All values of constant $a$ such that $f(z) = frac{z^2 - z +1}{z(z-1)^2}+frac{a}{sin z}$ has an antiderivative...
$begingroup$
I have to find all the possible $a in mathbb{C}$ for which the following function has an antiderivative in the domain $U = {z : |z|<2} setminus {0,1}$
$$f(z) = frac{z^2 - z +1}{z(z-1)^2}+frac{a}{sin z} = frac{1}{z}+frac{1}{(z-1)^2}+frac{a}{sin z}$$
Now, it's clear that in the domain $U, (z-1)^{-2} $ will have an antiderivative. So we're only bothered about $g(z) = f(z) - (z-1)^{-2}$
We would want the following
$$int_{gamma}g(z)dz = 0$$
where $gamma$ is a closed path in $V = {z : |z|<2}$ with $0, 1$ in it's interior; and by Residue Theorem we get
$$2 pi i + a cdot 2 pi i cdot 1=0 implies a = -1$$
Is this reasoning correct?
Now, I have the following question: What does the antiderivate of $g$ look like? Since we have the problematic $1/z$ term. Are we talking about something of the form
$$G(z) =int_{z_0}^{z} left(frac{1}{z}-frac{1}{sin z}right) dz$$
complex-analysis
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
$endgroup$
add a comment |
$begingroup$
I have to find all the possible $a in mathbb{C}$ for which the following function has an antiderivative in the domain $U = {z : |z|<2} setminus {0,1}$
$$f(z) = frac{z^2 - z +1}{z(z-1)^2}+frac{a}{sin z} = frac{1}{z}+frac{1}{(z-1)^2}+frac{a}{sin z}$$
Now, it's clear that in the domain $U, (z-1)^{-2} $ will have an antiderivative. So we're only bothered about $g(z) = f(z) - (z-1)^{-2}$
We would want the following
$$int_{gamma}g(z)dz = 0$$
where $gamma$ is a closed path in $V = {z : |z|<2}$ with $0, 1$ in it's interior; and by Residue Theorem we get
$$2 pi i + a cdot 2 pi i cdot 1=0 implies a = -1$$
Is this reasoning correct?
Now, I have the following question: What does the antiderivate of $g$ look like? Since we have the problematic $1/z$ term. Are we talking about something of the form
$$G(z) =int_{z_0}^{z} left(frac{1}{z}-frac{1}{sin z}right) dz$$
complex-analysis
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
$endgroup$
add a comment |
$begingroup$
I have to find all the possible $a in mathbb{C}$ for which the following function has an antiderivative in the domain $U = {z : |z|<2} setminus {0,1}$
$$f(z) = frac{z^2 - z +1}{z(z-1)^2}+frac{a}{sin z} = frac{1}{z}+frac{1}{(z-1)^2}+frac{a}{sin z}$$
Now, it's clear that in the domain $U, (z-1)^{-2} $ will have an antiderivative. So we're only bothered about $g(z) = f(z) - (z-1)^{-2}$
We would want the following
$$int_{gamma}g(z)dz = 0$$
where $gamma$ is a closed path in $V = {z : |z|<2}$ with $0, 1$ in it's interior; and by Residue Theorem we get
$$2 pi i + a cdot 2 pi i cdot 1=0 implies a = -1$$
Is this reasoning correct?
Now, I have the following question: What does the antiderivate of $g$ look like? Since we have the problematic $1/z$ term. Are we talking about something of the form
$$G(z) =int_{z_0}^{z} left(frac{1}{z}-frac{1}{sin z}right) dz$$
complex-analysis
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
$endgroup$
I have to find all the possible $a in mathbb{C}$ for which the following function has an antiderivative in the domain $U = {z : |z|<2} setminus {0,1}$
$$f(z) = frac{z^2 - z +1}{z(z-1)^2}+frac{a}{sin z} = frac{1}{z}+frac{1}{(z-1)^2}+frac{a}{sin z}$$
Now, it's clear that in the domain $U, (z-1)^{-2} $ will have an antiderivative. So we're only bothered about $g(z) = f(z) - (z-1)^{-2}$
We would want the following
$$int_{gamma}g(z)dz = 0$$
where $gamma$ is a closed path in $V = {z : |z|<2}$ with $0, 1$ in it's interior; and by Residue Theorem we get
$$2 pi i + a cdot 2 pi i cdot 1=0 implies a = -1$$
Is this reasoning correct?
Now, I have the following question: What does the antiderivate of $g$ look like? Since we have the problematic $1/z$ term. Are we talking about something of the form
$$G(z) =int_{z_0}^{z} left(frac{1}{z}-frac{1}{sin z}right) dz$$
complex-analysis
complex-analysis
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
asked Dec 18 '18 at 18:52
Naweed G. SeldonNaweed G. Seldon
1,314419
1,314419
add a comment |
add a comment |
1 Answer
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Yes. To be more precise, define$$begin{array}{rccc}varphicolon&D(0,2)setminus{0}&longrightarrow&mathbb C\&z&mapsto&begin{cases}dfrac1z-dfrac1{sin z}&text{ if }zneq0\0&text{ otherwise.}end{cases}end{array}$$Then you can define, for each $zin D(0,2)$, $G(z)$ as$$G(z)=int_{[0,z]}varphi(z),mathrm dz,$$where $[0,z]$ is the path$$begin{array}{ccc}[0,1]&longrightarrow&mathbb C\t&mapsto&tzend{array}$$and $G$ will be an antiderivative of your function. Actually, any other path in $D(0,2)$ going from $0$ to $z$ will do.
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes. To be more precise, define$$begin{array}{rccc}varphicolon&D(0,2)setminus{0}&longrightarrow&mathbb C\&z&mapsto&begin{cases}dfrac1z-dfrac1{sin z}&text{ if }zneq0\0&text{ otherwise.}end{cases}end{array}$$Then you can define, for each $zin D(0,2)$, $G(z)$ as$$G(z)=int_{[0,z]}varphi(z),mathrm dz,$$where $[0,z]$ is the path$$begin{array}{ccc}[0,1]&longrightarrow&mathbb C\t&mapsto&tzend{array}$$and $G$ will be an antiderivative of your function. Actually, any other path in $D(0,2)$ going from $0$ to $z$ will do.
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Yes. To be more precise, define$$begin{array}{rccc}varphicolon&D(0,2)setminus{0}&longrightarrow&mathbb C\&z&mapsto&begin{cases}dfrac1z-dfrac1{sin z}&text{ if }zneq0\0&text{ otherwise.}end{cases}end{array}$$Then you can define, for each $zin D(0,2)$, $G(z)$ as$$G(z)=int_{[0,z]}varphi(z),mathrm dz,$$where $[0,z]$ is the path$$begin{array}{ccc}[0,1]&longrightarrow&mathbb C\t&mapsto&tzend{array}$$and $G$ will be an antiderivative of your function. Actually, any other path in $D(0,2)$ going from $0$ to $z$ will do.
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Yes. To be more precise, define$$begin{array}{rccc}varphicolon&D(0,2)setminus{0}&longrightarrow&mathbb C\&z&mapsto&begin{cases}dfrac1z-dfrac1{sin z}&text{ if }zneq0\0&text{ otherwise.}end{cases}end{array}$$Then you can define, for each $zin D(0,2)$, $G(z)$ as$$G(z)=int_{[0,z]}varphi(z),mathrm dz,$$where $[0,z]$ is the path$$begin{array}{ccc}[0,1]&longrightarrow&mathbb C\t&mapsto&tzend{array}$$and $G$ will be an antiderivative of your function. Actually, any other path in $D(0,2)$ going from $0$ to $z$ will do.
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Yes. To be more precise, define$$begin{array}{rccc}varphicolon&D(0,2)setminus{0}&longrightarrow&mathbb C\&z&mapsto&begin{cases}dfrac1z-dfrac1{sin z}&text{ if }zneq0\0&text{ otherwise.}end{cases}end{array}$$Then you can define, for each $zin D(0,2)$, $G(z)$ as$$G(z)=int_{[0,z]}varphi(z),mathrm dz,$$where $[0,z]$ is the path$$begin{array}{ccc}[0,1]&longrightarrow&mathbb C\t&mapsto&tzend{array}$$and $G$ will be an antiderivative of your function. Actually, any other path in $D(0,2)$ going from $0$ to $z$ will do.
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
answered Dec 18 '18 at 19:02
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
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