1(z-2)^3sum_j=0^infty(-4)^jfrac1(z-2)^jtag1\
&=cdots
endalign*
Ad Q.3.2: The function under consideration
beginalign*
frac1(z^2-4)(z-2)
endalign*
has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.
answered 2 hours ago
Markus Scheuer
55.7k449135
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.
Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$.
beginalign*
D_1:&quad 0< |z-1|<2\
D_2:&quad |z-1|>2
endalign*
The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.
The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.
We have in $D_1$ the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1z-1cdotfrac12left(1+fracz-12right)\
&=frac12cdotfrac1z-1sum_j=0^infty(-1)^jleft(fracz-12right)^j\
&=-sum_j=-1^inftyleft(-frac12right)^j(z-1)^j
endalign*
and in $D_2$ we have the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1(z-1)^2cdotfrac11+frac2z-1\
&=frac1(z-1)^2sum_j=0^infty(-1)^jleft(frac2z-1right)^j\
&=frac14sum_j=-infty^-2left(-frac12right)^j(z-1)^j
endalign*
Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.
Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion
beginalign*
frac1(z^2-4)(z-2)&=frac1(z-2)^2cdotfrac1z+2\
&=frac1(z-2)^3cdotfrac11+frac4z-2\
&=frac1(z-2)^3sum_j=0^infty(-4)^jfrac1(z-2)^jtag1\
&=cdots
endalign*
Ad Q.3.2: The function under consideration
beginalign*
frac1(z^2-4)(z-2)
endalign*
has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.
answered 2 hours ago
Markus Scheuer
55.7k449135
Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.
Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$.
beginalign*
D_1:&quad 0< |z-1|<2\
D_2:&quad |z-1|>2
endalign*
The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.
The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.
We have in $D_1$ the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1z-1cdotfrac12left(1+fracz-12right)\
&=frac12cdotfrac1z-1sum_j=0^infty(-1)^jleft(fracz-12right)^j\
&=-sum_j=-1^inftyleft(-frac12right)^j(z-1)^j
endalign*
and in $D_2$ we have the representation
beginalign*
frac1(z-1)(z+1)&=frac1z-1cdotfrac12+(z-1)\
&=frac1(z-1)^2cdotfrac11+frac2z-1\
&=frac1(z-1)^2sum_j=0^infty(-1)^jleft(frac2z-1right)^j\
&=frac14sum_j=-infty^-2left(-frac12right)^j(z-1)^j
endalign*
Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.
Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion
beginalign*
frac1(z^2-4)(z-2)&=frac1(z-2)^2cdotfrac1z+2\
&=frac1(z-2)^3cdotfrac11+frac4z-2\
&=frac1(z-2)^3sum_j=0^infty(-4)^jfrac1(z-2)^jtag1\
&=cdots
endalign*
Ad Q.3.2: The function under consideration
beginalign*
frac1(z^2-4)(z-2)
endalign*
has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.
answered 2 hours ago
Markus Scheuer
55.7k449135
answered 2 hours ago
Markus Scheuer
55.7k449135
answered 2 hours ago
Markus Scheuer
55.7k449135
answered 2 hours ago
Markus Scheuer
55.7k449135
55.7k449135
add a comment |Â
add a comment |Â
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