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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 1 hour ago
José Carlos Santos
178k24139251
asked 1 hour 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 1 hour ago
José Carlos Santos
178k24139251
asked 1 hour 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 1 hour ago
José Carlos Santos
178k24139251
asked 1 hour 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 1 hour ago
José Carlos Santos
178k24139251
asked 1 hour ago
user221435user221435
285
edited 1 hour ago
José Carlos Santos
178k24139251
asked 1 hour ago
user221435user221435
285
edited 1 hour ago
José Carlos Santos
178k24139251
edited 1 hour ago
José Carlos Santos
178k24139251
edited 1 hour ago
José Carlos Santos
178k24139251
178k24139251
asked 1 hour ago
user221435user221435
285
asked 1 hour ago
user221435user221435
285
asked 1 hour ago
user221435user221435
285
285
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$
answered 1 hour 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
1 hour 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
1 hour ago
$begingroup$
Great, thank you
$endgroup$
– user221435
1 hour ago
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 1 hour ago
J. W. Tanner
5,2001520
answered 1 hour ago
Fareed AFFareed AF
1,025214
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$
answered 1 hour 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
1 hour 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
1 hour ago
$begingroup$
Great, thank you
$endgroup$
– user221435
1 hour ago
add a comment |
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$
answered 1 hour 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
1 hour 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
1 hour ago
$begingroup$
Great, thank you
$endgroup$
– user221435
1 hour ago
add a comment |
$begingroup$
It's because$$cos(x-360^circ)=cos(x)text{ and }sin(x-360^circ)=sin(x).$$
answered 1 hour 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 1 hour ago
José Carlos SantosJosé Carlos Santos
178k24139251
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
178k24139251
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
178k24139251
answered 1 hour 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
1 hour 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
1 hour ago
$begingroup$
Great, thank you
$endgroup$
– user221435
1 hour ago
add a 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
1 hour 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
1 hour ago
$begingroup$
Great, thank you
$endgroup$
– user221435
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
Great, thank you
$endgroup$
– user221435
1 hour ago
$begingroup$
Great, thank you
$endgroup$
– user221435
1 hour ago
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 1 hour ago
J. W. Tanner
5,2001520
answered 1 hour 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 1 hour ago
J. W. Tanner
5,2001520
answered 1 hour 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 1 hour ago
J. W. Tanner
5,2001520
answered 1 hour 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 1 hour ago
J. W. Tanner
5,2001520
answered 1 hour ago
Fareed AFFareed AF
1,025214
edited 1 hour ago
J. W. Tanner
5,2001520
edited 1 hour ago
J. W. Tanner
5,2001520
edited 1 hour ago
J. W. Tanner
5,2001520
5,2001520
answered 1 hour ago
Fareed AFFareed AF
1,025214
answered 1 hour ago
Fareed AFFareed AF
1,025214
answered 1 hour ago
Fareed AFFareed AF
1,025214
1,025214
add a comment |
add a comment |
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