Finding the area of a polygon with complex numbers.
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If we consider a regular polygon defined by the $n$-tuple $Z=(z_0, z_1, dots, z_{n-1})$, and set
$$A(Z)=frac{1}{2} Im left ( sum_{k=0}^{n-1} bar{z_k} z_{k+1} right )$$
Given that the number $Im(bar z_0 z_1)$ represents the signed area of the triangle $T=(0, z_0, z_1)$ (that's not necessarily positively oriented).
My question is: Can we argue that $A(Z)$ is the area of the polygon $Z$ purely based on the above, by considering the triangles $(0, z_k, z_{k+1})_k$, the answer I got suggested that since the areas are signed, the additional areas given by the expression $A(Z)$ 'cancel each other out', and we're left the desired answer (the center of $Z$ is not $0$ in general).
Surely this seems like a reasonable thing to say, but I'm looking for a showcase for why it's the true since I couldn't convince myself it is.
P.S. I do realize there is a similar question to this at Area of a complex polygon but I'm not trying to just prove the result, I need to deduce it from the areas of the $n$ triangles described above.
complex-numbers area polygons
asked Nov 21 at 13:46
FuzzyPixelz
337214
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up vote
0
down vote
favorite
If we consider a regular polygon defined by the $n$-tuple $Z=(z_0, z_1, dots, z_{n-1})$, and set
$$A(Z)=frac{1}{2} Im left ( sum_{k=0}^{n-1} bar{z_k} z_{k+1} right )$$
Given that the number $Im(bar z_0 z_1)$ represents the signed area of the triangle $T=(0, z_0, z_1)$ (that's not necessarily positively oriented).
My question is: Can we argue that $A(Z)$ is the area of the polygon $Z$ purely based on the above, by considering the triangles $(0, z_k, z_{k+1})_k$, the answer I got suggested that since the areas are signed, the additional areas given by the expression $A(Z)$ 'cancel each other out', and we're left the desired answer (the center of $Z$ is not $0$ in general).
Surely this seems like a reasonable thing to say, but I'm looking for a showcase for why it's the true since I couldn't convince myself it is.
P.S. I do realize there is a similar question to this at Area of a complex polygon but I'm not trying to just prove the result, I need to deduce it from the areas of the $n$ triangles described above.
complex-numbers area polygons
asked Nov 21 at 13:46
FuzzyPixelz
337214
You might find this interesting (both the whole video at the animation at 7m47s in): youtube.com/watch?v=0KjG8Pg6LGk&t=7m47s
– Akiva Weinberger
Nov 21 at 14:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we consider a regular polygon defined by the $n$-tuple $Z=(z_0, z_1, dots, z_{n-1})$, and set
$$A(Z)=frac{1}{2} Im left ( sum_{k=0}^{n-1} bar{z_k} z_{k+1} right )$$
Given that the number $Im(bar z_0 z_1)$ represents the signed area of the triangle $T=(0, z_0, z_1)$ (that's not necessarily positively oriented).
My question is: Can we argue that $A(Z)$ is the area of the polygon $Z$ purely based on the above, by considering the triangles $(0, z_k, z_{k+1})_k$, the answer I got suggested that since the areas are signed, the additional areas given by the expression $A(Z)$ 'cancel each other out', and we're left the desired answer (the center of $Z$ is not $0$ in general).
Surely this seems like a reasonable thing to say, but I'm looking for a showcase for why it's the true since I couldn't convince myself it is.
P.S. I do realize there is a similar question to this at Area of a complex polygon but I'm not trying to just prove the result, I need to deduce it from the areas of the $n$ triangles described above.
complex-numbers area polygons
asked Nov 21 at 13:46
FuzzyPixelz
337214
If we consider a regular polygon defined by the $n$-tuple $Z=(z_0, z_1, dots, z_{n-1})$, and set
$$A(Z)=frac{1}{2} Im left ( sum_{k=0}^{n-1} bar{z_k} z_{k+1} right )$$
Given that the number $Im(bar z_0 z_1)$ represents the signed area of the triangle $T=(0, z_0, z_1)$ (that's not necessarily positively oriented).
My question is: Can we argue that $A(Z)$ is the area of the polygon $Z$ purely based on the above, by considering the triangles $(0, z_k, z_{k+1})_k$, the answer I got suggested that since the areas are signed, the additional areas given by the expression $A(Z)$ 'cancel each other out', and we're left the desired answer (the center of $Z$ is not $0$ in general).
Surely this seems like a reasonable thing to say, but I'm looking for a showcase for why it's the true since I couldn't convince myself it is.
P.S. I do realize there is a similar question to this at Area of a complex polygon but I'm not trying to just prove the result, I need to deduce it from the areas of the $n$ triangles described above.
complex-numbers area polygons
complex-numbers area polygons
asked Nov 21 at 13:46
FuzzyPixelz
337214
asked Nov 21 at 13:46
FuzzyPixelz
337214
asked Nov 21 at 13:46
FuzzyPixelz
337214
asked Nov 21 at 13:46
FuzzyPixelz
337214
asked Nov 21 at 13:46
FuzzyPixelz
337214
337214
You might find this interesting (both the whole video at the animation at 7m47s in): youtube.com/watch?v=0KjG8Pg6LGk&t=7m47s
– Akiva Weinberger
Nov 21 at 14:00
add a comment |
You might find this interesting (both the whole video at the animation at 7m47s in): youtube.com/watch?v=0KjG8Pg6LGk&t=7m47s
– Akiva Weinberger
Nov 21 at 14:00
You might find this interesting (both the whole video at the animation at 7m47s in): youtube.com/watch?v=0KjG8Pg6LGk&t=7m47s
– Akiva Weinberger
Nov 21 at 14:00
You might find this interesting (both the whole video at the animation at 7m47s in): youtube.com/watch?v=0KjG8Pg6LGk&t=7m47s
– Akiva Weinberger
Nov 21 at 14:00
add a comment |
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You might find this interesting (both the whole video at the animation at 7m47s in): youtube.com/watch?v=0KjG8Pg6LGk&t=7m47s
– Akiva Weinberger
Nov 21 at 14:00