Find the function $v(x,y)$ that will make $f(z) = e^xsin(y) + iv(x,y)$ analytic, and such that $f(0)=0$
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Here's what I have done so far:
$$frac{partial u}{partial x} = e^xsin(y)=frac{partial v}{partial y}$$
$$frac{partial u}{partial y} = e^xcos(y)=-frac{partial v}{partial x}=-e^xcos(y)$$
So to find $v(x,y)$ do I integrate for $frac{partial v}{partial y}$ or $frac{partial v}{partial x}$ or both to find what the constant is?. I think I'm completely on the wrong track here. Or do I try to solve using harmonic functions.
complex-analysis
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
asked Nov 15 at 3:10
jaiidong
104
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up vote
0
down vote
favorite
Here's what I have done so far:
$$frac{partial u}{partial x} = e^xsin(y)=frac{partial v}{partial y}$$
$$frac{partial u}{partial y} = e^xcos(y)=-frac{partial v}{partial x}=-e^xcos(y)$$
So to find $v(x,y)$ do I integrate for $frac{partial v}{partial y}$ or $frac{partial v}{partial x}$ or both to find what the constant is?. I think I'm completely on the wrong track here. Or do I try to solve using harmonic functions.
complex-analysis
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
asked Nov 15 at 3:10
jaiidong
104
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here's what I have done so far:
$$frac{partial u}{partial x} = e^xsin(y)=frac{partial v}{partial y}$$
$$frac{partial u}{partial y} = e^xcos(y)=-frac{partial v}{partial x}=-e^xcos(y)$$
So to find $v(x,y)$ do I integrate for $frac{partial v}{partial y}$ or $frac{partial v}{partial x}$ or both to find what the constant is?. I think I'm completely on the wrong track here. Or do I try to solve using harmonic functions.
complex-analysis
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
asked Nov 15 at 3:10
jaiidong
104
Here's what I have done so far:
$$frac{partial u}{partial x} = e^xsin(y)=frac{partial v}{partial y}$$
$$frac{partial u}{partial y} = e^xcos(y)=-frac{partial v}{partial x}=-e^xcos(y)$$
So to find $v(x,y)$ do I integrate for $frac{partial v}{partial y}$ or $frac{partial v}{partial x}$ or both to find what the constant is?. I think I'm completely on the wrong track here. Or do I try to solve using harmonic functions.
complex-analysis
complex-analysis
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
asked Nov 15 at 3:10
jaiidong
104
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
asked Nov 15 at 3:10
jaiidong
104
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
edited Nov 15 at 3:19
Chinnapparaj R
4,6081725
4,6081725
asked Nov 15 at 3:10
jaiidong
104
asked Nov 15 at 3:10
jaiidong
104
asked Nov 15 at 3:10
jaiidong
104
104
add a comment |
add a comment |
1 Answer
1
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0
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From the first equation, you have
$$partial_y v = e^xsin(y) Rightarrow v = -e^xcos(y) + A(x)$$
From the second equation, you have
$$partial_x v = -e^xcos(y) Rightarrow v = -e^xcos(y) + B(y)$$
In order for these to both be true, we need
$$A(x) = B(y) = C$$
where $C$ is a constant, independent of $x$ and $y$. Thus, we have
$$ v = C-e^xcos(y) $$
and so,
$$ f(x+iy) = e^xsin(y) + i(C-e^xcos(y))$$
Substituting $z=0$ gives us
$$ f(0) = i(C-1) = 0 Rightarrow C=1$$
and thus, the function must be
$$ f(x+iy) = e^xsin(y)+i(1-e^xcos(y))$$
with $$v(x,y) = 1-e^xcos(y)$$
answered Nov 15 at 3:28
AlexanderJ93
5,279522
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
From the first equation, you have
$$partial_y v = e^xsin(y) Rightarrow v = -e^xcos(y) + A(x)$$
From the second equation, you have
$$partial_x v = -e^xcos(y) Rightarrow v = -e^xcos(y) + B(y)$$
In order for these to both be true, we need
$$A(x) = B(y) = C$$
where $C$ is a constant, independent of $x$ and $y$. Thus, we have
$$ v = C-e^xcos(y) $$
and so,
$$ f(x+iy) = e^xsin(y) + i(C-e^xcos(y))$$
Substituting $z=0$ gives us
$$ f(0) = i(C-1) = 0 Rightarrow C=1$$
and thus, the function must be
$$ f(x+iy) = e^xsin(y)+i(1-e^xcos(y))$$
with $$v(x,y) = 1-e^xcos(y)$$
answered Nov 15 at 3:28
AlexanderJ93
5,279522
add a comment |
up vote
0
down vote
accepted
From the first equation, you have
$$partial_y v = e^xsin(y) Rightarrow v = -e^xcos(y) + A(x)$$
From the second equation, you have
$$partial_x v = -e^xcos(y) Rightarrow v = -e^xcos(y) + B(y)$$
In order for these to both be true, we need
$$A(x) = B(y) = C$$
where $C$ is a constant, independent of $x$ and $y$. Thus, we have
$$ v = C-e^xcos(y) $$
and so,
$$ f(x+iy) = e^xsin(y) + i(C-e^xcos(y))$$
Substituting $z=0$ gives us
$$ f(0) = i(C-1) = 0 Rightarrow C=1$$
and thus, the function must be
$$ f(x+iy) = e^xsin(y)+i(1-e^xcos(y))$$
with $$v(x,y) = 1-e^xcos(y)$$
answered Nov 15 at 3:28
AlexanderJ93
5,279522
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
From the first equation, you have
$$partial_y v = e^xsin(y) Rightarrow v = -e^xcos(y) + A(x)$$
From the second equation, you have
$$partial_x v = -e^xcos(y) Rightarrow v = -e^xcos(y) + B(y)$$
In order for these to both be true, we need
$$A(x) = B(y) = C$$
where $C$ is a constant, independent of $x$ and $y$. Thus, we have
$$ v = C-e^xcos(y) $$
and so,
$$ f(x+iy) = e^xsin(y) + i(C-e^xcos(y))$$
Substituting $z=0$ gives us
$$ f(0) = i(C-1) = 0 Rightarrow C=1$$
and thus, the function must be
$$ f(x+iy) = e^xsin(y)+i(1-e^xcos(y))$$
with $$v(x,y) = 1-e^xcos(y)$$
answered Nov 15 at 3:28
AlexanderJ93
5,279522
From the first equation, you have
$$partial_y v = e^xsin(y) Rightarrow v = -e^xcos(y) + A(x)$$
From the second equation, you have
$$partial_x v = -e^xcos(y) Rightarrow v = -e^xcos(y) + B(y)$$
In order for these to both be true, we need
$$A(x) = B(y) = C$$
where $C$ is a constant, independent of $x$ and $y$. Thus, we have
$$ v = C-e^xcos(y) $$
and so,
$$ f(x+iy) = e^xsin(y) + i(C-e^xcos(y))$$
Substituting $z=0$ gives us
$$ f(0) = i(C-1) = 0 Rightarrow C=1$$
and thus, the function must be
$$ f(x+iy) = e^xsin(y)+i(1-e^xcos(y))$$
with $$v(x,y) = 1-e^xcos(y)$$
answered Nov 15 at 3:28
AlexanderJ93
5,279522
answered Nov 15 at 3:28
AlexanderJ93
5,279522
answered Nov 15 at 3:28
AlexanderJ93
5,279522
answered Nov 15 at 3:28
AlexanderJ93
5,279522
5,279522
add a comment |
add a comment |
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