Complex numbers in polar form

  • Thread starter seboastien
  • Start date
  • #1
53
0

Homework Statement


Compute the 4th roots of -16 in both Cartesian and polar form and plot their positions in the complex plane.


Homework Equations


z^1/n=(r^1/n)(e^i(theta)/n), (r^1/n)(e^i(theta)/n)(e^i2(pi)/n........


The Attempt at a Solution


How do I find the value of r, and theta??
 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0
Draw z=-16 in the complex plane. The distance from the origin to -16 in the complex plane is r and the angle between the positive real axis and the negative real axis rotating counter clock wise is [itex]\theta[/itex].
 
  • #3
53
0
How do I draw -16 in the complex plane, when I don't know r or theta?
 
  • #4
Cyosis
Homework Helper
1,495
0
Draw the the complex plane and put a dot where -16 is. Then calculate the distance and angle.
 
  • #5
53
0
where is -16?
 
  • #6
Cyosis
Homework Helper
1,495
0
Do you know where -16 is on the line of real numbers?
 
  • #7
53
0
are you saying that the argument is zero and that the modulus is 16?
 
  • #8
Cyosis
Homework Helper
1,495
0
The modulus is 16, but the argument is not 0. If the argument was 0 -16 would be placed on the positive real axis, which it clearly isn't.
 
  • #9
53
0
okay so you think the argument in pi
 
  • #10
53
0
that's not right
 
  • #11
53
0
why are you wasting my time?
 
  • #12
Cyosis
Homework Helper
1,495
0
Wasting your time? Why would that not be right? You may want to provide some arguments to why this is wrong.

Either way I can tell you that I am not wrong. Perhaps review the the relevant equation you posted before jumping the gun?
 
Last edited:
  • #13
53
0
Beacause the answer is apparently 2e^i((pi + 2kpi)/4) where K=0,1,2,3

that's why it wouldn't be right.
 
  • #14
607
0
Beacause the answer is apparently 2e^i((pi + 2kpi)/4) where K=0,1,2,3

that's why it wouldn't be right.

So ...take one of these numbers (say the k=0 one), convert it to Cartesian form, and take its 4th power. You can then check for yourself whether it is right.
 
  • #15
Cyosis
Homework Helper
1,495
0
Beacause the answer is apparently 2e^i((pi + 2kpi)/4) where K=0,1,2,3

that's why it wouldn't be right.

It is obvious that every multiple of 2pi added to the original argument will return you to that exact same spot, after all a circle is exactly 2pi radians.
 

Related Threads on Complex numbers in polar form

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
661
Replies
4
Views
1K
  • Last Post
Replies
3
Views
894
Replies
2
Views
6K
  • Last Post
Replies
8
Views
1K
Replies
7
Views
1K
Replies
2
Views
2K
Top