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Express a given continuous path $alpha(t):[a,b] rightarrow mathbbC setminus a$ in the form: $alpha(t)=a+r(t)e^itheta (t)$
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Let $a in mathbbC$. Given a continuous path $alpha(t):[b,c] rightarrow mathbbC setminus a$, how can we express the path in the form of $$alpha(t)=a+r(t)e^itheta (t)$$.
First of all, a theorem says for this given continuous path, there exist unique continuous functions $r(t),theta (t):[b,c] rightarrow mathbbR$ with $r(t)>0$ for all $t in [b,c]$ and $theta(b) in [0,2pi)$ such that
$$alpha(t)=a+r(t)e^itheta(t)$$
My question is, how do we find the expressions for $r(t)$ and $theta(t)$ if given such $alpha(t)$?
complex-analysis algebraic-topology homotopy-theory
asked Jul 24 at 4:16
bbw
31517
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up vote
2
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Let $a in mathbbC$. Given a continuous path $alpha(t):[b,c] rightarrow mathbbC setminus a$, how can we express the path in the form of $$alpha(t)=a+r(t)e^itheta (t)$$.
First of all, a theorem says for this given continuous path, there exist unique continuous functions $r(t),theta (t):[b,c] rightarrow mathbbR$ with $r(t)>0$ for all $t in [b,c]$ and $theta(b) in [0,2pi)$ such that
$$alpha(t)=a+r(t)e^itheta(t)$$
My question is, how do we find the expressions for $r(t)$ and $theta(t)$ if given such $alpha(t)$?
complex-analysis algebraic-topology homotopy-theory
asked Jul 24 at 4:16
bbw
31517
3
You are essentially just asking for the polar decomposition of $alpha(t)-a$.
– Ian
Jul 24 at 4:21
1
I'm confused by the question. You've expressed $alpha(t)$ in terms of the other two functions.
– Andres Mejia
Jul 24 at 5:02
My dear bbw, It seems to me you have perhaps ambiguated the symbol "$a$": you use it to refer both to an element of the complex plane $Bbb C$, and also to the greatest lower bound of the closed interval $[a, b] subset Bbb R$; of course, I suppose "$a$" could be both, but that interpretation feels a bit forced to me. Comnents?
– Robert Lewis
Jul 24 at 23:39
@RobertLewis Yes you are right, i have just fixed the notations.
– bbw
Jul 25 at 1:16
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $a in mathbbC$. Given a continuous path $alpha(t):[b,c] rightarrow mathbbC setminus a$, how can we express the path in the form of $$alpha(t)=a+r(t)e^itheta (t)$$.
First of all, a theorem says for this given continuous path, there exist unique continuous functions $r(t),theta (t):[b,c] rightarrow mathbbR$ with $r(t)>0$ for all $t in [b,c]$ and $theta(b) in [0,2pi)$ such that
$$alpha(t)=a+r(t)e^itheta(t)$$
My question is, how do we find the expressions for $r(t)$ and $theta(t)$ if given such $alpha(t)$?
complex-analysis algebraic-topology homotopy-theory
asked Jul 24 at 4:16
bbw
31517
Let $a in mathbbC$. Given a continuous path $alpha(t):[b,c] rightarrow mathbbC setminus a$, how can we express the path in the form of $$alpha(t)=a+r(t)e^itheta (t)$$.
First of all, a theorem says for this given continuous path, there exist unique continuous functions $r(t),theta (t):[b,c] rightarrow mathbbR$ with $r(t)>0$ for all $t in [b,c]$ and $theta(b) in [0,2pi)$ such that
$$alpha(t)=a+r(t)e^itheta(t)$$
My question is, how do we find the expressions for $r(t)$ and $theta(t)$ if given such $alpha(t)$?
complex-analysis algebraic-topology homotopy-theory
asked Jul 24 at 4:16
bbw
31517
edited Jul 25 at 1:15
asked Jul 24 at 4:16
bbw
31517
asked Jul 24 at 4:16
bbw
31517
asked Jul 24 at 4:16
bbw
31517
31517
3
You are essentially just asking for the polar decomposition of $alpha(t)-a$.
– Ian
Jul 24 at 4:21
1
I'm confused by the question. You've expressed $alpha(t)$ in terms of the other two functions.
– Andres Mejia
Jul 24 at 5:02
My dear bbw, It seems to me you have perhaps ambiguated the symbol "$a$": you use it to refer both to an element of the complex plane $Bbb C$, and also to the greatest lower bound of the closed interval $[a, b] subset Bbb R$; of course, I suppose "$a$" could be both, but that interpretation feels a bit forced to me. Comnents?
– Robert Lewis
Jul 24 at 23:39
@RobertLewis Yes you are right, i have just fixed the notations.
– bbw
Jul 25 at 1:16
add a comment |Â
3
You are essentially just asking for the polar decomposition of $alpha(t)-a$.
– Ian
Jul 24 at 4:21
1
I'm confused by the question. You've expressed $alpha(t)$ in terms of the other two functions.
– Andres Mejia
Jul 24 at 5:02
My dear bbw, It seems to me you have perhaps ambiguated the symbol "$a$": you use it to refer both to an element of the complex plane $Bbb C$, and also to the greatest lower bound of the closed interval $[a, b] subset Bbb R$; of course, I suppose "$a$" could be both, but that interpretation feels a bit forced to me. Comnents?
– Robert Lewis
Jul 24 at 23:39
@RobertLewis Yes you are right, i have just fixed the notations.
– bbw
Jul 25 at 1:16
3
3
You are essentially just asking for the polar decomposition of $alpha(t)-a$.
– Ian
Jul 24 at 4:21
You are essentially just asking for the polar decomposition of $alpha(t)-a$.
– Ian
Jul 24 at 4:21
1
1
I'm confused by the question. You've expressed $alpha(t)$ in terms of the other two functions.
– Andres Mejia
Jul 24 at 5:02
I'm confused by the question. You've expressed $alpha(t)$ in terms of the other two functions.
– Andres Mejia
Jul 24 at 5:02
My dear bbw, It seems to me you have perhaps ambiguated the symbol "$a$": you use it to refer both to an element of the complex plane $Bbb C$, and also to the greatest lower bound of the closed interval $[a, b] subset Bbb R$; of course, I suppose "$a$" could be both, but that interpretation feels a bit forced to me. Comnents?
– Robert Lewis
Jul 24 at 23:39
My dear bbw, It seems to me you have perhaps ambiguated the symbol "$a$": you use it to refer both to an element of the complex plane $Bbb C$, and also to the greatest lower bound of the closed interval $[a, b] subset Bbb R$; of course, I suppose "$a$" could be both, but that interpretation feels a bit forced to me. Comnents?
– Robert Lewis
Jul 24 at 23:39
@RobertLewis Yes you are right, i have just fixed the notations.
– bbw
Jul 25 at 1:16
@RobertLewis Yes you are right, i have just fixed the notations.
– bbw
Jul 25 at 1:16
add a comment |Â
3 Answers
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Writing $alpha (t)$ as $a+r(t)e^itheta t$ is just the polar form of $alpha (t) -a$, but it takes some work to get continuity of $theta (t)$. [ $r(t)=|alpha (t)-a|$ is certainly continuous]. A well known result in complex analysis says that in any open ball in $mathbb C$ not containing $0$ there is a continuous logarithm. Using this we get the following: for any given $s in [a,b]$ there is an open interval around $s$ in which we can choose a continuous $theta (t)$ such that $alpha (t)=a+r(t)e^itheta t$. By compactness we can find a finite number of these intervals that cover $[a,b]$. Starting from the left we can alter the choice of $theta (t)$ so that the function agrees on adjacent intervals: If we have $theta_1 (t)$ in one interval and $theta_2 (t)$ in an adjacent interval then $e^theta_1 (t)=e^theta_2 (t)$ in the intersection of these intervals, so $theta_1 (t)-theta_2 (t)$ is of the form $2npi $ for some integer $n$. By continuity this $n$ does not depend on $t$ and we can replace $theta_2 (t)$ by $theta_2 (t)+2npi$ to ensure that $theta_2 (t)=theta_1 (t)$ on the intersection. In a finite number of steps we can get one continuous function $theta (t)$.
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
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With the restrictions in your question it is in general impossible: You cannot always achieve $theta([b,c]) subset [0,2pi)$. Note that I used $[b,c]$ instead of $[a,b]$ because $a in mathbbC$.
If you start with any two continuous maps $r, theta : [b,c] to mathbbR$ such that $r([b,c]) subset (0,infty)$, you get a path $alpha_r,theta(t) = a + r(t)e^itheta(t)$. Now take for example $[b,c] = [0,2pi], r(t) = 1, theta(a) = t$. Then it is impossible to find a continuous $theta'$ with $theta'([0,2pi]) subset [0,2pi)$ such that $alpha_r,theta' = alpha_r,theta$, simply because not all points of $[0,2pi)$ can be in the image of $theta'$.
Conversely, let us start with $alpha$. As Kavi Rama Murthy explained in his answer, you necessarily have $r(t) = lvert alpha(t) - a rvert$ which is continuous. Consider the path $beta : [b,c] to mathbbC, beta(t) = fracalpha(t) - ar(t)$. This is a path in $S^1$ = unit circle around $0$. The map $e : mathbbR to S^1, e(t) = e^it$, is a covering map. This implies that there exists a unique continuous map $theta : [b,c] to mathbbR$ such that $e circ theta = beta$ and $theta(b) in [0,2 pi)$. The last condition is only "scaling". Then you get $alpha = alpha_r,theta$.
answered Jul 24 at 8:57
Paul Frost
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If
$alpha(t) = a + r(t) e^i theta(t), tag 1$
then
$alpha(t) - a = r(t) e^i theta(t), tag 2$
$bar alpha - bar a = r(t) e^-i theta(t); tag 3$
thus,
$vert alpha(t) - a vert^2 = (alpha(t) - a)(bar alpha(t) - bar a) = (r e^i theta(t)) (r(t) e^-i theta(t) = (r(t))^2, tag 4$
or
$r(t) = vert alpha(t) - a vert, tag 5$
which expresses $r(t)$ in terms of $alpha(t)$ and $a$.
So far, so good; obtaining a formula for $theta(t)$, however, is a little more difficult. Knowing $r(t)$ as we do, from (2) we may write
$e^i theta(t) = dfracalpha(t) - ar(t); tag 6$
it will make things a little easier to write if we encapsulate the right-hand side of (6) with a symbol of its own:
$beta(t) = dfracalpha(t) - ar(t); tag 7$
thus,
$e^i theta(t) = beta(t); tag 8$
we set
$beta(t) = sigma(t) + i omega(t), tag 9$
and recall that
$e^i theta(t) = cos theta(t) + i sin(theta(t)), tag10$
then conbining (8)-(10) we find
$cos theta(t) + i sin(theta(t)) = sigma(t) + i omega(t), tag11$
so that
$cos(theta(t)) = sigma(t), ; sin theta(t)) = omega(t); tag12$
it follows that, as long as $sigma(t) ne 0$, we may write
$tan theta(t) = dfracsin theta(t)cos theta(t) = dfracomega(t)sigma(t), tag13$
and/or with $omega theta(t) ne 0$,
$cot theta(t) = dfraccos theta(t)sin theta(t) = dfracsigma(t)omega(t); tag14$
since $tan$ and $cot$ are invertible on their respective domains of definition, which as the reader will recall overlap but are offset one from the other by $pi / 2$, (13) and (14) may be used to define a continuous $theta(t)$ uniquely, provided the initial value $theta(b)$ is known; but this value is available to us via
$alpha(b) = a + r(a) e^i theta(a) tag15$
by following the steps presented above in (1)-(14). Of course, (13)-(14) only determine $theta(t)$ within an additive term of $2pi n$, $n in Bbb Z$.
The preceding discussion demonstrates how $r(t)$ and $theta(t)$ may be found given $alpha(t) in Bbb C setminus a $; indeed, we have effectively given a high-level description of an algorithm for computing $r(t)$, $theta(t)$ from $alpha(t)$, which "works" for continuous $alpha$ and returns continuous $r(t)$ and $theta(t)$. In this sense, we have fulfilled the quest of the question; however, when $alpha(t)$ is differentiable in $t$, we may present a different method for determining $theta(t)$ which in some sense may be easier and more effective than the above; to wit, we write, from (9),
$i dot theta(t) e^i theta(t) = dot beta(t), tag16$
whence
$dot theta(t) = -i e^-i theta(t) dot beta(t) = -i dfracdot beta(t)beta(t); tag17$
from this equation we may, in principle at least, have $theta(t)$ by direct integration:
$theta(t) - theta(b) = displaystyle -i int_b^t dfracdot beta(s)beta(s) ; ds, tag18$
$theta(t) = theta(a) - i displaystyle int_b^t dfracdot beta(s)beta(s) ; ds. tag19$
Of course whichever of these two strategems we choose for finding $theta(t)$, we have to admit the possibility that the restriction $theta(t) in [0, 2 pi)$ must be relaxed if we are in fact to obtain a continuous solution.
Finally, it is I reckon worth pointing out that the integrand in (19) may in fact be expressed as $dot (lnbeta(s))$ provided we are prepared to allow the complex logarithm to enter in to our discussion.
answered Jul 27 at 20:36
Robert Lewis
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3 Answers
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
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up vote
0
down vote
Writing $alpha (t)$ as $a+r(t)e^itheta t$ is just the polar form of $alpha (t) -a$, but it takes some work to get continuity of $theta (t)$. [ $r(t)=|alpha (t)-a|$ is certainly continuous]. A well known result in complex analysis says that in any open ball in $mathbb C$ not containing $0$ there is a continuous logarithm. Using this we get the following: for any given $s in [a,b]$ there is an open interval around $s$ in which we can choose a continuous $theta (t)$ such that $alpha (t)=a+r(t)e^itheta t$. By compactness we can find a finite number of these intervals that cover $[a,b]$. Starting from the left we can alter the choice of $theta (t)$ so that the function agrees on adjacent intervals: If we have $theta_1 (t)$ in one interval and $theta_2 (t)$ in an adjacent interval then $e^theta_1 (t)=e^theta_2 (t)$ in the intersection of these intervals, so $theta_1 (t)-theta_2 (t)$ is of the form $2npi $ for some integer $n$. By continuity this $n$ does not depend on $t$ and we can replace $theta_2 (t)$ by $theta_2 (t)+2npi$ to ensure that $theta_2 (t)=theta_1 (t)$ on the intersection. In a finite number of steps we can get one continuous function $theta (t)$.
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
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up vote
0
down vote
Writing $alpha (t)$ as $a+r(t)e^itheta t$ is just the polar form of $alpha (t) -a$, but it takes some work to get continuity of $theta (t)$. [ $r(t)=|alpha (t)-a|$ is certainly continuous]. A well known result in complex analysis says that in any open ball in $mathbb C$ not containing $0$ there is a continuous logarithm. Using this we get the following: for any given $s in [a,b]$ there is an open interval around $s$ in which we can choose a continuous $theta (t)$ such that $alpha (t)=a+r(t)e^itheta t$. By compactness we can find a finite number of these intervals that cover $[a,b]$. Starting from the left we can alter the choice of $theta (t)$ so that the function agrees on adjacent intervals: If we have $theta_1 (t)$ in one interval and $theta_2 (t)$ in an adjacent interval then $e^theta_1 (t)=e^theta_2 (t)$ in the intersection of these intervals, so $theta_1 (t)-theta_2 (t)$ is of the form $2npi $ for some integer $n$. By continuity this $n$ does not depend on $t$ and we can replace $theta_2 (t)$ by $theta_2 (t)+2npi$ to ensure that $theta_2 (t)=theta_1 (t)$ on the intersection. In a finite number of steps we can get one continuous function $theta (t)$.
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
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up vote
0
down vote
up vote
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down vote
Writing $alpha (t)$ as $a+r(t)e^itheta t$ is just the polar form of $alpha (t) -a$, but it takes some work to get continuity of $theta (t)$. [ $r(t)=|alpha (t)-a|$ is certainly continuous]. A well known result in complex analysis says that in any open ball in $mathbb C$ not containing $0$ there is a continuous logarithm. Using this we get the following: for any given $s in [a,b]$ there is an open interval around $s$ in which we can choose a continuous $theta (t)$ such that $alpha (t)=a+r(t)e^itheta t$. By compactness we can find a finite number of these intervals that cover $[a,b]$. Starting from the left we can alter the choice of $theta (t)$ so that the function agrees on adjacent intervals: If we have $theta_1 (t)$ in one interval and $theta_2 (t)$ in an adjacent interval then $e^theta_1 (t)=e^theta_2 (t)$ in the intersection of these intervals, so $theta_1 (t)-theta_2 (t)$ is of the form $2npi $ for some integer $n$. By continuity this $n$ does not depend on $t$ and we can replace $theta_2 (t)$ by $theta_2 (t)+2npi$ to ensure that $theta_2 (t)=theta_1 (t)$ on the intersection. In a finite number of steps we can get one continuous function $theta (t)$.
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
Writing $alpha (t)$ as $a+r(t)e^itheta t$ is just the polar form of $alpha (t) -a$, but it takes some work to get continuity of $theta (t)$. [ $r(t)=|alpha (t)-a|$ is certainly continuous]. A well known result in complex analysis says that in any open ball in $mathbb C$ not containing $0$ there is a continuous logarithm. Using this we get the following: for any given $s in [a,b]$ there is an open interval around $s$ in which we can choose a continuous $theta (t)$ such that $alpha (t)=a+r(t)e^itheta t$. By compactness we can find a finite number of these intervals that cover $[a,b]$. Starting from the left we can alter the choice of $theta (t)$ so that the function agrees on adjacent intervals: If we have $theta_1 (t)$ in one interval and $theta_2 (t)$ in an adjacent interval then $e^theta_1 (t)=e^theta_2 (t)$ in the intersection of these intervals, so $theta_1 (t)-theta_2 (t)$ is of the form $2npi $ for some integer $n$. By continuity this $n$ does not depend on $t$ and we can replace $theta_2 (t)$ by $theta_2 (t)+2npi$ to ensure that $theta_2 (t)=theta_1 (t)$ on the intersection. In a finite number of steps we can get one continuous function $theta (t)$.
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
answered Jul 24 at 6:00
Kavi Rama Murthy
20.2k2829
20.2k2829
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up vote
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With the restrictions in your question it is in general impossible: You cannot always achieve $theta([b,c]) subset [0,2pi)$. Note that I used $[b,c]$ instead of $[a,b]$ because $a in mathbbC$.
If you start with any two continuous maps $r, theta : [b,c] to mathbbR$ such that $r([b,c]) subset (0,infty)$, you get a path $alpha_r,theta(t) = a + r(t)e^itheta(t)$. Now take for example $[b,c] = [0,2pi], r(t) = 1, theta(a) = t$. Then it is impossible to find a continuous $theta'$ with $theta'([0,2pi]) subset [0,2pi)$ such that $alpha_r,theta' = alpha_r,theta$, simply because not all points of $[0,2pi)$ can be in the image of $theta'$.
Conversely, let us start with $alpha$. As Kavi Rama Murthy explained in his answer, you necessarily have $r(t) = lvert alpha(t) - a rvert$ which is continuous. Consider the path $beta : [b,c] to mathbbC, beta(t) = fracalpha(t) - ar(t)$. This is a path in $S^1$ = unit circle around $0$. The map $e : mathbbR to S^1, e(t) = e^it$, is a covering map. This implies that there exists a unique continuous map $theta : [b,c] to mathbbR$ such that $e circ theta = beta$ and $theta(b) in [0,2 pi)$. The last condition is only "scaling". Then you get $alpha = alpha_r,theta$.
answered Jul 24 at 8:57
Paul Frost
3,623420
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up vote
0
down vote
With the restrictions in your question it is in general impossible: You cannot always achieve $theta([b,c]) subset [0,2pi)$. Note that I used $[b,c]$ instead of $[a,b]$ because $a in mathbbC$.
If you start with any two continuous maps $r, theta : [b,c] to mathbbR$ such that $r([b,c]) subset (0,infty)$, you get a path $alpha_r,theta(t) = a + r(t)e^itheta(t)$. Now take for example $[b,c] = [0,2pi], r(t) = 1, theta(a) = t$. Then it is impossible to find a continuous $theta'$ with $theta'([0,2pi]) subset [0,2pi)$ such that $alpha_r,theta' = alpha_r,theta$, simply because not all points of $[0,2pi)$ can be in the image of $theta'$.
Conversely, let us start with $alpha$. As Kavi Rama Murthy explained in his answer, you necessarily have $r(t) = lvert alpha(t) - a rvert$ which is continuous. Consider the path $beta : [b,c] to mathbbC, beta(t) = fracalpha(t) - ar(t)$. This is a path in $S^1$ = unit circle around $0$. The map $e : mathbbR to S^1, e(t) = e^it$, is a covering map. This implies that there exists a unique continuous map $theta : [b,c] to mathbbR$ such that $e circ theta = beta$ and $theta(b) in [0,2 pi)$. The last condition is only "scaling". Then you get $alpha = alpha_r,theta$.
answered Jul 24 at 8:57
Paul Frost
3,623420
add a comment |Â
up vote
0
down vote
up vote
0
down vote
With the restrictions in your question it is in general impossible: You cannot always achieve $theta([b,c]) subset [0,2pi)$. Note that I used $[b,c]$ instead of $[a,b]$ because $a in mathbbC$.
If you start with any two continuous maps $r, theta : [b,c] to mathbbR$ such that $r([b,c]) subset (0,infty)$, you get a path $alpha_r,theta(t) = a + r(t)e^itheta(t)$. Now take for example $[b,c] = [0,2pi], r(t) = 1, theta(a) = t$. Then it is impossible to find a continuous $theta'$ with $theta'([0,2pi]) subset [0,2pi)$ such that $alpha_r,theta' = alpha_r,theta$, simply because not all points of $[0,2pi)$ can be in the image of $theta'$.
Conversely, let us start with $alpha$. As Kavi Rama Murthy explained in his answer, you necessarily have $r(t) = lvert alpha(t) - a rvert$ which is continuous. Consider the path $beta : [b,c] to mathbbC, beta(t) = fracalpha(t) - ar(t)$. This is a path in $S^1$ = unit circle around $0$. The map $e : mathbbR to S^1, e(t) = e^it$, is a covering map. This implies that there exists a unique continuous map $theta : [b,c] to mathbbR$ such that $e circ theta = beta$ and $theta(b) in [0,2 pi)$. The last condition is only "scaling". Then you get $alpha = alpha_r,theta$.
answered Jul 24 at 8:57
Paul Frost
3,623420
With the restrictions in your question it is in general impossible: You cannot always achieve $theta([b,c]) subset [0,2pi)$. Note that I used $[b,c]$ instead of $[a,b]$ because $a in mathbbC$.
If you start with any two continuous maps $r, theta : [b,c] to mathbbR$ such that $r([b,c]) subset (0,infty)$, you get a path $alpha_r,theta(t) = a + r(t)e^itheta(t)$. Now take for example $[b,c] = [0,2pi], r(t) = 1, theta(a) = t$. Then it is impossible to find a continuous $theta'$ with $theta'([0,2pi]) subset [0,2pi)$ such that $alpha_r,theta' = alpha_r,theta$, simply because not all points of $[0,2pi)$ can be in the image of $theta'$.
Conversely, let us start with $alpha$. As Kavi Rama Murthy explained in his answer, you necessarily have $r(t) = lvert alpha(t) - a rvert$ which is continuous. Consider the path $beta : [b,c] to mathbbC, beta(t) = fracalpha(t) - ar(t)$. This is a path in $S^1$ = unit circle around $0$. The map $e : mathbbR to S^1, e(t) = e^it$, is a covering map. This implies that there exists a unique continuous map $theta : [b,c] to mathbbR$ such that $e circ theta = beta$ and $theta(b) in [0,2 pi)$. The last condition is only "scaling". Then you get $alpha = alpha_r,theta$.
answered Jul 24 at 8:57
Paul Frost
3,623420
edited Jul 24 at 9:24
answered Jul 24 at 8:57
Paul Frost
3,623420
answered Jul 24 at 8:57
Paul Frost
3,623420
answered Jul 24 at 8:57
Paul Frost
3,623420
3,623420
add a comment |Â
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If
$alpha(t) = a + r(t) e^i theta(t), tag 1$
then
$alpha(t) - a = r(t) e^i theta(t), tag 2$
$bar alpha - bar a = r(t) e^-i theta(t); tag 3$
thus,
$vert alpha(t) - a vert^2 = (alpha(t) - a)(bar alpha(t) - bar a) = (r e^i theta(t)) (r(t) e^-i theta(t) = (r(t))^2, tag 4$
or
$r(t) = vert alpha(t) - a vert, tag 5$
which expresses $r(t)$ in terms of $alpha(t)$ and $a$.
So far, so good; obtaining a formula for $theta(t)$, however, is a little more difficult. Knowing $r(t)$ as we do, from (2) we may write
$e^i theta(t) = dfracalpha(t) - ar(t); tag 6$
it will make things a little easier to write if we encapsulate the right-hand side of (6) with a symbol of its own:
$beta(t) = dfracalpha(t) - ar(t); tag 7$
thus,
$e^i theta(t) = beta(t); tag 8$
we set
$beta(t) = sigma(t) + i omega(t), tag 9$
and recall that
$e^i theta(t) = cos theta(t) + i sin(theta(t)), tag10$
then conbining (8)-(10) we find
$cos theta(t) + i sin(theta(t)) = sigma(t) + i omega(t), tag11$
so that
$cos(theta(t)) = sigma(t), ; sin theta(t)) = omega(t); tag12$
it follows that, as long as $sigma(t) ne 0$, we may write
$tan theta(t) = dfracsin theta(t)cos theta(t) = dfracomega(t)sigma(t), tag13$
and/or with $omega theta(t) ne 0$,
$cot theta(t) = dfraccos theta(t)sin theta(t) = dfracsigma(t)omega(t); tag14$
since $tan$ and $cot$ are invertible on their respective domains of definition, which as the reader will recall overlap but are offset one from the other by $pi / 2$, (13) and (14) may be used to define a continuous $theta(t)$ uniquely, provided the initial value $theta(b)$ is known; but this value is available to us via
$alpha(b) = a + r(a) e^i theta(a) tag15$
by following the steps presented above in (1)-(14). Of course, (13)-(14) only determine $theta(t)$ within an additive term of $2pi n$, $n in Bbb Z$.
The preceding discussion demonstrates how $r(t)$ and $theta(t)$ may be found given $alpha(t) in Bbb C setminus a $; indeed, we have effectively given a high-level description of an algorithm for computing $r(t)$, $theta(t)$ from $alpha(t)$, which "works" for continuous $alpha$ and returns continuous $r(t)$ and $theta(t)$. In this sense, we have fulfilled the quest of the question; however, when $alpha(t)$ is differentiable in $t$, we may present a different method for determining $theta(t)$ which in some sense may be easier and more effective than the above; to wit, we write, from (9),
$i dot theta(t) e^i theta(t) = dot beta(t), tag16$
whence
$dot theta(t) = -i e^-i theta(t) dot beta(t) = -i dfracdot beta(t)beta(t); tag17$
from this equation we may, in principle at least, have $theta(t)$ by direct integration:
$theta(t) - theta(b) = displaystyle -i int_b^t dfracdot beta(s)beta(s) ; ds, tag18$
$theta(t) = theta(a) - i displaystyle int_b^t dfracdot beta(s)beta(s) ; ds. tag19$
Of course whichever of these two strategems we choose for finding $theta(t)$, we have to admit the possibility that the restriction $theta(t) in [0, 2 pi)$ must be relaxed if we are in fact to obtain a continuous solution.
Finally, it is I reckon worth pointing out that the integrand in (19) may in fact be expressed as $dot (lnbeta(s))$ provided we are prepared to allow the complex logarithm to enter in to our discussion.
answered Jul 27 at 20:36
Robert Lewis
36.9k22255
add a comment |Â
up vote
0
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If
$alpha(t) = a + r(t) e^i theta(t), tag 1$
then
$alpha(t) - a = r(t) e^i theta(t), tag 2$
$bar alpha - bar a = r(t) e^-i theta(t); tag 3$
thus,
$vert alpha(t) - a vert^2 = (alpha(t) - a)(bar alpha(t) - bar a) = (r e^i theta(t)) (r(t) e^-i theta(t) = (r(t))^2, tag 4$
or
$r(t) = vert alpha(t) - a vert, tag 5$
which expresses $r(t)$ in terms of $alpha(t)$ and $a$.
So far, so good; obtaining a formula for $theta(t)$, however, is a little more difficult. Knowing $r(t)$ as we do, from (2) we may write
$e^i theta(t) = dfracalpha(t) - ar(t); tag 6$
it will make things a little easier to write if we encapsulate the right-hand side of (6) with a symbol of its own:
$beta(t) = dfracalpha(t) - ar(t); tag 7$
thus,
$e^i theta(t) = beta(t); tag 8$
we set
$beta(t) = sigma(t) + i omega(t), tag 9$
and recall that
$e^i theta(t) = cos theta(t) + i sin(theta(t)), tag10$
then conbining (8)-(10) we find
$cos theta(t) + i sin(theta(t)) = sigma(t) + i omega(t), tag11$
so that
$cos(theta(t)) = sigma(t), ; sin theta(t)) = omega(t); tag12$
it follows that, as long as $sigma(t) ne 0$, we may write
$tan theta(t) = dfracsin theta(t)cos theta(t) = dfracomega(t)sigma(t), tag13$
and/or with $omega theta(t) ne 0$,
$cot theta(t) = dfraccos theta(t)sin theta(t) = dfracsigma(t)omega(t); tag14$
since $tan$ and $cot$ are invertible on their respective domains of definition, which as the reader will recall overlap but are offset one from the other by $pi / 2$, (13) and (14) may be used to define a continuous $theta(t)$ uniquely, provided the initial value $theta(b)$ is known; but this value is available to us via
$alpha(b) = a + r(a) e^i theta(a) tag15$
by following the steps presented above in (1)-(14). Of course, (13)-(14) only determine $theta(t)$ within an additive term of $2pi n$, $n in Bbb Z$.
The preceding discussion demonstrates how $r(t)$ and $theta(t)$ may be found given $alpha(t) in Bbb C setminus a $; indeed, we have effectively given a high-level description of an algorithm for computing $r(t)$, $theta(t)$ from $alpha(t)$, which "works" for continuous $alpha$ and returns continuous $r(t)$ and $theta(t)$. In this sense, we have fulfilled the quest of the question; however, when $alpha(t)$ is differentiable in $t$, we may present a different method for determining $theta(t)$ which in some sense may be easier and more effective than the above; to wit, we write, from (9),
$i dot theta(t) e^i theta(t) = dot beta(t), tag16$
whence
$dot theta(t) = -i e^-i theta(t) dot beta(t) = -i dfracdot beta(t)beta(t); tag17$
from this equation we may, in principle at least, have $theta(t)$ by direct integration:
$theta(t) - theta(b) = displaystyle -i int_b^t dfracdot beta(s)beta(s) ; ds, tag18$
$theta(t) = theta(a) - i displaystyle int_b^t dfracdot beta(s)beta(s) ; ds. tag19$
Of course whichever of these two strategems we choose for finding $theta(t)$, we have to admit the possibility that the restriction $theta(t) in [0, 2 pi)$ must be relaxed if we are in fact to obtain a continuous solution.
Finally, it is I reckon worth pointing out that the integrand in (19) may in fact be expressed as $dot (lnbeta(s))$ provided we are prepared to allow the complex logarithm to enter in to our discussion.
answered Jul 27 at 20:36
Robert Lewis
36.9k22255
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If
$alpha(t) = a + r(t) e^i theta(t), tag 1$
then
$alpha(t) - a = r(t) e^i theta(t), tag 2$
$bar alpha - bar a = r(t) e^-i theta(t); tag 3$
thus,
$vert alpha(t) - a vert^2 = (alpha(t) - a)(bar alpha(t) - bar a) = (r e^i theta(t)) (r(t) e^-i theta(t) = (r(t))^2, tag 4$
or
$r(t) = vert alpha(t) - a vert, tag 5$
which expresses $r(t)$ in terms of $alpha(t)$ and $a$.
So far, so good; obtaining a formula for $theta(t)$, however, is a little more difficult. Knowing $r(t)$ as we do, from (2) we may write
$e^i theta(t) = dfracalpha(t) - ar(t); tag 6$
it will make things a little easier to write if we encapsulate the right-hand side of (6) with a symbol of its own:
$beta(t) = dfracalpha(t) - ar(t); tag 7$
thus,
$e^i theta(t) = beta(t); tag 8$
we set
$beta(t) = sigma(t) + i omega(t), tag 9$
and recall that
$e^i theta(t) = cos theta(t) + i sin(theta(t)), tag10$
then conbining (8)-(10) we find
$cos theta(t) + i sin(theta(t)) = sigma(t) + i omega(t), tag11$
so that
$cos(theta(t)) = sigma(t), ; sin theta(t)) = omega(t); tag12$
it follows that, as long as $sigma(t) ne 0$, we may write
$tan theta(t) = dfracsin theta(t)cos theta(t) = dfracomega(t)sigma(t), tag13$
and/or with $omega theta(t) ne 0$,
$cot theta(t) = dfraccos theta(t)sin theta(t) = dfracsigma(t)omega(t); tag14$
since $tan$ and $cot$ are invertible on their respective domains of definition, which as the reader will recall overlap but are offset one from the other by $pi / 2$, (13) and (14) may be used to define a continuous $theta(t)$ uniquely, provided the initial value $theta(b)$ is known; but this value is available to us via
$alpha(b) = a + r(a) e^i theta(a) tag15$
by following the steps presented above in (1)-(14). Of course, (13)-(14) only determine $theta(t)$ within an additive term of $2pi n$, $n in Bbb Z$.
The preceding discussion demonstrates how $r(t)$ and $theta(t)$ may be found given $alpha(t) in Bbb C setminus a $; indeed, we have effectively given a high-level description of an algorithm for computing $r(t)$, $theta(t)$ from $alpha(t)$, which "works" for continuous $alpha$ and returns continuous $r(t)$ and $theta(t)$. In this sense, we have fulfilled the quest of the question; however, when $alpha(t)$ is differentiable in $t$, we may present a different method for determining $theta(t)$ which in some sense may be easier and more effective than the above; to wit, we write, from (9),
$i dot theta(t) e^i theta(t) = dot beta(t), tag16$
whence
$dot theta(t) = -i e^-i theta(t) dot beta(t) = -i dfracdot beta(t)beta(t); tag17$
from this equation we may, in principle at least, have $theta(t)$ by direct integration:
$theta(t) - theta(b) = displaystyle -i int_b^t dfracdot beta(s)beta(s) ; ds, tag18$
$theta(t) = theta(a) - i displaystyle int_b^t dfracdot beta(s)beta(s) ; ds. tag19$
Of course whichever of these two strategems we choose for finding $theta(t)$, we have to admit the possibility that the restriction $theta(t) in [0, 2 pi)$ must be relaxed if we are in fact to obtain a continuous solution.
Finally, it is I reckon worth pointing out that the integrand in (19) may in fact be expressed as $dot (lnbeta(s))$ provided we are prepared to allow the complex logarithm to enter in to our discussion.
answered Jul 27 at 20:36
Robert Lewis
36.9k22255
If
$alpha(t) = a + r(t) e^i theta(t), tag 1$
then
$alpha(t) - a = r(t) e^i theta(t), tag 2$
$bar alpha - bar a = r(t) e^-i theta(t); tag 3$
thus,
$vert alpha(t) - a vert^2 = (alpha(t) - a)(bar alpha(t) - bar a) = (r e^i theta(t)) (r(t) e^-i theta(t) = (r(t))^2, tag 4$
or
$r(t) = vert alpha(t) - a vert, tag 5$
which expresses $r(t)$ in terms of $alpha(t)$ and $a$.
So far, so good; obtaining a formula for $theta(t)$, however, is a little more difficult. Knowing $r(t)$ as we do, from (2) we may write
$e^i theta(t) = dfracalpha(t) - ar(t); tag 6$
it will make things a little easier to write if we encapsulate the right-hand side of (6) with a symbol of its own:
$beta(t) = dfracalpha(t) - ar(t); tag 7$
thus,
$e^i theta(t) = beta(t); tag 8$
we set
$beta(t) = sigma(t) + i omega(t), tag 9$
and recall that
$e^i theta(t) = cos theta(t) + i sin(theta(t)), tag10$
then conbining (8)-(10) we find
$cos theta(t) + i sin(theta(t)) = sigma(t) + i omega(t), tag11$
so that
$cos(theta(t)) = sigma(t), ; sin theta(t)) = omega(t); tag12$
it follows that, as long as $sigma(t) ne 0$, we may write
$tan theta(t) = dfracsin theta(t)cos theta(t) = dfracomega(t)sigma(t), tag13$
and/or with $omega theta(t) ne 0$,
$cot theta(t) = dfraccos theta(t)sin theta(t) = dfracsigma(t)omega(t); tag14$
since $tan$ and $cot$ are invertible on their respective domains of definition, which as the reader will recall overlap but are offset one from the other by $pi / 2$, (13) and (14) may be used to define a continuous $theta(t)$ uniquely, provided the initial value $theta(b)$ is known; but this value is available to us via
$alpha(b) = a + r(a) e^i theta(a) tag15$
by following the steps presented above in (1)-(14). Of course, (13)-(14) only determine $theta(t)$ within an additive term of $2pi n$, $n in Bbb Z$.
The preceding discussion demonstrates how $r(t)$ and $theta(t)$ may be found given $alpha(t) in Bbb C setminus a $; indeed, we have effectively given a high-level description of an algorithm for computing $r(t)$, $theta(t)$ from $alpha(t)$, which "works" for continuous $alpha$ and returns continuous $r(t)$ and $theta(t)$. In this sense, we have fulfilled the quest of the question; however, when $alpha(t)$ is differentiable in $t$, we may present a different method for determining $theta(t)$ which in some sense may be easier and more effective than the above; to wit, we write, from (9),
$i dot theta(t) e^i theta(t) = dot beta(t), tag16$
whence
$dot theta(t) = -i e^-i theta(t) dot beta(t) = -i dfracdot beta(t)beta(t); tag17$
from this equation we may, in principle at least, have $theta(t)$ by direct integration:
$theta(t) - theta(b) = displaystyle -i int_b^t dfracdot beta(s)beta(s) ; ds, tag18$
$theta(t) = theta(a) - i displaystyle int_b^t dfracdot beta(s)beta(s) ; ds. tag19$
Of course whichever of these two strategems we choose for finding $theta(t)$, we have to admit the possibility that the restriction $theta(t) in [0, 2 pi)$ must be relaxed if we are in fact to obtain a continuous solution.
Finally, it is I reckon worth pointing out that the integrand in (19) may in fact be expressed as $dot (lnbeta(s))$ provided we are prepared to allow the complex logarithm to enter in to our discussion.
answered Jul 27 at 20:36
Robert Lewis
36.9k22255
edited Jul 27 at 20:42
answered Jul 27 at 20:36
Robert Lewis
36.9k22255
answered Jul 27 at 20:36
Robert Lewis
36.9k22255
answered Jul 27 at 20:36
Robert Lewis
36.9k22255
36.9k22255
add a comment |Â
add a comment |Â
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3
You are essentially just asking for the polar decomposition of $alpha(t)-a$.
– Ian
Jul 24 at 4:21
1
I'm confused by the question. You've expressed $alpha(t)$ in terms of the other two functions.
– Andres Mejia
Jul 24 at 5:02
My dear bbw, It seems to me you have perhaps ambiguated the symbol "$a$": you use it to refer both to an element of the complex plane $Bbb C$, and also to the greatest lower bound of the closed interval $[a, b] subset Bbb R$; of course, I suppose "$a$" could be both, but that interpretation feels a bit forced to me. Comnents?
– Robert Lewis
Jul 24 at 23:39
@RobertLewis Yes you are right, i have just fixed the notations.
– bbw
Jul 25 at 1:16