Complex numbers z=-3-4i polar form Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraExpressing $e^z$ where $z=a+bi$ in polar form.Polar form of Complex numbersConverting to polar formComplex number polar form equationWhy is e used for polar form of complex numbers?Show $-27$ in polar form.writing in polar form (complex numbers)Are all complex exponentials in polar form?Complex numbers polar form changeDifferentiating polar functions using complex numbers
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Complex numbers z=-3-4i polar form
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraExpressing $e^z$ where $z=a+bi$ in polar form.Polar form of Complex numbersConverting to polar formComplex number polar form equationWhy is e used for polar form of complex numbers?Show $-27$ in polar form.writing in polar form (complex numbers)Are all complex exponentials in polar form?Complex numbers polar form changeDifferentiating polar functions using complex numbers
$begingroup$
$z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.
complex-numbers polar-coordinates
edited 3 hours ago
José Carlos Santos
178k24139251
asked 3 hours ago
user221435user221435
285
$endgroup$
add a comment |
$begingroup$
$z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.
complex-numbers polar-coordinates
edited 3 hours ago
José Carlos Santos
178k24139251
asked 3 hours ago
user221435user221435
285
$endgroup$
add a comment |
$begingroup$
$z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.
complex-numbers polar-coordinates
edited 3 hours ago
José Carlos Santos
178k24139251
asked 3 hours ago
user221435user221435
285
$endgroup$
$z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.
complex-numbers polar-coordinates
complex-numbers polar-coordinates
edited 3 hours ago
José Carlos Santos
178k24139251
asked 3 hours ago
user221435user221435
285
edited 3 hours ago
José Carlos Santos
178k24139251
asked 3 hours ago
user221435user221435
285
edited 3 hours ago
José Carlos Santos
178k24139251
edited 3 hours ago
José Carlos Santos
178k24139251
edited 3 hours ago
José Carlos Santos
178k24139251
178k24139251
asked 3 hours ago
user221435user221435
285
asked 3 hours ago
user221435user221435
285
asked 3 hours ago
user221435user221435
285
285
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
$begingroup$
Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
$endgroup$
– user221435
3 hours ago
2
$begingroup$
We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
Great, thank you
$endgroup$
– user221435
3 hours ago
$begingroup$
Which makes mathematicians very different from 3D modelers and animators, where it really matters if the object turns 1 time (360°) or 10 times (3600°). The end result is the same, but all the pictures in between will be different.
$endgroup$
– Thomas Weller
45 mins ago
1
$begingroup$
@ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
$endgroup$
– DreamConspiracy
24 mins ago
|
show 1 more comment
$begingroup$
It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.
The principal angle is an angle between $-180^circ$ and $+180^circ$
edited 2 hours ago
J. W. Tanner
5,2001520
answered 3 hours ago
Fareed AFFareed AF
1,025214
$endgroup$
add a comment |
$begingroup$
The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.
By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.
answered 15 mins ago
ArthurArthur
123k7122211
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
$begingroup$
Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
$endgroup$
– user221435
3 hours ago
2
$begingroup$
We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
Great, thank you
$endgroup$
– user221435
3 hours ago
$begingroup$
Which makes mathematicians very different from 3D modelers and animators, where it really matters if the object turns 1 time (360°) or 10 times (3600°). The end result is the same, but all the pictures in between will be different.
$endgroup$
– Thomas Weller
45 mins ago
1
$begingroup$
@ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
$endgroup$
– DreamConspiracy
24 mins ago
|
show 1 more comment
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
$begingroup$
Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
$endgroup$
– user221435
3 hours ago
2
$begingroup$
We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
Great, thank you
$endgroup$
– user221435
3 hours ago
$begingroup$
Which makes mathematicians very different from 3D modelers and animators, where it really matters if the object turns 1 time (360°) or 10 times (3600°). The end result is the same, but all the pictures in between will be different.
$endgroup$
– Thomas Weller
45 mins ago
1
$begingroup$
@ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
$endgroup$
– DreamConspiracy
24 mins ago
|
show 1 more comment
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
It's because$$cos(x-360^circ)=cos(x)text and sin(x-360^circ)=sin(x).$$
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
178k24139251
178k24139251
$begingroup$
Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
$endgroup$
– user221435
3 hours ago
2
$begingroup$
We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
Great, thank you
$endgroup$
– user221435
3 hours ago
$begingroup$
Which makes mathematicians very different from 3D modelers and animators, where it really matters if the object turns 1 time (360°) or 10 times (3600°). The end result is the same, but all the pictures in between will be different.
$endgroup$
– Thomas Weller
45 mins ago
1
$begingroup$
@ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
$endgroup$
– DreamConspiracy
24 mins ago
|
show 1 more comment
$begingroup$
Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
$endgroup$
– user221435
3 hours ago
2
$begingroup$
We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
Great, thank you
$endgroup$
– user221435
3 hours ago
$begingroup$
Which makes mathematicians very different from 3D modelers and animators, where it really matters if the object turns 1 time (360°) or 10 times (3600°). The end result is the same, but all the pictures in between will be different.
$endgroup$
– Thomas Weller
45 mins ago
1
$begingroup$
@ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
$endgroup$
– DreamConspiracy
24 mins ago
$begingroup$
Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
$endgroup$
– user221435
3 hours ago
$begingroup$
Usually I don't do this when I get the degree, why is it that in this case I needed to do this?
$endgroup$
– user221435
3 hours ago
2
2
$begingroup$
We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
We don't need to do this, but we usually try to work with angles in the range $[0^circ,360^circ)$.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
Great, thank you
$endgroup$
– user221435
3 hours ago
$begingroup$
Great, thank you
$endgroup$
– user221435
3 hours ago
$begingroup$
Which makes mathematicians very different from 3D modelers and animators, where it really matters if the object turns 1 time (360°) or 10 times (3600°). The end result is the same, but all the pictures in between will be different.
$endgroup$
– Thomas Weller
45 mins ago
$begingroup$
Which makes mathematicians very different from 3D modelers and animators, where it really matters if the object turns 1 time (360°) or 10 times (3600°). The end result is the same, but all the pictures in between will be different.
$endgroup$
– Thomas Weller
45 mins ago
1
1
$begingroup$
@ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
$endgroup$
– DreamConspiracy
24 mins ago
$begingroup$
@ThomasWeller this is not exactly right. In the case above, the value measured in degrees is an angle, which should always be in the range given. In modelers and animators, the value given is something like angular distance (ie the some of the absolute values of changes in angle over time) which happens to be measured in the same units as angles (degrees, radians), but is not the same type of value. Because of this, applying trigonometric functions to angular distances is not meaningful (without further assumptions)
$endgroup$
– DreamConspiracy
24 mins ago
|
show 1 more comment
$begingroup$
It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.
The principal angle is an angle between $-180^circ$ and $+180^circ$
edited 2 hours ago
J. W. Tanner
5,2001520
answered 3 hours ago
Fareed AFFareed AF
1,025214
$endgroup$
add a comment |
$begingroup$
It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.
The principal angle is an angle between $-180^circ$ and $+180^circ$
edited 2 hours ago
J. W. Tanner
5,2001520
answered 3 hours ago
Fareed AFFareed AF
1,025214
$endgroup$
add a comment |
$begingroup$
It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.
The principal angle is an angle between $-180^circ$ and $+180^circ$
edited 2 hours ago
J. W. Tanner
5,2001520
answered 3 hours ago
Fareed AFFareed AF
1,025214
$endgroup$
It is just to get the principal value of the angle, since if you rotate by an angle $466^circ$ you'll get to the same position as rotating $106^circ$ so we usually take the smallest angle that is needed to arrive at the desired position.
The principal angle is an angle between $-180^circ$ and $+180^circ$
edited 2 hours ago
J. W. Tanner
5,2001520
answered 3 hours ago
Fareed AFFareed AF
1,025214
edited 2 hours ago
J. W. Tanner
5,2001520
edited 2 hours ago
J. W. Tanner
5,2001520
edited 2 hours ago
J. W. Tanner
5,2001520
5,2001520
answered 3 hours ago
Fareed AFFareed AF
1,025214
answered 3 hours ago
Fareed AFFareed AF
1,025214
answered 3 hours ago
Fareed AFFareed AF
1,025214
1,025214
add a comment |
add a comment |
$begingroup$
The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.
By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.
answered 15 mins ago
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.
By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.
answered 15 mins ago
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.
By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.
answered 15 mins ago
ArthurArthur
123k7122211
$endgroup$
The polar form of a complex number is given by a distance from the origin and an angle against the positive real axis ("$x$-axis"). Increasing or decreasing this angle by $360^circ$ will result in the same point. So adding or subtracting multiples of $360^circ$ from the angle component of a set of polar coordinates will not change which point those coordinates represent.
By convention, we usually like this angle to be either in the range $[0^circ, 360^circ)$ or $(-180^circ, 180^circ]$. This is not a requirement by any means (unless explicitly stated in the exercise), but it's easier to tell by a glance what direction from the origin is represented by an angle of $270^circ$ than by $2430^circ$. So there is some merit to keeping the numbers small.
answered 15 mins ago
ArthurArthur
123k7122211
answered 15 mins ago
ArthurArthur
123k7122211
answered 15 mins ago
ArthurArthur
123k7122211
answered 15 mins ago
ArthurArthur
123k7122211
123k7122211
add a comment |
add a comment |
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