Can the functions from reals to complex numbers be treated as functions from complex numbers to complex numbers?Is this out-of-context theorem true?Question regarding assumptions in Morera's theorem and conditions for existence of antiderivative.Showing a complex function defined as an integral is holomorphicGeometric interpretation of complex path integralHolomorphic function definition. Am I missing something very obvious?Help to prove this is an equivalence relationHow can I prove this $int_gammaf=-int_-gammaf$Is this complex analysis proof rigorous?Can we prove that $overlinez^2$ has no antiderivative without using Morera's Theorem?$lim_rtoinftyint_c-ir^c+irz^-lambda(log(z))^-1dz$
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Can the functions from reals to complex numbers be treated as functions from complex numbers to complex numbers?
Is this out-of-context theorem true?Question regarding assumptions in Morera's theorem and conditions for existence of antiderivative.Showing a complex function defined as an integral is holomorphicGeometric interpretation of complex path integralHolomorphic function definition. Am I missing something very obvious?Help to prove this is an equivalence relationHow can I prove this $int_gammaf=-int_-gammaf$Is this complex analysis proof rigorous?Can we prove that $overlinez^2$ has no antiderivative without using Morera's Theorem?$lim_rtoinftyint_c-ir^c+irz^-lambda(log(z))^-1dz$
$begingroup$
I have recently started to learn complex analysis. This question popped into my head when I was reading the path integrals. Since we have to consider the derivative of path parametrization in the definition of path integral, and this parameterization has as its domain a closed real interval.
Writing it out, if the path $gamma(t): [a,b] rightarrow mathbbC$ is given by $gamma(t) = x(t) + i y(t)$, then why do we define its derivative w.r.t. $t$ as $x'(t) + iy'(t)$? Is it a consequence of treating it as a function from $mathbbC text to mathbbC$? I don't think that's true because it means it is precomposed with projection function $p(z) = Re(z)$, which isn't holomorphic. So is it so just by definition?
complex-analysis
asked 2 days ago
studentstudent
497
$endgroup$
add a comment |
$begingroup$
I have recently started to learn complex analysis. This question popped into my head when I was reading the path integrals. Since we have to consider the derivative of path parametrization in the definition of path integral, and this parameterization has as its domain a closed real interval.
Writing it out, if the path $gamma(t): [a,b] rightarrow mathbbC$ is given by $gamma(t) = x(t) + i y(t)$, then why do we define its derivative w.r.t. $t$ as $x'(t) + iy'(t)$? Is it a consequence of treating it as a function from $mathbbC text to mathbbC$? I don't think that's true because it means it is precomposed with projection function $p(z) = Re(z)$, which isn't holomorphic. So is it so just by definition?
complex-analysis
asked 2 days ago
studentstudent
497
$endgroup$
2
$begingroup$
It's just like the derivative of a function $f:mathbbRtomathbbR^2$
$endgroup$
– saulspatz
2 days ago
$begingroup$
But why do we have it like this? Is it just by definition? I ask this because being holomorphic and being differentiable in the multi-real sense are not the same thing. So this seems like an arbitrary thing to do: to throw in this thing in complex integrals and holomorphic functions.
$endgroup$
– student
2 days ago
add a comment |
$begingroup$
I have recently started to learn complex analysis. This question popped into my head when I was reading the path integrals. Since we have to consider the derivative of path parametrization in the definition of path integral, and this parameterization has as its domain a closed real interval.
Writing it out, if the path $gamma(t): [a,b] rightarrow mathbbC$ is given by $gamma(t) = x(t) + i y(t)$, then why do we define its derivative w.r.t. $t$ as $x'(t) + iy'(t)$? Is it a consequence of treating it as a function from $mathbbC text to mathbbC$? I don't think that's true because it means it is precomposed with projection function $p(z) = Re(z)$, which isn't holomorphic. So is it so just by definition?
complex-analysis
asked 2 days ago
studentstudent
497
$endgroup$
I have recently started to learn complex analysis. This question popped into my head when I was reading the path integrals. Since we have to consider the derivative of path parametrization in the definition of path integral, and this parameterization has as its domain a closed real interval.
Writing it out, if the path $gamma(t): [a,b] rightarrow mathbbC$ is given by $gamma(t) = x(t) + i y(t)$, then why do we define its derivative w.r.t. $t$ as $x'(t) + iy'(t)$? Is it a consequence of treating it as a function from $mathbbC text to mathbbC$? I don't think that's true because it means it is precomposed with projection function $p(z) = Re(z)$, which isn't holomorphic. So is it so just by definition?
complex-analysis
complex-analysis
asked 2 days ago
studentstudent
497
asked 2 days ago
studentstudent
497
asked 2 days ago
studentstudent
497
asked 2 days ago
studentstudent
497
asked 2 days ago
studentstudent
497
497
2
$begingroup$
It's just like the derivative of a function $f:mathbbRtomathbbR^2$
$endgroup$
– saulspatz
2 days ago
$begingroup$
But why do we have it like this? Is it just by definition? I ask this because being holomorphic and being differentiable in the multi-real sense are not the same thing. So this seems like an arbitrary thing to do: to throw in this thing in complex integrals and holomorphic functions.
$endgroup$
– student
2 days ago
add a comment |
2
$begingroup$
It's just like the derivative of a function $f:mathbbRtomathbbR^2$
$endgroup$
– saulspatz
2 days ago
$begingroup$
But why do we have it like this? Is it just by definition? I ask this because being holomorphic and being differentiable in the multi-real sense are not the same thing. So this seems like an arbitrary thing to do: to throw in this thing in complex integrals and holomorphic functions.
$endgroup$
– student
2 days ago
2
2
$begingroup$
It's just like the derivative of a function $f:mathbbRtomathbbR^2$
$endgroup$
– saulspatz
2 days ago
$begingroup$
It's just like the derivative of a function $f:mathbbRtomathbbR^2$
$endgroup$
– saulspatz
2 days ago
$begingroup$
But why do we have it like this? Is it just by definition? I ask this because being holomorphic and being differentiable in the multi-real sense are not the same thing. So this seems like an arbitrary thing to do: to throw in this thing in complex integrals and holomorphic functions.
$endgroup$
– student
2 days ago
$begingroup$
But why do we have it like this? Is it just by definition? I ask this because being holomorphic and being differentiable in the multi-real sense are not the same thing. So this seems like an arbitrary thing to do: to throw in this thing in complex integrals and holomorphic functions.
$endgroup$
– student
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $gamma(t) = x(t)+iy(t),$ then $$gamma(t+h)-gamma(t)over h = x(t+h-x(t)over h+iy(t+h)-y(t)over h$$ so taking limits gives $$gamma'(t)=x'(t)+iy'(t)$$
On the path $gamma,$ we are just looking at a function of a real variable, so holomorphy doesn't enter into it. Of course, when you start talking about the properties of the integral, you have to have holomorphy to get all the powerful theorems of complex variables.
answered 2 days ago
saulspatzsaulspatz
17.2k31435
$endgroup$
add a comment |
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$begingroup$
If $gamma(t) = x(t)+iy(t),$ then $$gamma(t+h)-gamma(t)over h = x(t+h-x(t)over h+iy(t+h)-y(t)over h$$ so taking limits gives $$gamma'(t)=x'(t)+iy'(t)$$
On the path $gamma,$ we are just looking at a function of a real variable, so holomorphy doesn't enter into it. Of course, when you start talking about the properties of the integral, you have to have holomorphy to get all the powerful theorems of complex variables.
answered 2 days ago
saulspatzsaulspatz
17.2k31435
$endgroup$
add a comment |
$begingroup$
If $gamma(t) = x(t)+iy(t),$ then $$gamma(t+h)-gamma(t)over h = x(t+h-x(t)over h+iy(t+h)-y(t)over h$$ so taking limits gives $$gamma'(t)=x'(t)+iy'(t)$$
On the path $gamma,$ we are just looking at a function of a real variable, so holomorphy doesn't enter into it. Of course, when you start talking about the properties of the integral, you have to have holomorphy to get all the powerful theorems of complex variables.
answered 2 days ago
saulspatzsaulspatz
17.2k31435
$endgroup$
add a comment |
$begingroup$
If $gamma(t) = x(t)+iy(t),$ then $$gamma(t+h)-gamma(t)over h = x(t+h-x(t)over h+iy(t+h)-y(t)over h$$ so taking limits gives $$gamma'(t)=x'(t)+iy'(t)$$
On the path $gamma,$ we are just looking at a function of a real variable, so holomorphy doesn't enter into it. Of course, when you start talking about the properties of the integral, you have to have holomorphy to get all the powerful theorems of complex variables.
answered 2 days ago
saulspatzsaulspatz
17.2k31435
$endgroup$
If $gamma(t) = x(t)+iy(t),$ then $$gamma(t+h)-gamma(t)over h = x(t+h-x(t)over h+iy(t+h)-y(t)over h$$ so taking limits gives $$gamma'(t)=x'(t)+iy'(t)$$
On the path $gamma,$ we are just looking at a function of a real variable, so holomorphy doesn't enter into it. Of course, when you start talking about the properties of the integral, you have to have holomorphy to get all the powerful theorems of complex variables.
answered 2 days ago
saulspatzsaulspatz
17.2k31435
edited 2 days ago
answered 2 days ago
saulspatzsaulspatz
17.2k31435
answered 2 days ago
saulspatzsaulspatz
17.2k31435
answered 2 days ago
saulspatzsaulspatz
17.2k31435
17.2k31435
add a comment |
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$begingroup$
It's just like the derivative of a function $f:mathbbRtomathbbR^2$
$endgroup$
– saulspatz
2 days ago
$begingroup$
But why do we have it like this? Is it just by definition? I ask this because being holomorphic and being differentiable in the multi-real sense are not the same thing. So this seems like an arbitrary thing to do: to throw in this thing in complex integrals and holomorphic functions.
$endgroup$
– student
2 days ago