Problems with putting $r=e^{-theta cot(theta)}$ into rectangular form
$begingroup$
Recently, I found a polar curve which contains all(?) complex numbers $z$, which satisfy $Im(z^z)=0$; or in other words, all complex numbers which raised to itself give a real number.
The curve is described by the equation $r=e^{-theta cot(theta)}$. The following image shows the graph of the equation for $0leq thetaleq 2pi$:
The function has infinite amount of lines going away into the infinity, as $theta$ is increased below $0$ or above $2pi$. As far as I have been able to find, only the section $0leq thetaleq 2pi$ actually satisfies $Im(re^{,itheta, re^{,itheta}})=Im(z^z)=0$. I may be wrong, because I haven't looked at this closely, but that's not the important part to the question.
It is easy to observe, that the curves formed on the graph form an interesting curve, one which would be nice to analyze in the rectangular form. However, the farthest I have been able to convert between systems is this:
$$x^2+y^2=e^{-{2xover y}arctanbig({yover x}big)}$$
There are however many problems with this equation:
1. The rectangular equation is lossy
When I graphed the resultant rectangular equation, not surprisingly I found that the graph is only true for the first quadrant, the rest were mere artifact reflections of it:
Now, I'll admit that that's a beautiful eye, but this is not really what I wanted to get. ;)
2. Equation is (at least in my effort so far) insolvable for $x$ or $y$
Not as big of a problem, but it would be nice to solve for one of the variables.
Most importantly I want to fix problem number 1. I feel like I should know how to fix this issue, but I'm having a huge brainfart. Can anyone point me in the right direction?
EDIT: I have considered by now, that the nature of $arctan$ or the square root may cause this problem, but I have not been able to somehow change the expression to one that views the second quadrant correctly at the least.
complex-numbers polar-coordinates
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
$endgroup$
add a comment |
$begingroup$
Recently, I found a polar curve which contains all(?) complex numbers $z$, which satisfy $Im(z^z)=0$; or in other words, all complex numbers which raised to itself give a real number.
The curve is described by the equation $r=e^{-theta cot(theta)}$. The following image shows the graph of the equation for $0leq thetaleq 2pi$:
The function has infinite amount of lines going away into the infinity, as $theta$ is increased below $0$ or above $2pi$. As far as I have been able to find, only the section $0leq thetaleq 2pi$ actually satisfies $Im(re^{,itheta, re^{,itheta}})=Im(z^z)=0$. I may be wrong, because I haven't looked at this closely, but that's not the important part to the question.
It is easy to observe, that the curves formed on the graph form an interesting curve, one which would be nice to analyze in the rectangular form. However, the farthest I have been able to convert between systems is this:
$$x^2+y^2=e^{-{2xover y}arctanbig({yover x}big)}$$
There are however many problems with this equation:
1. The rectangular equation is lossy
When I graphed the resultant rectangular equation, not surprisingly I found that the graph is only true for the first quadrant, the rest were mere artifact reflections of it:
Now, I'll admit that that's a beautiful eye, but this is not really what I wanted to get. ;)
2. Equation is (at least in my effort so far) insolvable for $x$ or $y$
Not as big of a problem, but it would be nice to solve for one of the variables.
Most importantly I want to fix problem number 1. I feel like I should know how to fix this issue, but I'm having a huge brainfart. Can anyone point me in the right direction?
EDIT: I have considered by now, that the nature of $arctan$ or the square root may cause this problem, but I have not been able to somehow change the expression to one that views the second quadrant correctly at the least.
complex-numbers polar-coordinates
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
$endgroup$
$begingroup$
Your equation is lossy because $arctan$ give only one of the values in $[0, 2pi)$ that have the given tangent.
$endgroup$
– Todor Markov
Dec 24 '18 at 11:30
$begingroup$
@TodorMarkov I figured ;) I just don't know what to do with that fact. Perhaps I could throw a sgn function where appropriate, or make this piecewise somehow?
$endgroup$
– KKZiomek
Dec 24 '18 at 14:53
$begingroup$
@ KKZiomek More generally the equation $Im(z^z)=0$ leads to $2pi n = r log(r) sin(theta)+ r theta cos(theta) $ with $nin Z$
$endgroup$
– Dr. Wolfgang Hintze
Dec 29 '18 at 10:55
$begingroup$
@Dr.WolfgangHintze yup, I just explored the principal branch
$endgroup$
– KKZiomek
Dec 31 '18 at 19:34
$begingroup$
@KKZiomek I made the comment because you stessed the "all" with a question mark in "I found a polar curve which contains all(?) complex numbers $z$ which satisfy $Im(z)=0$". I think you could extend your interesting analysis to the general case as well.
$endgroup$
– Dr. Wolfgang Hintze
Jan 1 at 15:21
add a comment |
$begingroup$
Recently, I found a polar curve which contains all(?) complex numbers $z$, which satisfy $Im(z^z)=0$; or in other words, all complex numbers which raised to itself give a real number.
The curve is described by the equation $r=e^{-theta cot(theta)}$. The following image shows the graph of the equation for $0leq thetaleq 2pi$:
The function has infinite amount of lines going away into the infinity, as $theta$ is increased below $0$ or above $2pi$. As far as I have been able to find, only the section $0leq thetaleq 2pi$ actually satisfies $Im(re^{,itheta, re^{,itheta}})=Im(z^z)=0$. I may be wrong, because I haven't looked at this closely, but that's not the important part to the question.
It is easy to observe, that the curves formed on the graph form an interesting curve, one which would be nice to analyze in the rectangular form. However, the farthest I have been able to convert between systems is this:
$$x^2+y^2=e^{-{2xover y}arctanbig({yover x}big)}$$
There are however many problems with this equation:
1. The rectangular equation is lossy
When I graphed the resultant rectangular equation, not surprisingly I found that the graph is only true for the first quadrant, the rest were mere artifact reflections of it:
Now, I'll admit that that's a beautiful eye, but this is not really what I wanted to get. ;)
2. Equation is (at least in my effort so far) insolvable for $x$ or $y$
Not as big of a problem, but it would be nice to solve for one of the variables.
Most importantly I want to fix problem number 1. I feel like I should know how to fix this issue, but I'm having a huge brainfart. Can anyone point me in the right direction?
EDIT: I have considered by now, that the nature of $arctan$ or the square root may cause this problem, but I have not been able to somehow change the expression to one that views the second quadrant correctly at the least.
complex-numbers polar-coordinates
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
$endgroup$
Recently, I found a polar curve which contains all(?) complex numbers $z$, which satisfy $Im(z^z)=0$; or in other words, all complex numbers which raised to itself give a real number.
The curve is described by the equation $r=e^{-theta cot(theta)}$. The following image shows the graph of the equation for $0leq thetaleq 2pi$:
The function has infinite amount of lines going away into the infinity, as $theta$ is increased below $0$ or above $2pi$. As far as I have been able to find, only the section $0leq thetaleq 2pi$ actually satisfies $Im(re^{,itheta, re^{,itheta}})=Im(z^z)=0$. I may be wrong, because I haven't looked at this closely, but that's not the important part to the question.
It is easy to observe, that the curves formed on the graph form an interesting curve, one which would be nice to analyze in the rectangular form. However, the farthest I have been able to convert between systems is this:
$$x^2+y^2=e^{-{2xover y}arctanbig({yover x}big)}$$
There are however many problems with this equation:
1. The rectangular equation is lossy
When I graphed the resultant rectangular equation, not surprisingly I found that the graph is only true for the first quadrant, the rest were mere artifact reflections of it:
Now, I'll admit that that's a beautiful eye, but this is not really what I wanted to get. ;)
2. Equation is (at least in my effort so far) insolvable for $x$ or $y$
Not as big of a problem, but it would be nice to solve for one of the variables.
Most importantly I want to fix problem number 1. I feel like I should know how to fix this issue, but I'm having a huge brainfart. Can anyone point me in the right direction?
EDIT: I have considered by now, that the nature of $arctan$ or the square root may cause this problem, but I have not been able to somehow change the expression to one that views the second quadrant correctly at the least.
complex-numbers polar-coordinates
complex-numbers polar-coordinates
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
edited Dec 23 '18 at 16:18
KKZiomek
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
asked Dec 21 '18 at 23:49
KKZiomekKKZiomek
2,2021639
2,2021639
$begingroup$
Your equation is lossy because $arctan$ give only one of the values in $[0, 2pi)$ that have the given tangent.
$endgroup$
– Todor Markov
Dec 24 '18 at 11:30
$begingroup$
@TodorMarkov I figured ;) I just don't know what to do with that fact. Perhaps I could throw a sgn function where appropriate, or make this piecewise somehow?
$endgroup$
– KKZiomek
Dec 24 '18 at 14:53
$begingroup$
@ KKZiomek More generally the equation $Im(z^z)=0$ leads to $2pi n = r log(r) sin(theta)+ r theta cos(theta) $ with $nin Z$
$endgroup$
– Dr. Wolfgang Hintze
Dec 29 '18 at 10:55
$begingroup$
@Dr.WolfgangHintze yup, I just explored the principal branch
$endgroup$
– KKZiomek
Dec 31 '18 at 19:34
$begingroup$
@KKZiomek I made the comment because you stessed the "all" with a question mark in "I found a polar curve which contains all(?) complex numbers $z$ which satisfy $Im(z)=0$". I think you could extend your interesting analysis to the general case as well.
$endgroup$
– Dr. Wolfgang Hintze
Jan 1 at 15:21
add a comment |
$begingroup$
Your equation is lossy because $arctan$ give only one of the values in $[0, 2pi)$ that have the given tangent.
$endgroup$
– Todor Markov
Dec 24 '18 at 11:30
$begingroup$
@TodorMarkov I figured ;) I just don't know what to do with that fact. Perhaps I could throw a sgn function where appropriate, or make this piecewise somehow?
$endgroup$
– KKZiomek
Dec 24 '18 at 14:53
$begingroup$
@ KKZiomek More generally the equation $Im(z^z)=0$ leads to $2pi n = r log(r) sin(theta)+ r theta cos(theta) $ with $nin Z$
$endgroup$
– Dr. Wolfgang Hintze
Dec 29 '18 at 10:55
$begingroup$
@Dr.WolfgangHintze yup, I just explored the principal branch
$endgroup$
– KKZiomek
Dec 31 '18 at 19:34
$begingroup$
@KKZiomek I made the comment because you stessed the "all" with a question mark in "I found a polar curve which contains all(?) complex numbers $z$ which satisfy $Im(z)=0$". I think you could extend your interesting analysis to the general case as well.
$endgroup$
– Dr. Wolfgang Hintze
Jan 1 at 15:21
$begingroup$
Your equation is lossy because $arctan$ give only one of the values in $[0, 2pi)$ that have the given tangent.
$endgroup$
– Todor Markov
Dec 24 '18 at 11:30
$begingroup$
Your equation is lossy because $arctan$ give only one of the values in $[0, 2pi)$ that have the given tangent.
$endgroup$
– Todor Markov
Dec 24 '18 at 11:30
$begingroup$
@TodorMarkov I figured ;) I just don't know what to do with that fact. Perhaps I could throw a sgn function where appropriate, or make this piecewise somehow?
$endgroup$
– KKZiomek
Dec 24 '18 at 14:53
$begingroup$
@TodorMarkov I figured ;) I just don't know what to do with that fact. Perhaps I could throw a sgn function where appropriate, or make this piecewise somehow?
$endgroup$
– KKZiomek
Dec 24 '18 at 14:53
$begingroup$
@ KKZiomek More generally the equation $Im(z^z)=0$ leads to $2pi n = r log(r) sin(theta)+ r theta cos(theta) $ with $nin Z$
$endgroup$
– Dr. Wolfgang Hintze
Dec 29 '18 at 10:55
$begingroup$
@ KKZiomek More generally the equation $Im(z^z)=0$ leads to $2pi n = r log(r) sin(theta)+ r theta cos(theta) $ with $nin Z$
$endgroup$
– Dr. Wolfgang Hintze
Dec 29 '18 at 10:55
$begingroup$
@Dr.WolfgangHintze yup, I just explored the principal branch
$endgroup$
– KKZiomek
Dec 31 '18 at 19:34
$begingroup$
@Dr.WolfgangHintze yup, I just explored the principal branch
$endgroup$
– KKZiomek
Dec 31 '18 at 19:34
$begingroup$
@KKZiomek I made the comment because you stessed the "all" with a question mark in "I found a polar curve which contains all(?) complex numbers $z$ which satisfy $Im(z)=0$". I think you could extend your interesting analysis to the general case as well.
$endgroup$
– Dr. Wolfgang Hintze
Jan 1 at 15:21
$begingroup$
@KKZiomek I made the comment because you stessed the "all" with a question mark in "I found a polar curve which contains all(?) complex numbers $z$ which satisfy $Im(z)=0$". I think you could extend your interesting analysis to the general case as well.
$endgroup$
– Dr. Wolfgang Hintze
Jan 1 at 15:21
add a comment |
1 Answer
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$begingroup$
I explored further the idea, that the $arctan$ may be causing said problems. I figured that since the codomain of inverse tangent is $Y={y:-{piover 2}<yleq {piover 2}}$, and since $tan(x+pi)=tan x$, then all values from the second quadrant, should actually spit out values of respective angles from the fourth quadrant (quadrant that is $pi$ away from the second quadrant).
So I went on, and subtracted $pi$ from the value of inverse tangent in the rectangular formula, to get the resulting angle from the second quadrant to be converted into the fourth quadrant. This is the result:
So, as can be seen, the new (blue) formula, made it possible to produce a reflection of the second quadrant in the first quadrant.
I, then, negated the argument of the arctangent, and negated the whole exponent, to reflect the curve along the y axis, and I was able to get an overlap over the actual curve for the polar equation:
Notice, that the blue curve also provided correct resemblance of the actual third quadrant. That is because the third quadrant is also $pi$ bigger than the first quadrant, which falls in the codomain of the inverse tangent. Hence, subtracting $pi$ from it, also resulted in the correct curve.
Subtracting $2pi$ instead of $pi$ also corrected the fourth quadrant:
Putting this all together, I have managed to scramble a piecewise rectangular definition of the curve ($0leqthetaleq 2pi$):
$$C:
begin{cases}
x^2+y^2=e^{{2xover y}arctanbig(-{yover x}big)}&text{if}, 0,leqthetaleq {piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-pibig)}&text{if}, {piover 2},leqthetaleq{3piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-2pibig)}&text{if}, {3piover 2},leqthetaleq 2pi\
end{cases}$$
In general, subtracting $pi$ successively "unlocks" the next two quadrants of the curve.
So, I have figured out the answer to my own question :)
P.S. What will happen to the bounty I set for this question now?
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
$endgroup$
add a comment |
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$begingroup$
I explored further the idea, that the $arctan$ may be causing said problems. I figured that since the codomain of inverse tangent is $Y={y:-{piover 2}<yleq {piover 2}}$, and since $tan(x+pi)=tan x$, then all values from the second quadrant, should actually spit out values of respective angles from the fourth quadrant (quadrant that is $pi$ away from the second quadrant).
So I went on, and subtracted $pi$ from the value of inverse tangent in the rectangular formula, to get the resulting angle from the second quadrant to be converted into the fourth quadrant. This is the result:
So, as can be seen, the new (blue) formula, made it possible to produce a reflection of the second quadrant in the first quadrant.
I, then, negated the argument of the arctangent, and negated the whole exponent, to reflect the curve along the y axis, and I was able to get an overlap over the actual curve for the polar equation:
Notice, that the blue curve also provided correct resemblance of the actual third quadrant. That is because the third quadrant is also $pi$ bigger than the first quadrant, which falls in the codomain of the inverse tangent. Hence, subtracting $pi$ from it, also resulted in the correct curve.
Subtracting $2pi$ instead of $pi$ also corrected the fourth quadrant:
Putting this all together, I have managed to scramble a piecewise rectangular definition of the curve ($0leqthetaleq 2pi$):
$$C:
begin{cases}
x^2+y^2=e^{{2xover y}arctanbig(-{yover x}big)}&text{if}, 0,leqthetaleq {piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-pibig)}&text{if}, {piover 2},leqthetaleq{3piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-2pibig)}&text{if}, {3piover 2},leqthetaleq 2pi\
end{cases}$$
In general, subtracting $pi$ successively "unlocks" the next two quadrants of the curve.
So, I have figured out the answer to my own question :)
P.S. What will happen to the bounty I set for this question now?
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
$endgroup$
add a comment |
$begingroup$
I explored further the idea, that the $arctan$ may be causing said problems. I figured that since the codomain of inverse tangent is $Y={y:-{piover 2}<yleq {piover 2}}$, and since $tan(x+pi)=tan x$, then all values from the second quadrant, should actually spit out values of respective angles from the fourth quadrant (quadrant that is $pi$ away from the second quadrant).
So I went on, and subtracted $pi$ from the value of inverse tangent in the rectangular formula, to get the resulting angle from the second quadrant to be converted into the fourth quadrant. This is the result:
So, as can be seen, the new (blue) formula, made it possible to produce a reflection of the second quadrant in the first quadrant.
I, then, negated the argument of the arctangent, and negated the whole exponent, to reflect the curve along the y axis, and I was able to get an overlap over the actual curve for the polar equation:
Notice, that the blue curve also provided correct resemblance of the actual third quadrant. That is because the third quadrant is also $pi$ bigger than the first quadrant, which falls in the codomain of the inverse tangent. Hence, subtracting $pi$ from it, also resulted in the correct curve.
Subtracting $2pi$ instead of $pi$ also corrected the fourth quadrant:
Putting this all together, I have managed to scramble a piecewise rectangular definition of the curve ($0leqthetaleq 2pi$):
$$C:
begin{cases}
x^2+y^2=e^{{2xover y}arctanbig(-{yover x}big)}&text{if}, 0,leqthetaleq {piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-pibig)}&text{if}, {piover 2},leqthetaleq{3piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-2pibig)}&text{if}, {3piover 2},leqthetaleq 2pi\
end{cases}$$
In general, subtracting $pi$ successively "unlocks" the next two quadrants of the curve.
So, I have figured out the answer to my own question :)
P.S. What will happen to the bounty I set for this question now?
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
$endgroup$
add a comment |
$begingroup$
I explored further the idea, that the $arctan$ may be causing said problems. I figured that since the codomain of inverse tangent is $Y={y:-{piover 2}<yleq {piover 2}}$, and since $tan(x+pi)=tan x$, then all values from the second quadrant, should actually spit out values of respective angles from the fourth quadrant (quadrant that is $pi$ away from the second quadrant).
So I went on, and subtracted $pi$ from the value of inverse tangent in the rectangular formula, to get the resulting angle from the second quadrant to be converted into the fourth quadrant. This is the result:
So, as can be seen, the new (blue) formula, made it possible to produce a reflection of the second quadrant in the first quadrant.
I, then, negated the argument of the arctangent, and negated the whole exponent, to reflect the curve along the y axis, and I was able to get an overlap over the actual curve for the polar equation:
Notice, that the blue curve also provided correct resemblance of the actual third quadrant. That is because the third quadrant is also $pi$ bigger than the first quadrant, which falls in the codomain of the inverse tangent. Hence, subtracting $pi$ from it, also resulted in the correct curve.
Subtracting $2pi$ instead of $pi$ also corrected the fourth quadrant:
Putting this all together, I have managed to scramble a piecewise rectangular definition of the curve ($0leqthetaleq 2pi$):
$$C:
begin{cases}
x^2+y^2=e^{{2xover y}arctanbig(-{yover x}big)}&text{if}, 0,leqthetaleq {piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-pibig)}&text{if}, {piover 2},leqthetaleq{3piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-2pibig)}&text{if}, {3piover 2},leqthetaleq 2pi\
end{cases}$$
In general, subtracting $pi$ successively "unlocks" the next two quadrants of the curve.
So, I have figured out the answer to my own question :)
P.S. What will happen to the bounty I set for this question now?
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
$endgroup$
I explored further the idea, that the $arctan$ may be causing said problems. I figured that since the codomain of inverse tangent is $Y={y:-{piover 2}<yleq {piover 2}}$, and since $tan(x+pi)=tan x$, then all values from the second quadrant, should actually spit out values of respective angles from the fourth quadrant (quadrant that is $pi$ away from the second quadrant).
So I went on, and subtracted $pi$ from the value of inverse tangent in the rectangular formula, to get the resulting angle from the second quadrant to be converted into the fourth quadrant. This is the result:
So, as can be seen, the new (blue) formula, made it possible to produce a reflection of the second quadrant in the first quadrant.
I, then, negated the argument of the arctangent, and negated the whole exponent, to reflect the curve along the y axis, and I was able to get an overlap over the actual curve for the polar equation:
Notice, that the blue curve also provided correct resemblance of the actual third quadrant. That is because the third quadrant is also $pi$ bigger than the first quadrant, which falls in the codomain of the inverse tangent. Hence, subtracting $pi$ from it, also resulted in the correct curve.
Subtracting $2pi$ instead of $pi$ also corrected the fourth quadrant:
Putting this all together, I have managed to scramble a piecewise rectangular definition of the curve ($0leqthetaleq 2pi$):
$$C:
begin{cases}
x^2+y^2=e^{{2xover y}arctanbig(-{yover x}big)}&text{if}, 0,leqthetaleq {piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-pibig)}&text{if}, {piover 2},leqthetaleq{3piover 2}\
x^2+y^2=e^{{2xover y}big(arctanbig(-{yover x}big)-2pibig)}&text{if}, {3piover 2},leqthetaleq 2pi\
end{cases}$$
In general, subtracting $pi$ successively "unlocks" the next two quadrants of the curve.
So, I have figured out the answer to my own question :)
P.S. What will happen to the bounty I set for this question now?
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
answered Dec 24 '18 at 16:34
KKZiomekKKZiomek
2,2021639
2,2021639
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Your equation is lossy because $arctan$ give only one of the values in $[0, 2pi)$ that have the given tangent.
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– Todor Markov
Dec 24 '18 at 11:30
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@TodorMarkov I figured ;) I just don't know what to do with that fact. Perhaps I could throw a sgn function where appropriate, or make this piecewise somehow?
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– KKZiomek
Dec 24 '18 at 14:53
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@ KKZiomek More generally the equation $Im(z^z)=0$ leads to $2pi n = r log(r) sin(theta)+ r theta cos(theta) $ with $nin Z$
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– Dr. Wolfgang Hintze
Dec 29 '18 at 10:55
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@Dr.WolfgangHintze yup, I just explored the principal branch
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– KKZiomek
Dec 31 '18 at 19:34
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@KKZiomek I made the comment because you stessed the "all" with a question mark in "I found a polar curve which contains all(?) complex numbers $z$ which satisfy $Im(z)=0$". I think you could extend your interesting analysis to the general case as well.
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– Dr. Wolfgang Hintze
Jan 1 at 15:21