Mixing
Derivative of an Area Preserving Operation [closed]
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Let z be in the complex domain.
Then define $hatz = e^jthetaz$, where $theta$ is a constant.
If the derivative is taken, by using the fact that rotation won't change the area (area preserving operation), can we say that $dhatz = dz $?
Thank you very much.
derivatives
asked Jul 15 at 23:53
NoOne
32
closed as off-topic by amWhy, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, John Ma Jul 16 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Trần Thúc Minh TrÃÂ, Parcly Taxel, John Ma
add a comment |Â
up vote
0
down vote
favorite
Let z be in the complex domain.
Then define $hatz = e^jthetaz$, where $theta$ is a constant.
If the derivative is taken, by using the fact that rotation won't change the area (area preserving operation), can we say that $dhatz = dz $?
Thank you very much.
derivatives
asked Jul 15 at 23:53
NoOne
32
closed as off-topic by amWhy, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, John Ma Jul 16 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Trần Thúc Minh TrÃÂ, Parcly Taxel, John Ma
Is $theta$ constant or supposed to be the argument of $z$? If constant, do you know how to compare dy and dx in y=(-1)*x?
– Mark S.
Jul 16 at 0:01
Yes, $theta$ is a constant. In this case $y = -x rightarrow y=e^-jpix rightarrow dy = -dx$, isn't it?. Thank you.
– NoOne
Jul 16 at 0:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let z be in the complex domain.
Then define $hatz = e^jthetaz$, where $theta$ is a constant.
If the derivative is taken, by using the fact that rotation won't change the area (area preserving operation), can we say that $dhatz = dz $?
Thank you very much.
derivatives
asked Jul 15 at 23:53
NoOne
32
Let z be in the complex domain.
Then define $hatz = e^jthetaz$, where $theta$ is a constant.
If the derivative is taken, by using the fact that rotation won't change the area (area preserving operation), can we say that $dhatz = dz $?
Thank you very much.
derivatives
asked Jul 15 at 23:53
NoOne
32
edited Jul 16 at 0:12
asked Jul 15 at 23:53
NoOne
32
asked Jul 15 at 23:53
NoOne
32
asked Jul 15 at 23:53
NoOne
32
32
closed as off-topic by amWhy, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, John Ma Jul 16 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Trần Thúc Minh TrÃÂ, Parcly Taxel, John Ma
closed as off-topic by amWhy, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, John Ma Jul 16 at 12:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Trần Thúc Minh TrÃÂ, Parcly Taxel, John Ma
Is $theta$ constant or supposed to be the argument of $z$? If constant, do you know how to compare dy and dx in y=(-1)*x?
– Mark S.
Jul 16 at 0:01
Yes, $theta$ is a constant. In this case $y = -x rightarrow y=e^-jpix rightarrow dy = -dx$, isn't it?. Thank you.
– NoOne
Jul 16 at 0:12
add a comment |Â
Is $theta$ constant or supposed to be the argument of $z$? If constant, do you know how to compare dy and dx in y=(-1)*x?
– Mark S.
Jul 16 at 0:01
Yes, $theta$ is a constant. In this case $y = -x rightarrow y=e^-jpix rightarrow dy = -dx$, isn't it?. Thank you.
– NoOne
Jul 16 at 0:12
Is $theta$ constant or supposed to be the argument of $z$? If constant, do you know how to compare dy and dx in y=(-1)*x?
– Mark S.
Jul 16 at 0:01
Is $theta$ constant or supposed to be the argument of $z$? If constant, do you know how to compare dy and dx in y=(-1)*x?
– Mark S.
Jul 16 at 0:01
Yes, $theta$ is a constant. In this case $y = -x rightarrow y=e^-jpix rightarrow dy = -dx$, isn't it?. Thank you.
– NoOne
Jul 16 at 0:12
Yes, $theta$ is a constant. In this case $y = -x rightarrow y=e^-jpix rightarrow dy = -dx$, isn't it?. Thank you.
– NoOne
Jul 16 at 0:12
add a comment |Â
1 Answer
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1
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Treat this as an extended comment, not really an answer.
I think you are confused here. Let $Z=Cz$ with $Cin mathbbC$. Then $dZ = Cdz$ and this has nothing to do with "area". This is a simple differential (more along the lines of length, than area).
If you are actually interested in area, then $dx dy$ is what you are looking for. Then writing $z=x+iy$, $Z=X+iY$ and $C=a+ib$, you get
$$
X+iY=(ax-by)+i(ay+bx)
$$
Then using a Jacobian
$$
dX dY = |beginvmatrixa & -b\
b & aendvmatrix| dxdy = (a^2+b^2) dxdy = |C|^2 dxdy
$$
Clearly, if $C=e^itheta$, then the area element remains unchanged (as it should be).
Alternatively, if you insist on using $dz$ in relation to area, the area element is $dz doverlinez$. From which you can again see that $dZdoverlineZ=|C|^2 dz doverlinez$.
answered Jul 16 at 1:53
Hamed
4,401421
Thank you very much.
– NoOne
Jul 16 at 2:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Treat this as an extended comment, not really an answer.
I think you are confused here. Let $Z=Cz$ with $Cin mathbbC$. Then $dZ = Cdz$ and this has nothing to do with "area". This is a simple differential (more along the lines of length, than area).
If you are actually interested in area, then $dx dy$ is what you are looking for. Then writing $z=x+iy$, $Z=X+iY$ and $C=a+ib$, you get
$$
X+iY=(ax-by)+i(ay+bx)
$$
Then using a Jacobian
$$
dX dY = |beginvmatrixa & -b\
b & aendvmatrix| dxdy = (a^2+b^2) dxdy = |C|^2 dxdy
$$
Clearly, if $C=e^itheta$, then the area element remains unchanged (as it should be).
Alternatively, if you insist on using $dz$ in relation to area, the area element is $dz doverlinez$. From which you can again see that $dZdoverlineZ=|C|^2 dz doverlinez$.
answered Jul 16 at 1:53
Hamed
4,401421
Thank you very much.
– NoOne
Jul 16 at 2:10
add a comment |Â
up vote
1
down vote
accepted
Treat this as an extended comment, not really an answer.
I think you are confused here. Let $Z=Cz$ with $Cin mathbbC$. Then $dZ = Cdz$ and this has nothing to do with "area". This is a simple differential (more along the lines of length, than area).
If you are actually interested in area, then $dx dy$ is what you are looking for. Then writing $z=x+iy$, $Z=X+iY$ and $C=a+ib$, you get
$$
X+iY=(ax-by)+i(ay+bx)
$$
Then using a Jacobian
$$
dX dY = |beginvmatrixa & -b\
b & aendvmatrix| dxdy = (a^2+b^2) dxdy = |C|^2 dxdy
$$
Clearly, if $C=e^itheta$, then the area element remains unchanged (as it should be).
Alternatively, if you insist on using $dz$ in relation to area, the area element is $dz doverlinez$. From which you can again see that $dZdoverlineZ=|C|^2 dz doverlinez$.
answered Jul 16 at 1:53
Hamed
4,401421
Thank you very much.
– NoOne
Jul 16 at 2:10
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Treat this as an extended comment, not really an answer.
I think you are confused here. Let $Z=Cz$ with $Cin mathbbC$. Then $dZ = Cdz$ and this has nothing to do with "area". This is a simple differential (more along the lines of length, than area).
If you are actually interested in area, then $dx dy$ is what you are looking for. Then writing $z=x+iy$, $Z=X+iY$ and $C=a+ib$, you get
$$
X+iY=(ax-by)+i(ay+bx)
$$
Then using a Jacobian
$$
dX dY = |beginvmatrixa & -b\
b & aendvmatrix| dxdy = (a^2+b^2) dxdy = |C|^2 dxdy
$$
Clearly, if $C=e^itheta$, then the area element remains unchanged (as it should be).
Alternatively, if you insist on using $dz$ in relation to area, the area element is $dz doverlinez$. From which you can again see that $dZdoverlineZ=|C|^2 dz doverlinez$.
answered Jul 16 at 1:53
Hamed
4,401421
Treat this as an extended comment, not really an answer.
I think you are confused here. Let $Z=Cz$ with $Cin mathbbC$. Then $dZ = Cdz$ and this has nothing to do with "area". This is a simple differential (more along the lines of length, than area).
If you are actually interested in area, then $dx dy$ is what you are looking for. Then writing $z=x+iy$, $Z=X+iY$ and $C=a+ib$, you get
$$
X+iY=(ax-by)+i(ay+bx)
$$
Then using a Jacobian
$$
dX dY = |beginvmatrixa & -b\
b & aendvmatrix| dxdy = (a^2+b^2) dxdy = |C|^2 dxdy
$$
Clearly, if $C=e^itheta$, then the area element remains unchanged (as it should be).
Alternatively, if you insist on using $dz$ in relation to area, the area element is $dz doverlinez$. From which you can again see that $dZdoverlineZ=|C|^2 dz doverlinez$.
answered Jul 16 at 1:53
Hamed
4,401421
answered Jul 16 at 1:53
Hamed
4,401421
answered Jul 16 at 1:53
Hamed
4,401421
answered Jul 16 at 1:53
Hamed
4,401421
4,401421
Thank you very much.
– NoOne
Jul 16 at 2:10
add a comment |Â
Thank you very much.
– NoOne
Jul 16 at 2:10
Thank you very much.
– NoOne
Jul 16 at 2:10
Thank you very much.
– NoOne
Jul 16 at 2:10
add a comment |Â
Is $theta$ constant or supposed to be the argument of $z$? If constant, do you know how to compare dy and dx in y=(-1)*x?
– Mark S.
Jul 16 at 0:01
Yes, $theta$ is a constant. In this case $y = -x rightarrow y=e^-jpix rightarrow dy = -dx$, isn't it?. Thank you.
– NoOne
Jul 16 at 0:12