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!







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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.
    – Hans Lundmark
    Aug 18 at 8:58














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!







share|cite|improve this question




















  • 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.
    – Hans Lundmark
    Aug 18 at 8:58












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!







share|cite|improve this question












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!









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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.
    – 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.
    – 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.
– 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.
– Hans Lundmark
Aug 18 at 8:58










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 $$-3i=3e^-idfracpi2$$






share|cite|improve this answer
















  • 1




    Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
    – Truth-seek
    Aug 18 at 7:24











  • Sure! All of these discussions are true in limit....
    – Mostafa Ayaz
    Aug 18 at 8:55










  • I love this idea of using limits! Thanks for the insightful response!
    – numericalorange
    Aug 18 at 21:31










  • You're welcome ^____^
    – Mostafa Ayaz
    Aug 19 at 20:05

















up vote
2
down vote













Try to draw a picture. I think you will be able to see the angle: enter image description here






share|cite|improve this answer




















  • Okay, sounds good. :) Thanks a lot.
    – numericalorange
    Aug 18 at 21:31

















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, $z=3e^-ifracpi2$.






share|cite|improve this answer


















  • 1




    That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
    – Deepak
    Aug 18 at 6:52











  • exactly. thanks
    – Ronald
    Aug 18 at 6:53










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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 $$-3i=3e^-idfracpi2$$






share|cite|improve this answer
















  • 1




    Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
    – Truth-seek
    Aug 18 at 7:24











  • Sure! All of these discussions are true in limit....
    – Mostafa Ayaz
    Aug 18 at 8:55










  • I love this idea of using limits! Thanks for the insightful response!
    – numericalorange
    Aug 18 at 21:31










  • You're welcome ^____^
    – Mostafa Ayaz
    Aug 19 at 20:05














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 $$-3i=3e^-idfracpi2$$






share|cite|improve this answer
















  • 1




    Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
    – Truth-seek
    Aug 18 at 7:24











  • Sure! All of these discussions are true in limit....
    – Mostafa Ayaz
    Aug 18 at 8:55










  • I love this idea of using limits! Thanks for the insightful response!
    – numericalorange
    Aug 18 at 21:31










  • You're welcome ^____^
    – Mostafa Ayaz
    Aug 19 at 20:05












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 $$-3i=3e^-idfracpi2$$






share|cite|improve this answer












According to a definition we always have $$-pi<thetale pi$$here we can write $$theta=tan^-1dfrac-30=-dfracpi2\r=3$$therefore $$-3i=3e^-idfracpi2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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 $dfrac-pi2$
    – Truth-seek
    Aug 18 at 7:24











  • Sure! All of these discussions are true in limit....
    – Mostafa Ayaz
    Aug 18 at 8:55










  • I love this idea of using limits! Thanks for the insightful response!
    – numericalorange
    Aug 18 at 21:31










  • You're welcome ^____^
    – Mostafa Ayaz
    Aug 19 at 20:05












  • 1




    Can you just add something like: the closer $x$ gets to $0$ without becoming $0$, the closer $tan^-1dfrac-3x$ gets to $dfrac-pi2$
    – Truth-seek
    Aug 18 at 7:24











  • Sure! All of these discussions are true in limit....
    – Mostafa Ayaz
    Aug 18 at 8:55










  • I love this idea of using limits! Thanks for the insightful response!
    – numericalorange
    Aug 18 at 21:31










  • You're welcome ^____^
    – 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 $dfrac-pi2$
– 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 $dfrac-pi2$
– Truth-seek
Aug 18 at 7:24













Sure! All of these discussions are true in limit....
– Mostafa Ayaz
Aug 18 at 8:55




Sure! All of these discussions are true in limit....
– Mostafa Ayaz
Aug 18 at 8:55












I love this idea of using limits! Thanks for the insightful response!
– numericalorange
Aug 18 at 21:31




I love this idea of using limits! Thanks for the insightful response!
– numericalorange
Aug 18 at 21:31












You're welcome ^____^
– Mostafa Ayaz
Aug 19 at 20:05




You're welcome ^____^
– Mostafa Ayaz
Aug 19 at 20:05










up vote
2
down vote













Try to draw a picture. I think you will be able to see the angle: enter image description here






share|cite|improve this answer




















  • Okay, sounds good. :) Thanks a lot.
    – numericalorange
    Aug 18 at 21:31














up vote
2
down vote













Try to draw a picture. I think you will be able to see the angle: enter image description here






share|cite|improve this answer




















  • Okay, sounds good. :) Thanks a lot.
    – numericalorange
    Aug 18 at 21:31












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: enter image description here






share|cite|improve this answer












Try to draw a picture. I think you will be able to see the angle: enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 18 at 6:39









Botond

3,9532632




3,9532632











  • Okay, sounds good. :) Thanks a lot.
    – numericalorange
    Aug 18 at 21:31
















  • Okay, sounds good. :) Thanks a lot.
    – numericalorange
    Aug 18 at 21:31















Okay, sounds good. :) Thanks a lot.
– numericalorange
Aug 18 at 21:31




Okay, sounds good. :) Thanks a lot.
– numericalorange
Aug 18 at 21:31










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, $z=3e^-ifracpi2$.






share|cite|improve this answer


















  • 1




    That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
    – Deepak
    Aug 18 at 6:52











  • exactly. thanks
    – Ronald
    Aug 18 at 6:53














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, $z=3e^-ifracpi2$.






share|cite|improve this answer


















  • 1




    That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
    – Deepak
    Aug 18 at 6:52











  • exactly. thanks
    – Ronald
    Aug 18 at 6:53












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, $z=3e^-ifracpi2$.






share|cite|improve this answer














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, $z=3e^-ifracpi2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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).
    – Deepak
    Aug 18 at 6:52











  • exactly. thanks
    – Ronald
    Aug 18 at 6:53












  • 1




    That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
    – Deepak
    Aug 18 at 6:52











  • exactly. thanks
    – Ronald
    Aug 18 at 6:53







1




1




That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
– Deepak
Aug 18 at 6:52





That should be $z = 3e^-ifracpi2$ (you're missing an "$i$" in the exponent).
– Deepak
Aug 18 at 6:52













exactly. thanks
– Ronald
Aug 18 at 6:53




exactly. thanks
– Ronald
Aug 18 at 6:53












 

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