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How to subtract complex numbers in polar form?
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I tried to subtract two complex numbers in polar form without transforming them into the cartesian form. Therefore I used the approach made by Mark Viola in the following link.
Adding two polar vectors
I managed to get the following result.
$$e^i(phi-phi_1)=fracr_1-r_2e^i(phi_2-phi_1)sqrtr_1^2+r_2^2-2r_1r_2cos (phi_2-phi_1) tag 1$$
At this point I do not know how to achieve the final equation like mentioned in the link. I post the final equation for adding two complex numbers in polar form:
$$phi=phi_1+operatornamearctan2left(r_2sin(phi_2-phi_1),r_1+r_2cos(phi_2-phi_1)right) tag 2$$
Any help or hints which leading to the final equation are appreciated.
complex-numbers vectors polar-coordinates
asked Jul 23 at 16:18
Pavel
133
add a comment |Â
up vote
1
down vote
favorite
I tried to subtract two complex numbers in polar form without transforming them into the cartesian form. Therefore I used the approach made by Mark Viola in the following link.
Adding two polar vectors
I managed to get the following result.
$$e^i(phi-phi_1)=fracr_1-r_2e^i(phi_2-phi_1)sqrtr_1^2+r_2^2-2r_1r_2cos (phi_2-phi_1) tag 1$$
At this point I do not know how to achieve the final equation like mentioned in the link. I post the final equation for adding two complex numbers in polar form:
$$phi=phi_1+operatornamearctan2left(r_2sin(phi_2-phi_1),r_1+r_2cos(phi_2-phi_1)right) tag 2$$
Any help or hints which leading to the final equation are appreciated.
complex-numbers vectors polar-coordinates
asked Jul 23 at 16:18
Pavel
133
1
Hint: to compute $z_1 - z_2$, you can compute $z_1 + (-z_2)$. So if you can negate $z_2$ (in polar form), then you can apply the addition formula from the other page to get the result you want. Hint: to negate $(r, theta)$, try $(r, theta + pi)$.
– John Hughes
Jul 23 at 16:26
Thanks for the hint and the quick answer, I will try your approach.
– Pavel
Jul 25 at 19:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I tried to subtract two complex numbers in polar form without transforming them into the cartesian form. Therefore I used the approach made by Mark Viola in the following link.
Adding two polar vectors
I managed to get the following result.
$$e^i(phi-phi_1)=fracr_1-r_2e^i(phi_2-phi_1)sqrtr_1^2+r_2^2-2r_1r_2cos (phi_2-phi_1) tag 1$$
At this point I do not know how to achieve the final equation like mentioned in the link. I post the final equation for adding two complex numbers in polar form:
$$phi=phi_1+operatornamearctan2left(r_2sin(phi_2-phi_1),r_1+r_2cos(phi_2-phi_1)right) tag 2$$
Any help or hints which leading to the final equation are appreciated.
complex-numbers vectors polar-coordinates
asked Jul 23 at 16:18
Pavel
133
I tried to subtract two complex numbers in polar form without transforming them into the cartesian form. Therefore I used the approach made by Mark Viola in the following link.
Adding two polar vectors
I managed to get the following result.
$$e^i(phi-phi_1)=fracr_1-r_2e^i(phi_2-phi_1)sqrtr_1^2+r_2^2-2r_1r_2cos (phi_2-phi_1) tag 1$$
At this point I do not know how to achieve the final equation like mentioned in the link. I post the final equation for adding two complex numbers in polar form:
$$phi=phi_1+operatornamearctan2left(r_2sin(phi_2-phi_1),r_1+r_2cos(phi_2-phi_1)right) tag 2$$
Any help or hints which leading to the final equation are appreciated.
complex-numbers vectors polar-coordinates
asked Jul 23 at 16:18
Pavel
133
edited Jul 23 at 16:29
asked Jul 23 at 16:18
Pavel
133
asked Jul 23 at 16:18
Pavel
133
asked Jul 23 at 16:18
Pavel
133
133
1
Hint: to compute $z_1 - z_2$, you can compute $z_1 + (-z_2)$. So if you can negate $z_2$ (in polar form), then you can apply the addition formula from the other page to get the result you want. Hint: to negate $(r, theta)$, try $(r, theta + pi)$.
– John Hughes
Jul 23 at 16:26
Thanks for the hint and the quick answer, I will try your approach.
– Pavel
Jul 25 at 19:12
add a comment |Â
1
Hint: to compute $z_1 - z_2$, you can compute $z_1 + (-z_2)$. So if you can negate $z_2$ (in polar form), then you can apply the addition formula from the other page to get the result you want. Hint: to negate $(r, theta)$, try $(r, theta + pi)$.
– John Hughes
Jul 23 at 16:26
Thanks for the hint and the quick answer, I will try your approach.
– Pavel
Jul 25 at 19:12
1
1
Hint: to compute $z_1 - z_2$, you can compute $z_1 + (-z_2)$. So if you can negate $z_2$ (in polar form), then you can apply the addition formula from the other page to get the result you want. Hint: to negate $(r, theta)$, try $(r, theta + pi)$.
– John Hughes
Jul 23 at 16:26
Hint: to compute $z_1 - z_2$, you can compute $z_1 + (-z_2)$. So if you can negate $z_2$ (in polar form), then you can apply the addition formula from the other page to get the result you want. Hint: to negate $(r, theta)$, try $(r, theta + pi)$.
– John Hughes
Jul 23 at 16:26
Thanks for the hint and the quick answer, I will try your approach.
– Pavel
Jul 25 at 19:12
Thanks for the hint and the quick answer, I will try your approach.
– Pavel
Jul 25 at 19:12
add a comment |Â
1 Answer
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oldest
votes
up vote
1
down vote
accepted
In fact, you can't avoid the conversion from polar to Cartesian and back to polar, even if done in a single go (any expression you will find will be of a comparable complexity).
The combined equations are
$$r^2=(r_1cosphi_1-r_2cosphi_2)^2+(r_1sinphi_1-r_2sinphi_2)^2
\=r_1^2-2r_1r_2cos(phi_1-phi_2)+r_2^2$$
and
$$tanphi=fracr_1sinphi_1-r_2sinphi_2r_1cosphi_1-r_2cosphi_2.$$
I don't think there is a nice simplification of the expression of the argument. And computationally, the simplification of the modulus is not so attractive as it involves an additional evaluation of a circular function.
answered Jul 23 at 16:42
Yves Daoust
111k665203
Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $phi_2$ and $phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again
– Pavel
Jul 25 at 19:10
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
In fact, you can't avoid the conversion from polar to Cartesian and back to polar, even if done in a single go (any expression you will find will be of a comparable complexity).
The combined equations are
$$r^2=(r_1cosphi_1-r_2cosphi_2)^2+(r_1sinphi_1-r_2sinphi_2)^2
\=r_1^2-2r_1r_2cos(phi_1-phi_2)+r_2^2$$
and
$$tanphi=fracr_1sinphi_1-r_2sinphi_2r_1cosphi_1-r_2cosphi_2.$$
I don't think there is a nice simplification of the expression of the argument. And computationally, the simplification of the modulus is not so attractive as it involves an additional evaluation of a circular function.
answered Jul 23 at 16:42
Yves Daoust
111k665203
Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $phi_2$ and $phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again
– Pavel
Jul 25 at 19:10
add a comment |Â
up vote
1
down vote
accepted
In fact, you can't avoid the conversion from polar to Cartesian and back to polar, even if done in a single go (any expression you will find will be of a comparable complexity).
The combined equations are
$$r^2=(r_1cosphi_1-r_2cosphi_2)^2+(r_1sinphi_1-r_2sinphi_2)^2
\=r_1^2-2r_1r_2cos(phi_1-phi_2)+r_2^2$$
and
$$tanphi=fracr_1sinphi_1-r_2sinphi_2r_1cosphi_1-r_2cosphi_2.$$
I don't think there is a nice simplification of the expression of the argument. And computationally, the simplification of the modulus is not so attractive as it involves an additional evaluation of a circular function.
answered Jul 23 at 16:42
Yves Daoust
111k665203
Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $phi_2$ and $phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again
– Pavel
Jul 25 at 19:10
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In fact, you can't avoid the conversion from polar to Cartesian and back to polar, even if done in a single go (any expression you will find will be of a comparable complexity).
The combined equations are
$$r^2=(r_1cosphi_1-r_2cosphi_2)^2+(r_1sinphi_1-r_2sinphi_2)^2
\=r_1^2-2r_1r_2cos(phi_1-phi_2)+r_2^2$$
and
$$tanphi=fracr_1sinphi_1-r_2sinphi_2r_1cosphi_1-r_2cosphi_2.$$
I don't think there is a nice simplification of the expression of the argument. And computationally, the simplification of the modulus is not so attractive as it involves an additional evaluation of a circular function.
answered Jul 23 at 16:42
Yves Daoust
111k665203
In fact, you can't avoid the conversion from polar to Cartesian and back to polar, even if done in a single go (any expression you will find will be of a comparable complexity).
The combined equations are
$$r^2=(r_1cosphi_1-r_2cosphi_2)^2+(r_1sinphi_1-r_2sinphi_2)^2
\=r_1^2-2r_1r_2cos(phi_1-phi_2)+r_2^2$$
and
$$tanphi=fracr_1sinphi_1-r_2sinphi_2r_1cosphi_1-r_2cosphi_2.$$
I don't think there is a nice simplification of the expression of the argument. And computationally, the simplification of the modulus is not so attractive as it involves an additional evaluation of a circular function.
answered Jul 23 at 16:42
Yves Daoust
111k665203
answered Jul 23 at 16:42
Yves Daoust
111k665203
answered Jul 23 at 16:42
Yves Daoust
111k665203
answered Jul 23 at 16:42
Yves Daoust
111k665203
111k665203
Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $phi_2$ and $phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again
– Pavel
Jul 25 at 19:10
add a comment |Â
Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $phi_2$ and $phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again
– Pavel
Jul 25 at 19:10
Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $phi_2$ and $phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again
– Pavel
Jul 25 at 19:10
Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $phi_2$ and $phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again
– Pavel
Jul 25 at 19:10
add a comment |Â
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1
Hint: to compute $z_1 - z_2$, you can compute $z_1 + (-z_2)$. So if you can negate $z_2$ (in polar form), then you can apply the addition formula from the other page to get the result you want. Hint: to negate $(r, theta)$, try $(r, theta + pi)$.
– John Hughes
Jul 23 at 16:26
Thanks for the hint and the quick answer, I will try your approach.
– Pavel
Jul 25 at 19:12