# Complex numbers

1. Sep 24, 2007

### Oblio

The phase of a complex number is z=re$$^{i\theta}$$

This first example is a simple z=1+i, but where does the r come from for this?

2. Sep 24, 2007

### bel

The "r" is the distance from the origin. Thus, $$z=re^{i\psi}$$
$$z= r (cos(\psi)+i sin (\psi) )$$.
Hence, for $$z=1+i$$, $$r= \sqrt{1^2 + 1^2}= \sqrt{2}$$.

Last edited: Sep 24, 2007
3. Sep 24, 2007

### Oblio

If it was 1+4i

Would it be:
$$r= \sqrt{1^2 + 4^2}$$?

4. Sep 24, 2007

### learningphysics

yeah, that's right.

5. Sep 24, 2007

### Oblio

Is a reasonable answer for this phase then:
/sqrt{2}e ^i(theta) ?

6. Sep 24, 2007

### Oblio

since its just 'i' and no angle is given

7. Sep 24, 2007

### learningphysics

wait... phase usually refers to the angle... what exactly does the question ask you to find?

"The phase of a complex number is z=re$$^{i\theta}$$"

that doesn't seem right... did you write this out word for word?

8. Sep 24, 2007

### Oblio

No, phase is just the angle but there doesn't seem to be any angle so.. yeah...
exact words are: phase is the value of theta when z is expressed as z=re^i(theta).

Can you find an angle with just 1+i ?

9. Sep 24, 2007

### learningphysics

yes, you can. arctan(1/1). Draw the point on the complex plane... the x-coordinate is the real... y-coordinate is the complex part... what angle does the line from the origin to the point make with the positive x-axis...

That's the angle you need.

10. Sep 24, 2007

### HallsofIvy

Staff Emeritus
No, for any complex number, both r and $\theta$ are both given. Your original question was about r only, that's why nothing was said about $\theta$.

In the "complex plane" or Argand diagram, any complex number x+ iy can be associated with the point (x,y). $r (cos(\theta)+ i sin(\theta))= re^{i\theta}$ is just that point given in polar coordinates. Since $x^2+ y^2= r^2 cos^2(theta)+ r^2 sin^2(\theta)= r^2$, $r= \sqrt{x^2+ y^2}$. Since $y/x= r^2 sin^2(\theta)/[r^2 cos^2(\theta)= tan(\theta)$, $\theta= arctan(y/x)$.

In your first example, 1+i, $r= \sqrt{1^2+ 1^2}= \sqrt{2}$ and $\theta= arctan(1/1)= arctan(1)= \pi/4$.

In your second example, 1+ 4i, $r= \sqrt{1^2+ 4^2}= \sqrt{17}$ and $\theta= arctan(4/1)= arectan(4)= 1.3 radians approximately. Am puzzled by your saying "The phase of a complex number is [itex]z=re^{i\theta}$. Normally the "phase" is given as an angle. I would have thought the "phase" of the number $x+ iy= r e^{i\theta}$ would be just $\theta$.

11. Sep 24, 2007

### Oblio

ahhh, 1 and 1 is a 45 degrees.

12. Sep 24, 2007

### Oblio

when looking at a more complicated one like sqrt[2e] ^ -i(pi)/4
is the entire exponent my imaginery part?
and the base is real?

13. Sep 24, 2007

### Oblio

when theyre like this, all hope of visualizing it is gone.

14. Sep 24, 2007

### learningphysics

The e is inside the square root?

15. Sep 24, 2007

### Oblio

hmm
the books a little iffy..

I'll say no its not actually.

16. Sep 24, 2007

### Proggle

They're very easy to visualize, it's just like a point in the plane described in polar coordinates.

In your last example, sqrt(2) is the distance from the origin to the point in the plane and -pi/4 is the angle (in radians) starting from the real axis (or the x axis in the analogy).

17. Sep 24, 2007

### learningphysics

Then it shouldn't be hard to visualize... r = sqrt(2). theta = -pi/4

18. Sep 24, 2007

### Oblio

lol yeah but its sqrt[2] multiplied by e and raised to -i(pi)/4 ! :P

Am I correct in saying the exponent is imaginery?