# Complex numbers in polar form

1. Aug 19, 2009

### seboastien

1. The problem statement, all variables and given/known data
Compute the 4th roots of -16 in both Cartesian and polar form and plot their positions in the complex plane.

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

3. The attempt at a solution
How do I find the value of r, and theta??

2. Aug 19, 2009

### Cyosis

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 $\theta$.

3. Aug 19, 2009

### seboastien

How do I draw -16 in the complex plane, when I don't know r or theta?

4. Aug 19, 2009

### Cyosis

Draw the the complex plane and put a dot where -16 is. Then calculate the distance and angle.

5. Aug 19, 2009

### seboastien

where is -16?

6. Aug 19, 2009

### Cyosis

Do you know where -16 is on the line of real numbers?

7. Aug 19, 2009

### seboastien

are you saying that the argument is zero and that the modulus is 16?

8. Aug 19, 2009

### Cyosis

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. Aug 19, 2009

### seboastien

okay so you think the argument in pi

10. Aug 19, 2009

### seboastien

that's not right

11. Aug 19, 2009

### seboastien

why are you wasting my time?

12. Aug 19, 2009

### Cyosis

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: Aug 19, 2009
13. Aug 19, 2009

### seboastien

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. Aug 19, 2009

### g_edgar

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. Aug 19, 2009

### Cyosis

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.