What does this set mean of $mathbbC $ in complex plane? [closed]
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A = $z in mathbbC ; Im(dfracz-ab)>0, b neq0$;
complex-analysis complex-numbers
asked Aug 17 at 2:30
Ricardo Freire
1788
closed as unclear what you're asking by Xander Henderson, Leucippus, Siong Thye Goh, Theoretical Economist, Arnaud D. Aug 17 at 10:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-1
down vote
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A = $z in mathbbC ; Im(dfracz-ab)>0, b neq0$;
complex-analysis complex-numbers
asked Aug 17 at 2:30
Ricardo Freire
1788
closed as unclear what you're asking by Xander Henderson, Leucippus, Siong Thye Goh, Theoretical Economist, Arnaud D. Aug 17 at 10:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
what's specifically $a$?
– Gonzalo Benavides
Aug 17 at 2:49
this exercise does not say anything about $a$.
– Ricardo Freire
Aug 17 at 3:06
Indeed, as @GonzaloBenavides asks, precisely what are $a$ and $b$? Both real? Given beforehand? I hope you don’t mean for $a$ and $b$ to be the real and imaginary parts of $z$.
– Lubin
Aug 17 at 3:08
Well, I suppose that if $a=b=1$, it should be easy to answer the question.
– Lubin
Aug 17 at 3:12
@RicardoFreire you just mean $a$ and $b$ to be complex numbers, right? Is this exercise in a section or chapter about convex envelopes of open sets in the plane?
– Van Latimer
Aug 17 at 18:46
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
A = $z in mathbbC ; Im(dfracz-ab)>0, b neq0$;
complex-analysis complex-numbers
asked Aug 17 at 2:30
Ricardo Freire
1788
A = $z in mathbbC ; Im(dfracz-ab)>0, b neq0$;
complex-analysis complex-numbers
asked Aug 17 at 2:30
Ricardo Freire
1788
asked Aug 17 at 2:30
Ricardo Freire
1788
asked Aug 17 at 2:30
Ricardo Freire
1788
asked Aug 17 at 2:30
Ricardo Freire
1788
1788
closed as unclear what you're asking by Xander Henderson, Leucippus, Siong Thye Goh, Theoretical Economist, Arnaud D. Aug 17 at 10:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Xander Henderson, Leucippus, Siong Thye Goh, Theoretical Economist, Arnaud D. Aug 17 at 10:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
what's specifically $a$?
– Gonzalo Benavides
Aug 17 at 2:49
this exercise does not say anything about $a$.
– Ricardo Freire
Aug 17 at 3:06
Indeed, as @GonzaloBenavides asks, precisely what are $a$ and $b$? Both real? Given beforehand? I hope you don’t mean for $a$ and $b$ to be the real and imaginary parts of $z$.
– Lubin
Aug 17 at 3:08
Well, I suppose that if $a=b=1$, it should be easy to answer the question.
– Lubin
Aug 17 at 3:12
@RicardoFreire you just mean $a$ and $b$ to be complex numbers, right? Is this exercise in a section or chapter about convex envelopes of open sets in the plane?
– Van Latimer
Aug 17 at 18:46
add a comment |Â
1
what's specifically $a$?
– Gonzalo Benavides
Aug 17 at 2:49
this exercise does not say anything about $a$.
– Ricardo Freire
Aug 17 at 3:06
Indeed, as @GonzaloBenavides asks, precisely what are $a$ and $b$? Both real? Given beforehand? I hope you don’t mean for $a$ and $b$ to be the real and imaginary parts of $z$.
– Lubin
Aug 17 at 3:08
Well, I suppose that if $a=b=1$, it should be easy to answer the question.
– Lubin
Aug 17 at 3:12
@RicardoFreire you just mean $a$ and $b$ to be complex numbers, right? Is this exercise in a section or chapter about convex envelopes of open sets in the plane?
– Van Latimer
Aug 17 at 18:46
1
1
what's specifically $a$?
– Gonzalo Benavides
Aug 17 at 2:49
what's specifically $a$?
– Gonzalo Benavides
Aug 17 at 2:49
this exercise does not say anything about $a$.
– Ricardo Freire
Aug 17 at 3:06
this exercise does not say anything about $a$.
– Ricardo Freire
Aug 17 at 3:06
Indeed, as @GonzaloBenavides asks, precisely what are $a$ and $b$? Both real? Given beforehand? I hope you don’t mean for $a$ and $b$ to be the real and imaginary parts of $z$.
– Lubin
Aug 17 at 3:08
Indeed, as @GonzaloBenavides asks, precisely what are $a$ and $b$? Both real? Given beforehand? I hope you don’t mean for $a$ and $b$ to be the real and imaginary parts of $z$.
– Lubin
Aug 17 at 3:08
Well, I suppose that if $a=b=1$, it should be easy to answer the question.
– Lubin
Aug 17 at 3:12
Well, I suppose that if $a=b=1$, it should be easy to answer the question.
– Lubin
Aug 17 at 3:12
@RicardoFreire you just mean $a$ and $b$ to be complex numbers, right? Is this exercise in a section or chapter about convex envelopes of open sets in the plane?
– Van Latimer
Aug 17 at 18:46
@RicardoFreire you just mean $a$ and $b$ to be complex numbers, right? Is this exercise in a section or chapter about convex envelopes of open sets in the plane?
– Van Latimer
Aug 17 at 18:46
add a comment |Â
1 Answer
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This type of set shows up when talking about convex envelopes and things like that. Given complex numbers $a$ and $b$, imagine the line in the complex plane starting at $a$ and pointing in the direction of $b$. This set then gives all complex number to one side of this line. How is this? Given a complex number $z$, imagine $z-a$ as the vector from $a$ to $z$. Then $argfracz-ab =arg(z-a) - arg b$ (make sure you use a branch of arg so that this quantity is positive). If then $0 <argfracz-ab< pi$, that is, if $fracz-ab$ has positive imaginary part, we have that the vector $z-a$ is gotten to from the vector $b$ (which we are imagining as having its tail at $a$) by rotating counter clockwise by between 0 and 180 degrees. This is precisely to say that $z$ is on the side of the line counterclockwise from $b$ (the left if rotate our paper so that $b$ points north).
answered Aug 17 at 3:51
Van Latimer
1298
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
This type of set shows up when talking about convex envelopes and things like that. Given complex numbers $a$ and $b$, imagine the line in the complex plane starting at $a$ and pointing in the direction of $b$. This set then gives all complex number to one side of this line. How is this? Given a complex number $z$, imagine $z-a$ as the vector from $a$ to $z$. Then $argfracz-ab =arg(z-a) - arg b$ (make sure you use a branch of arg so that this quantity is positive). If then $0 <argfracz-ab< pi$, that is, if $fracz-ab$ has positive imaginary part, we have that the vector $z-a$ is gotten to from the vector $b$ (which we are imagining as having its tail at $a$) by rotating counter clockwise by between 0 and 180 degrees. This is precisely to say that $z$ is on the side of the line counterclockwise from $b$ (the left if rotate our paper so that $b$ points north).
answered Aug 17 at 3:51
Van Latimer
1298
add a comment |Â
up vote
1
down vote
This type of set shows up when talking about convex envelopes and things like that. Given complex numbers $a$ and $b$, imagine the line in the complex plane starting at $a$ and pointing in the direction of $b$. This set then gives all complex number to one side of this line. How is this? Given a complex number $z$, imagine $z-a$ as the vector from $a$ to $z$. Then $argfracz-ab =arg(z-a) - arg b$ (make sure you use a branch of arg so that this quantity is positive). If then $0 <argfracz-ab< pi$, that is, if $fracz-ab$ has positive imaginary part, we have that the vector $z-a$ is gotten to from the vector $b$ (which we are imagining as having its tail at $a$) by rotating counter clockwise by between 0 and 180 degrees. This is precisely to say that $z$ is on the side of the line counterclockwise from $b$ (the left if rotate our paper so that $b$ points north).
answered Aug 17 at 3:51
Van Latimer
1298
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This type of set shows up when talking about convex envelopes and things like that. Given complex numbers $a$ and $b$, imagine the line in the complex plane starting at $a$ and pointing in the direction of $b$. This set then gives all complex number to one side of this line. How is this? Given a complex number $z$, imagine $z-a$ as the vector from $a$ to $z$. Then $argfracz-ab =arg(z-a) - arg b$ (make sure you use a branch of arg so that this quantity is positive). If then $0 <argfracz-ab< pi$, that is, if $fracz-ab$ has positive imaginary part, we have that the vector $z-a$ is gotten to from the vector $b$ (which we are imagining as having its tail at $a$) by rotating counter clockwise by between 0 and 180 degrees. This is precisely to say that $z$ is on the side of the line counterclockwise from $b$ (the left if rotate our paper so that $b$ points north).
answered Aug 17 at 3:51
Van Latimer
1298
This type of set shows up when talking about convex envelopes and things like that. Given complex numbers $a$ and $b$, imagine the line in the complex plane starting at $a$ and pointing in the direction of $b$. This set then gives all complex number to one side of this line. How is this? Given a complex number $z$, imagine $z-a$ as the vector from $a$ to $z$. Then $argfracz-ab =arg(z-a) - arg b$ (make sure you use a branch of arg so that this quantity is positive). If then $0 <argfracz-ab< pi$, that is, if $fracz-ab$ has positive imaginary part, we have that the vector $z-a$ is gotten to from the vector $b$ (which we are imagining as having its tail at $a$) by rotating counter clockwise by between 0 and 180 degrees. This is precisely to say that $z$ is on the side of the line counterclockwise from $b$ (the left if rotate our paper so that $b$ points north).
answered Aug 17 at 3:51
Van Latimer
1298
edited Aug 17 at 18:47
answered Aug 17 at 3:51
Van Latimer
1298
answered Aug 17 at 3:51
Van Latimer
1298
answered Aug 17 at 3:51
Van Latimer
1298
1298
add a comment |Â
add a comment |Â
1
what's specifically $a$?
– Gonzalo Benavides
Aug 17 at 2:49
this exercise does not say anything about $a$.
– Ricardo Freire
Aug 17 at 3:06
Indeed, as @GonzaloBenavides asks, precisely what are $a$ and $b$? Both real? Given beforehand? I hope you don’t mean for $a$ and $b$ to be the real and imaginary parts of $z$.
– Lubin
Aug 17 at 3:08
Well, I suppose that if $a=b=1$, it should be easy to answer the question.
– Lubin
Aug 17 at 3:12
@RicardoFreire you just mean $a$ and $b$ to be complex numbers, right? Is this exercise in a section or chapter about convex envelopes of open sets in the plane?
– Van Latimer
Aug 17 at 18:46