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?