How do you expand a taylor series about a complex number?
0
$begingroup$
Normally a Taylor series is constructed along real numbers. However, for practical purposes mathematics often heralds that commonly known continuous functions in the real plane are equivalent to their own Taylor series in the complex plane as well.
Suppose I want to construct a Taylor series for $f(x)=sqrt{x}$ (the real variable), but, I want to expand about the complex number, say, $(1+i)^{2}$. Is the process for constructing the series about that complex number any different than constructing a Taylor series about a regular real number? Otherwise, what do I need to do differently?
real-analysis complex-analysis taylor-expansion
asked Dec 8 '18 at 22:10
user608672user608672
64
$endgroup$
|
show 1 more comment
0
$begingroup$
Normally a Taylor series is constructed along real numbers. However, for practical purposes mathematics often heralds that commonly known continuous functions in the real plane are equivalent to their own Taylor series in the complex plane as well.
Suppose I want to construct a Taylor series for $f(x)=sqrt{x}$ (the real variable), but, I want to expand about the complex number, say, $(1+i)^{2}$. Is the process for constructing the series about that complex number any different than constructing a Taylor series about a regular real number? Otherwise, what do I need to do differently?
real-analysis complex-analysis taylor-expansion
asked Dec 8 '18 at 22:10
user608672user608672
64
$endgroup$
$begingroup$
There's nothing new. You do it like for the reals.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:12
$begingroup$
Okay, and how is that justified? How do I know that's true? Is there a domain condition I need to state? Do I need to actually start with $z$ and then take the limit as the $iy$ component goes to zero? I feel like complex analysis would have been discovered a much longer time ago if it was that straightforward.
$endgroup$
– user608672
Dec 8 '18 at 22:13
$begingroup$
It's worth noting that if $f$ is not holomorphic then the Laurent series is necessary to converge at isolated singularities. However, if $f$ is holomorphic then they are the same.
$endgroup$
– Dando18
Dec 8 '18 at 22:15
$begingroup$
I can't can't explain that in a comment. Read a Complex Analysis textbook.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:16
$begingroup$
I vaguely remember evaluating path integrals to be zero in the complex plane and then evaluating singularities to be anything but, though it's been a while and I don't remember most of it. So if "f" is holomorphic, then I don't need to worry about that Laurent series? There's something missing here, because if I assume "x" as the input value, that's only a real variable, it seems like I should be starting with z and then proving something for just x. I think the premise of what I proposed is wrong because if I expand about $1+i$ then I'm already assuming a function of a complex variable.
$endgroup$
– user608672
Dec 8 '18 at 22:19
|
show 1 more comment
0
0
0
$begingroup$
Normally a Taylor series is constructed along real numbers. However, for practical purposes mathematics often heralds that commonly known continuous functions in the real plane are equivalent to their own Taylor series in the complex plane as well.
Suppose I want to construct a Taylor series for $f(x)=sqrt{x}$ (the real variable), but, I want to expand about the complex number, say, $(1+i)^{2}$. Is the process for constructing the series about that complex number any different than constructing a Taylor series about a regular real number? Otherwise, what do I need to do differently?
real-analysis complex-analysis taylor-expansion
asked Dec 8 '18 at 22:10
user608672user608672
64
$endgroup$
Normally a Taylor series is constructed along real numbers. However, for practical purposes mathematics often heralds that commonly known continuous functions in the real plane are equivalent to their own Taylor series in the complex plane as well.
Suppose I want to construct a Taylor series for $f(x)=sqrt{x}$ (the real variable), but, I want to expand about the complex number, say, $(1+i)^{2}$. Is the process for constructing the series about that complex number any different than constructing a Taylor series about a regular real number? Otherwise, what do I need to do differently?
real-analysis complex-analysis taylor-expansion
real-analysis complex-analysis taylor-expansion
asked Dec 8 '18 at 22:10
user608672user608672
64
asked Dec 8 '18 at 22:10
user608672user608672
64
edited Dec 8 '18 at 22:12
user608672
asked Dec 8 '18 at 22:10
user608672user608672
64
asked Dec 8 '18 at 22:10
user608672user608672
64
asked Dec 8 '18 at 22:10
user608672user608672
64
64
$begingroup$
There's nothing new. You do it like for the reals.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:12
$begingroup$
Okay, and how is that justified? How do I know that's true? Is there a domain condition I need to state? Do I need to actually start with $z$ and then take the limit as the $iy$ component goes to zero? I feel like complex analysis would have been discovered a much longer time ago if it was that straightforward.
$endgroup$
– user608672
Dec 8 '18 at 22:13
$begingroup$
It's worth noting that if $f$ is not holomorphic then the Laurent series is necessary to converge at isolated singularities. However, if $f$ is holomorphic then they are the same.
$endgroup$
– Dando18
Dec 8 '18 at 22:15
$begingroup$
I can't can't explain that in a comment. Read a Complex Analysis textbook.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:16
$begingroup$
I vaguely remember evaluating path integrals to be zero in the complex plane and then evaluating singularities to be anything but, though it's been a while and I don't remember most of it. So if "f" is holomorphic, then I don't need to worry about that Laurent series? There's something missing here, because if I assume "x" as the input value, that's only a real variable, it seems like I should be starting with z and then proving something for just x. I think the premise of what I proposed is wrong because if I expand about $1+i$ then I'm already assuming a function of a complex variable.
$endgroup$
– user608672
Dec 8 '18 at 22:19
|
show 1 more comment
$begingroup$
There's nothing new. You do it like for the reals.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:12
$begingroup$
Okay, and how is that justified? How do I know that's true? Is there a domain condition I need to state? Do I need to actually start with $z$ and then take the limit as the $iy$ component goes to zero? I feel like complex analysis would have been discovered a much longer time ago if it was that straightforward.
$endgroup$
– user608672
Dec 8 '18 at 22:13
$begingroup$
It's worth noting that if $f$ is not holomorphic then the Laurent series is necessary to converge at isolated singularities. However, if $f$ is holomorphic then they are the same.
$endgroup$
– Dando18
Dec 8 '18 at 22:15
$begingroup$
I can't can't explain that in a comment. Read a Complex Analysis textbook.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:16
$begingroup$
I vaguely remember evaluating path integrals to be zero in the complex plane and then evaluating singularities to be anything but, though it's been a while and I don't remember most of it. So if "f" is holomorphic, then I don't need to worry about that Laurent series? There's something missing here, because if I assume "x" as the input value, that's only a real variable, it seems like I should be starting with z and then proving something for just x. I think the premise of what I proposed is wrong because if I expand about $1+i$ then I'm already assuming a function of a complex variable.
$endgroup$
– user608672
Dec 8 '18 at 22:19
$begingroup$
There's nothing new. You do it like for the reals.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:12
$begingroup$
There's nothing new. You do it like for the reals.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:12
$begingroup$
Okay, and how is that justified? How do I know that's true? Is there a domain condition I need to state? Do I need to actually start with $z$ and then take the limit as the $iy$ component goes to zero? I feel like complex analysis would have been discovered a much longer time ago if it was that straightforward.
$endgroup$
– user608672
Dec 8 '18 at 22:13
$begingroup$
Okay, and how is that justified? How do I know that's true? Is there a domain condition I need to state? Do I need to actually start with $z$ and then take the limit as the $iy$ component goes to zero? I feel like complex analysis would have been discovered a much longer time ago if it was that straightforward.
$endgroup$
– user608672
Dec 8 '18 at 22:13
$begingroup$
It's worth noting that if $f$ is not holomorphic then the Laurent series is necessary to converge at isolated singularities. However, if $f$ is holomorphic then they are the same.
$endgroup$
– Dando18
Dec 8 '18 at 22:15
$begingroup$
It's worth noting that if $f$ is not holomorphic then the Laurent series is necessary to converge at isolated singularities. However, if $f$ is holomorphic then they are the same.
$endgroup$
– Dando18
Dec 8 '18 at 22:15
$begingroup$
I can't can't explain that in a comment. Read a Complex Analysis textbook.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:16
$begingroup$
I can't can't explain that in a comment. Read a Complex Analysis textbook.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 22:16
$begingroup$
I vaguely remember evaluating path integrals to be zero in the complex plane and then evaluating singularities to be anything but, though it's been a while and I don't remember most of it. So if "f" is holomorphic, then I don't need to worry about that Laurent series? There's something missing here, because if I assume "x" as the input value, that's only a real variable, it seems like I should be starting with z and then proving something for just x. I think the premise of what I proposed is wrong because if I expand about $1+i$ then I'm already assuming a function of a complex variable.
$endgroup$
– user608672
Dec 8 '18 at 22:19
$begingroup$
I vaguely remember evaluating path integrals to be zero in the complex plane and then evaluating singularities to be anything but, though it's been a while and I don't remember most of it. So if "f" is holomorphic, then I don't need to worry about that Laurent series? There's something missing here, because if I assume "x" as the input value, that's only a real variable, it seems like I should be starting with z and then proving something for just x. I think the premise of what I proposed is wrong because if I expand about $1+i$ then I'm already assuming a function of a complex variable.
$endgroup$
– user608672
Dec 8 '18 at 22:19
|
show 1 more comment
1 Answer
1
active
oldest
votes
1
$begingroup$
The term you are looking for is analytic function. The procedure is no different for the complex case.
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
$endgroup$
$begingroup$
I understand the convention that if a function is infinitely differentiable then it's equivalent to its own Taylor series in the complex plane. That article says "There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others." I want to know the ways they are different so I can be confident in what I'm doing.
$endgroup$
– user608672
Dec 8 '18 at 22:25
1
$begingroup$
It's good to see though that different people are on the same page about it being the same process. However, what if I had a function that I could only define as a Taylor series, like if it was a solution to a differential equation. What would need to be proven about it? Would I have to prove it's holomorphicity? Or is the fact that it is already a series for real variables good enough?
$endgroup$
– user608672
Dec 8 '18 at 22:26
$begingroup$
Look at the example about $1/(1+x^2)$ below. That is a real analytic function that is not constant, and yet bounded on the whole real line. On the complex plane, every bounded analytic function is constant. But I wouldn't say this is a particularly interesting statement since it is "unnatural" to consider the function $1/(1+x^2)$ only on the real line. All analytic functions on the real line "naturally" want to be extended to the complex plane (via a concept known as analytic continuation). There are no differences in what you can do really.
$endgroup$
– zoidberg
Dec 8 '18 at 22:30
$begingroup$
What do you mean what would need to be proven about it? For instance, if you were trying to show that your function $f(z)$ on the complex plane has a power series everywhere, then showing that it's true only on the real line would not be sufficient. The function $f(z) = 1/(1+z^2)$ is analytic everywhere on the real line, but not the entire complex plane (it has poles at $pm i$).
$endgroup$
– zoidberg
Dec 8 '18 at 22:35
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
1
$begingroup$
The term you are looking for is analytic function. The procedure is no different for the complex case.
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
$endgroup$
$begingroup$
I understand the convention that if a function is infinitely differentiable then it's equivalent to its own Taylor series in the complex plane. That article says "There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others." I want to know the ways they are different so I can be confident in what I'm doing.
$endgroup$
– user608672
Dec 8 '18 at 22:25
1
$begingroup$
It's good to see though that different people are on the same page about it being the same process. However, what if I had a function that I could only define as a Taylor series, like if it was a solution to a differential equation. What would need to be proven about it? Would I have to prove it's holomorphicity? Or is the fact that it is already a series for real variables good enough?
$endgroup$
– user608672
Dec 8 '18 at 22:26
$begingroup$
Look at the example about $1/(1+x^2)$ below. That is a real analytic function that is not constant, and yet bounded on the whole real line. On the complex plane, every bounded analytic function is constant. But I wouldn't say this is a particularly interesting statement since it is "unnatural" to consider the function $1/(1+x^2)$ only on the real line. All analytic functions on the real line "naturally" want to be extended to the complex plane (via a concept known as analytic continuation). There are no differences in what you can do really.
$endgroup$
– zoidberg
Dec 8 '18 at 22:30
$begingroup$
What do you mean what would need to be proven about it? For instance, if you were trying to show that your function $f(z)$ on the complex plane has a power series everywhere, then showing that it's true only on the real line would not be sufficient. The function $f(z) = 1/(1+z^2)$ is analytic everywhere on the real line, but not the entire complex plane (it has poles at $pm i$).
$endgroup$
– zoidberg
Dec 8 '18 at 22:35
add a comment |
1
$begingroup$
The term you are looking for is analytic function. The procedure is no different for the complex case.
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
$endgroup$
$begingroup$
I understand the convention that if a function is infinitely differentiable then it's equivalent to its own Taylor series in the complex plane. That article says "There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others." I want to know the ways they are different so I can be confident in what I'm doing.
$endgroup$
– user608672
Dec 8 '18 at 22:25
1
$begingroup$
It's good to see though that different people are on the same page about it being the same process. However, what if I had a function that I could only define as a Taylor series, like if it was a solution to a differential equation. What would need to be proven about it? Would I have to prove it's holomorphicity? Or is the fact that it is already a series for real variables good enough?
$endgroup$
– user608672
Dec 8 '18 at 22:26
$begingroup$
Look at the example about $1/(1+x^2)$ below. That is a real analytic function that is not constant, and yet bounded on the whole real line. On the complex plane, every bounded analytic function is constant. But I wouldn't say this is a particularly interesting statement since it is "unnatural" to consider the function $1/(1+x^2)$ only on the real line. All analytic functions on the real line "naturally" want to be extended to the complex plane (via a concept known as analytic continuation). There are no differences in what you can do really.
$endgroup$
– zoidberg
Dec 8 '18 at 22:30
$begingroup$
What do you mean what would need to be proven about it? For instance, if you were trying to show that your function $f(z)$ on the complex plane has a power series everywhere, then showing that it's true only on the real line would not be sufficient. The function $f(z) = 1/(1+z^2)$ is analytic everywhere on the real line, but not the entire complex plane (it has poles at $pm i$).
$endgroup$
– zoidberg
Dec 8 '18 at 22:35
add a comment |
1
1
1
$begingroup$
The term you are looking for is analytic function. The procedure is no different for the complex case.
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
$endgroup$
The term you are looking for is analytic function. The procedure is no different for the complex case.
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
edited Dec 8 '18 at 22:25
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
answered Dec 8 '18 at 22:23
zoidbergzoidberg
1,070113
1,070113
$begingroup$
I understand the convention that if a function is infinitely differentiable then it's equivalent to its own Taylor series in the complex plane. That article says "There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others." I want to know the ways they are different so I can be confident in what I'm doing.
$endgroup$
– user608672
Dec 8 '18 at 22:25
1
$begingroup$
It's good to see though that different people are on the same page about it being the same process. However, what if I had a function that I could only define as a Taylor series, like if it was a solution to a differential equation. What would need to be proven about it? Would I have to prove it's holomorphicity? Or is the fact that it is already a series for real variables good enough?
$endgroup$
– user608672
Dec 8 '18 at 22:26
$begingroup$
Look at the example about $1/(1+x^2)$ below. That is a real analytic function that is not constant, and yet bounded on the whole real line. On the complex plane, every bounded analytic function is constant. But I wouldn't say this is a particularly interesting statement since it is "unnatural" to consider the function $1/(1+x^2)$ only on the real line. All analytic functions on the real line "naturally" want to be extended to the complex plane (via a concept known as analytic continuation). There are no differences in what you can do really.
$endgroup$
– zoidberg
Dec 8 '18 at 22:30
$begingroup$
What do you mean what would need to be proven about it? For instance, if you were trying to show that your function $f(z)$ on the complex plane has a power series everywhere, then showing that it's true only on the real line would not be sufficient. The function $f(z) = 1/(1+z^2)$ is analytic everywhere on the real line, but not the entire complex plane (it has poles at $pm i$).
$endgroup$
– zoidberg
Dec 8 '18 at 22:35
add a comment |
$begingroup$
I understand the convention that if a function is infinitely differentiable then it's equivalent to its own Taylor series in the complex plane. That article says "There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others." I want to know the ways they are different so I can be confident in what I'm doing.
$endgroup$
– user608672
Dec 8 '18 at 22:25
1
$begingroup$
It's good to see though that different people are on the same page about it being the same process. However, what if I had a function that I could only define as a Taylor series, like if it was a solution to a differential equation. What would need to be proven about it? Would I have to prove it's holomorphicity? Or is the fact that it is already a series for real variables good enough?
$endgroup$
– user608672
Dec 8 '18 at 22:26
$begingroup$
Look at the example about $1/(1+x^2)$ below. That is a real analytic function that is not constant, and yet bounded on the whole real line. On the complex plane, every bounded analytic function is constant. But I wouldn't say this is a particularly interesting statement since it is "unnatural" to consider the function $1/(1+x^2)$ only on the real line. All analytic functions on the real line "naturally" want to be extended to the complex plane (via a concept known as analytic continuation). There are no differences in what you can do really.
$endgroup$
– zoidberg
Dec 8 '18 at 22:30
$begingroup$
What do you mean what would need to be proven about it? For instance, if you were trying to show that your function $f(z)$ on the complex plane has a power series everywhere, then showing that it's true only on the real line would not be sufficient. The function $f(z) = 1/(1+z^2)$ is analytic everywhere on the real line, but not the entire complex plane (it has poles at $pm i$).
$endgroup$
– zoidberg
Dec 8 '18 at 22:35
$begingroup$
I understand the convention that if a function is infinitely differentiable then it's equivalent to its own Taylor series in the complex plane. That article says "There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others." I want to know the ways they are different so I can be confident in what I'm doing.
$endgroup$
– user608672
Dec 8 '18 at 22:25
$begingroup$
I understand the convention that if a function is infinitely differentiable then it's equivalent to its own Taylor series in the complex plane. That article says "There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others." I want to know the ways they are different so I can be confident in what I'm doing.
$endgroup$
– user608672
Dec 8 '18 at 22:25
1
1
$begingroup$
It's good to see though that different people are on the same page about it being the same process. However, what if I had a function that I could only define as a Taylor series, like if it was a solution to a differential equation. What would need to be proven about it? Would I have to prove it's holomorphicity? Or is the fact that it is already a series for real variables good enough?
$endgroup$
– user608672
Dec 8 '18 at 22:26
$begingroup$
It's good to see though that different people are on the same page about it being the same process. However, what if I had a function that I could only define as a Taylor series, like if it was a solution to a differential equation. What would need to be proven about it? Would I have to prove it's holomorphicity? Or is the fact that it is already a series for real variables good enough?
$endgroup$
– user608672
Dec 8 '18 at 22:26
$begingroup$
Look at the example about $1/(1+x^2)$ below. That is a real analytic function that is not constant, and yet bounded on the whole real line. On the complex plane, every bounded analytic function is constant. But I wouldn't say this is a particularly interesting statement since it is "unnatural" to consider the function $1/(1+x^2)$ only on the real line. All analytic functions on the real line "naturally" want to be extended to the complex plane (via a concept known as analytic continuation). There are no differences in what you can do really.
$endgroup$
– zoidberg
Dec 8 '18 at 22:30
$begingroup$
Look at the example about $1/(1+x^2)$ below. That is a real analytic function that is not constant, and yet bounded on the whole real line. On the complex plane, every bounded analytic function is constant. But I wouldn't say this is a particularly interesting statement since it is "unnatural" to consider the function $1/(1+x^2)$ only on the real line. All analytic functions on the real line "naturally" want to be extended to the complex plane (via a concept known as analytic continuation). There are no differences in what you can do really.
$endgroup$
– zoidberg
Dec 8 '18 at 22:30
$begingroup$
What do you mean what would need to be proven about it? For instance, if you were trying to show that your function $f(z)$ on the complex plane has a power series everywhere, then showing that it's true only on the real line would not be sufficient. The function $f(z) = 1/(1+z^2)$ is analytic everywhere on the real line, but not the entire complex plane (it has poles at $pm i$).
$endgroup$
– zoidberg
Dec 8 '18 at 22:35
$begingroup$
What do you mean what would need to be proven about it? For instance, if you were trying to show that your function $f(z)$ on the complex plane has a power series everywhere, then showing that it's true only on the real line would not be sufficient. The function $f(z) = 1/(1+z^2)$ is analytic everywhere on the real line, but not the entire complex plane (it has poles at $pm i$).
$endgroup$
– zoidberg
Dec 8 '18 at 22:35
add a comment |
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There's nothing new. You do it like for the reals.
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– José Carlos Santos
Dec 8 '18 at 22:12
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Okay, and how is that justified? How do I know that's true? Is there a domain condition I need to state? Do I need to actually start with $z$ and then take the limit as the $iy$ component goes to zero? I feel like complex analysis would have been discovered a much longer time ago if it was that straightforward.
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– user608672
Dec 8 '18 at 22:13
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It's worth noting that if $f$ is not holomorphic then the Laurent series is necessary to converge at isolated singularities. However, if $f$ is holomorphic then they are the same.
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– Dando18
Dec 8 '18 at 22:15
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I can't can't explain that in a comment. Read a Complex Analysis textbook.
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– José Carlos Santos
Dec 8 '18 at 22:16
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I vaguely remember evaluating path integrals to be zero in the complex plane and then evaluating singularities to be anything but, though it's been a while and I don't remember most of it. So if "f" is holomorphic, then I don't need to worry about that Laurent series? There's something missing here, because if I assume "x" as the input value, that's only a real variable, it seems like I should be starting with z and then proving something for just x. I think the premise of what I proposed is wrong because if I expand about $1+i$ then I'm already assuming a function of a complex variable.
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– user608672
Dec 8 '18 at 22:19