Write $-3i$ in polar coordinates.
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can someone please help me with a trivial question?
Write $-3i$ in polar coordinates.
So $z=a+bi=rcistheta$ with $r=sqrta^2+b^2$ and $theta = arctanfracba$. However, what if $a=0$ such as the case for $-3i$? I am confused!
complex-numbers polar-coordinates
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up vote
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can someone please help me with a trivial question?
Write $-3i$ in polar coordinates.
So $z=a+bi=rcistheta$ with $r=sqrta^2+b^2$ and $theta = arctanfracba$. However, what if $a=0$ such as the case for $-3i$? I am confused!
complex-numbers polar-coordinates
Warning: the formula $theta = arctan (b/a)$ is only correct if $a>0$. (It's not only $a=0$ that would cause trouble, also $a<0$.) Just draw a picture instead.
â Hans Lundmark
Aug 18 at 8:58
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
can someone please help me with a trivial question?
Write $-3i$ in polar coordinates.
So $z=a+bi=rcistheta$ with $r=sqrta^2+b^2$ and $theta = arctanfracba$. However, what if $a=0$ such as the case for $-3i$? I am confused!
complex-numbers polar-coordinates
can someone please help me with a trivial question?
Write $-3i$ in polar coordinates.
So $z=a+bi=rcistheta$ with $r=sqrta^2+b^2$ and $theta = arctanfracba$. However, what if $a=0$ such as the case for $-3i$? I am confused!
complex-numbers polar-coordinates
asked Aug 18 at 6:16
numericalorange
1,276110
1,276110
Warning: the formula $theta = arctan (b/a)$ is only correct if $a>0$. (It's not only $a=0$ that would cause trouble, also $a<0$.) Just draw a picture instead.
â Hans Lundmark
Aug 18 at 8:58
add a comment |Â
Warning: the formula $theta = arctan (b/a)$ is only correct if $a>0$. (It's not only $a=0$ that would cause trouble, also $a<0$.) Just draw a picture instead.
â Hans Lundmark
Aug 18 at 8:58
Warning: the formula $theta = arctan (b/a)$ is only correct if $a>0$. (It's not only $a=0$ that would cause trouble, also $a<0$.) Just draw a picture instead.
â Hans Lundmark
Aug 18 at 8:58
Warning: the formula $theta = arctan (b/a)$ is only correct if $a>0$. (It's not only $a=0$ that would cause trouble, also $a<0$.) Just draw a picture instead.
â Hans Lundmark
Aug 18 at 8:58
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
According to a definition we always have $$-pi<thetale pi$$here we can write $$theta=tan^-1dfrac-30=-dfracpi2\r=3$$therefore $$-3i=3e^-idfracpi2$$
1
Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
â Truth-seek
Aug 18 at 7:24
Sure! All of these discussions are true in limit....
â Mostafa Ayaz
Aug 18 at 8:55
I love this idea of using limits! Thanks for the insightful response!
â numericalorange
Aug 18 at 21:31
You're welcome ^____^
â Mostafa Ayaz
Aug 19 at 20:05
add a comment |Â
up vote
2
down vote
Try to draw a picture. I think you will be able to see the angle:
Okay, sounds good. :) Thanks a lot.
â numericalorange
Aug 18 at 21:31
add a comment |Â
up vote
2
down vote
Think of the complex number $z=x+iy$ as a vector from the origin to the point $(x,y)$. Then, it can be characterized by the length $r=|z|$ of the this vector and the angle between the axis $x>0$ and the vector (calculated counterclockwise). Thus, $z=re^itheta$ where $r=sqrtx^2+y^2=3$ and the angle is $-fracpi2$. i.e, $z=3e^-ifracpi2$.
1
That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
â Deepak
Aug 18 at 6:52
exactly. thanks
â Ronald
Aug 18 at 6:53
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
According to a definition we always have $$-pi<thetale pi$$here we can write $$theta=tan^-1dfrac-30=-dfracpi2\r=3$$therefore $$-3i=3e^-idfracpi2$$
1
Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
â Truth-seek
Aug 18 at 7:24
Sure! All of these discussions are true in limit....
â Mostafa Ayaz
Aug 18 at 8:55
I love this idea of using limits! Thanks for the insightful response!
â numericalorange
Aug 18 at 21:31
You're welcome ^____^
â Mostafa Ayaz
Aug 19 at 20:05
add a comment |Â
up vote
2
down vote
accepted
According to a definition we always have $$-pi<thetale pi$$here we can write $$theta=tan^-1dfrac-30=-dfracpi2\r=3$$therefore $$-3i=3e^-idfracpi2$$
1
Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
â Truth-seek
Aug 18 at 7:24
Sure! All of these discussions are true in limit....
â Mostafa Ayaz
Aug 18 at 8:55
I love this idea of using limits! Thanks for the insightful response!
â numericalorange
Aug 18 at 21:31
You're welcome ^____^
â Mostafa Ayaz
Aug 19 at 20:05
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
According to a definition we always have $$-pi<thetale pi$$here we can write $$theta=tan^-1dfrac-30=-dfracpi2\r=3$$therefore $$-3i=3e^-idfracpi2$$
According to a definition we always have $$-pi<thetale pi$$here we can write $$theta=tan^-1dfrac-30=-dfracpi2\r=3$$therefore $$-3i=3e^-idfracpi2$$
answered Aug 18 at 6:56
Mostafa Ayaz
9,6633730
9,6633730
1
Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
â Truth-seek
Aug 18 at 7:24
Sure! All of these discussions are true in limit....
â Mostafa Ayaz
Aug 18 at 8:55
I love this idea of using limits! Thanks for the insightful response!
â numericalorange
Aug 18 at 21:31
You're welcome ^____^
â Mostafa Ayaz
Aug 19 at 20:05
add a comment |Â
1
Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
â Truth-seek
Aug 18 at 7:24
Sure! All of these discussions are true in limit....
â Mostafa Ayaz
Aug 18 at 8:55
I love this idea of using limits! Thanks for the insightful response!
â numericalorange
Aug 18 at 21:31
You're welcome ^____^
â Mostafa Ayaz
Aug 19 at 20:05
1
1
Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
â Truth-seek
Aug 18 at 7:24
Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
â Truth-seek
Aug 18 at 7:24
Sure! All of these discussions are true in limit....
â Mostafa Ayaz
Aug 18 at 8:55
Sure! All of these discussions are true in limit....
â Mostafa Ayaz
Aug 18 at 8:55
I love this idea of using limits! Thanks for the insightful response!
â numericalorange
Aug 18 at 21:31
I love this idea of using limits! Thanks for the insightful response!
â numericalorange
Aug 18 at 21:31
You're welcome ^____^
â Mostafa Ayaz
Aug 19 at 20:05
You're welcome ^____^
â Mostafa Ayaz
Aug 19 at 20:05
add a comment |Â
up vote
2
down vote
Try to draw a picture. I think you will be able to see the angle:
Okay, sounds good. :) Thanks a lot.
â numericalorange
Aug 18 at 21:31
add a comment |Â
up vote
2
down vote
Try to draw a picture. I think you will be able to see the angle:
Okay, sounds good. :) Thanks a lot.
â numericalorange
Aug 18 at 21:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Try to draw a picture. I think you will be able to see the angle:
Try to draw a picture. I think you will be able to see the angle:
answered Aug 18 at 6:39
Botond
3,9532632
3,9532632
Okay, sounds good. :) Thanks a lot.
â numericalorange
Aug 18 at 21:31
add a comment |Â
Okay, sounds good. :) Thanks a lot.
â numericalorange
Aug 18 at 21:31
Okay, sounds good. :) Thanks a lot.
â numericalorange
Aug 18 at 21:31
Okay, sounds good. :) Thanks a lot.
â numericalorange
Aug 18 at 21:31
add a comment |Â
up vote
2
down vote
Think of the complex number $z=x+iy$ as a vector from the origin to the point $(x,y)$. Then, it can be characterized by the length $r=|z|$ of the this vector and the angle between the axis $x>0$ and the vector (calculated counterclockwise). Thus, $z=re^itheta$ where $r=sqrtx^2+y^2=3$ and the angle is $-fracpi2$. i.e, $z=3e^-ifracpi2$.
1
That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
â Deepak
Aug 18 at 6:52
exactly. thanks
â Ronald
Aug 18 at 6:53
add a comment |Â
up vote
2
down vote
Think of the complex number $z=x+iy$ as a vector from the origin to the point $(x,y)$. Then, it can be characterized by the length $r=|z|$ of the this vector and the angle between the axis $x>0$ and the vector (calculated counterclockwise). Thus, $z=re^itheta$ where $r=sqrtx^2+y^2=3$ and the angle is $-fracpi2$. i.e, $z=3e^-ifracpi2$.
1
That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
â Deepak
Aug 18 at 6:52
exactly. thanks
â Ronald
Aug 18 at 6:53
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Think of the complex number $z=x+iy$ as a vector from the origin to the point $(x,y)$. Then, it can be characterized by the length $r=|z|$ of the this vector and the angle between the axis $x>0$ and the vector (calculated counterclockwise). Thus, $z=re^itheta$ where $r=sqrtx^2+y^2=3$ and the angle is $-fracpi2$. i.e, $z=3e^-ifracpi2$.
Think of the complex number $z=x+iy$ as a vector from the origin to the point $(x,y)$. Then, it can be characterized by the length $r=|z|$ of the this vector and the angle between the axis $x>0$ and the vector (calculated counterclockwise). Thus, $z=re^itheta$ where $r=sqrtx^2+y^2=3$ and the angle is $-fracpi2$. i.e, $z=3e^-ifracpi2$.
edited Aug 18 at 6:53
answered Aug 18 at 6:47
Ronald
1,6361821
1,6361821
1
That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
â Deepak
Aug 18 at 6:52
exactly. thanks
â Ronald
Aug 18 at 6:53
add a comment |Â
1
That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
â Deepak
Aug 18 at 6:52
exactly. thanks
â Ronald
Aug 18 at 6:53
1
1
That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
â Deepak
Aug 18 at 6:52
That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
â Deepak
Aug 18 at 6:52
exactly. thanks
â Ronald
Aug 18 at 6:53
exactly. thanks
â Ronald
Aug 18 at 6:53
add a comment |Â
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Warning: the formula $theta = arctan (b/a)$ is only correct if $a>0$. (It's not only $a=0$ that would cause trouble, also $a<0$.) Just draw a picture instead.
â Hans Lundmark
Aug 18 at 8:58