# Physical definition of a complex angle

1. Aug 28, 2006

### Thomas2054

I may be on a wild goose chase here, but I am looking for a physical definition of a complex angle, i.e., an angle of the form a + bi. Is there such a beast or is a complex angle purely a mathematical construct?

Thanks.

Thomas

2. Aug 28, 2006

### HallsofIvy

Why should there be a "physical definition"? Mathematics is not physics.

3. Aug 28, 2006

### Thomas2054

I agree that there does not have to be a definition, I am wondering if there is. As an engineer I like physical definitions, but can live without them. Since complex angles are used in physics I can understand that there may be a physical definition.

Thomas

4. Aug 28, 2006

### Swapnil

By physical, do you mean geometrical? If yes, then you can graph the point a=bi in the real-imaginary plane where your x-axis is the real axis and your y-axis is your imaginary axis.

I am not sure if this is what you wanted...

5. Aug 28, 2006

### Staff: Mentor

There are a number of ways that complex numbers are used visually in EE and physics. One common way is to orient the real axis vertically and the imaginary axis pointing out of the page coming at you, and put the + time axis going horizontally to the right. Now imagine a point on the Re-Im plane that is tracing out the circle $$A e^{j\theta}$$ as the point travels to the right in time. The point traces out a spiral in that 3-D space, and when viewed in just 2-D with RE on the vertical axis and time on the vertical horizontal axis, you see a sine wave traced out. This visualization can help you picture how some circuits work over time.

Another very helpful visualization involving complex spaces is used in stability analysis gain-phase work. You put the $$\sigma$$ and $$j \omega$$ axes on the horizontal plane, and extend the gain axis vertically out of that plane. You then plot the zeros and poles in the 2-D horizontal plane, and observe what that does to the overall gain contour in 3-D vertically. Very cool stuff. I'll post a link to a picture if I can find one handy.

Last edited: Aug 28, 2006
6. Aug 28, 2006

### Staff: Mentor

One other complex number visualization that helped me a lot is for travelling waves on a string. The vertical axis is the vertical displacement of the string (obviously), and the complex axis pointing out of the page at you is the tension in the string. Antinodes in stading waves on a string always bugged me, until I visualized the tension waveform.....

7. Apr 9, 2009

### jwhipple

Hello there. I found this old thread when I was looking for the same information. This turned up in a Google search, and it looked like a promising lead, but nobody seems to have given a simple answer, so I made up my own. Caveat: I'm neither a physicist nor a mathematician. Anyway, in case you're still around and still interested, consider this:

Suppose you have a complex number a such that a = x + yi = r*e^i*theta.

Draw a ray from the origin through point a. Follow the unit circle counterclockwise from the positive x-axis to the point where it intersects the ray. The length of this arc is theta, the real angle. Now take a half-turn to the left and measure the distance from the intersection point to point a. This is the messy part: You must measure this distance using the logs of the distances from the origin to the two points, ln 1 for the intersection point and ln r for point a, and subtracting: (ln 1) - (ln r). This simplifies to -ln r, and it is the imaginary portion of the complex angle.

That's a physical definition of sorts, but not very satisfying, I fear. So I have come up with what I think is a better answer:

Let us move to what I call the exponential plane. (Maybe proper mathematicians call it something else, but I haven't found another name for it . . .) I mean a Cartesian plane where each point (x', y') represents e^(x' + iy'). In other words, complex number a, where a = x + yi = r*e^i*theta = e^(ln r + i*theta), which is represented by point (x, y) on the complex plane, is represented on the exponential plane by ln a, which is equal to ln r + i*theta, or in other words, by point (x', y'), where x' = ln r and y' = i*theta. (I know the ln function is multivalued, but I believe that doesn’t affect this particular discussion. We can just use the principal value of ln and leave it at that, no?)

Adding the real and imaginary portions of the complex angle -- let's call it "xang" for short -- from the "messy" part above, we get xang a = theta + i*(-ln r). See the connection? In the exponential plane, the complex angle of a can be measured physically by (1) following the y-axis from the origin to point (0, y') to get the real part of the angle and (2) taking a half-turn to the left and measuring the horizontal distance from the y-axis to point (x', y') to get the imaginary part of the angle. With this orientation, moving up/down gives positive/negative real values, and moving left/right gives positive/negative imaginary values for the complex angle.

Finally, let me note in passing that xang a = (ln a)/i

This may not be original, though I haven't seen it elsewhere. Then again, it may not be right . . .

FWIW

Jeremy Whipple
a translator in Tokyo

Last edited: Apr 10, 2009
8. Apr 10, 2009

### gammamcc

The best justification I have seen is the famous formula by Euler
e^(ix)=cosx +isinx (can prove via complex Taylor series of LHS).
Cos and sin terms justify the real and imaginary axes and associated plots of the unit circle in the complex plane just like in trig - the formula justifies the use of separate axes beyond the mere idea of linear independence (as in linear algebra) of real and imaginary numbers.

9. Apr 14, 2009

### jwhipple

Sorry, that obviously (?) should have been "y' = theta."

Jeremy Whipple

10. Apr 14, 2009

### rochfor1

Complex angle? You mean complex number...complex vector? a + b i really isn't an angle, although it can be associated to one (its argument)

11. Apr 14, 2009

### jwhipple

Are you sure? Instead of "really isn't an angle," I think it's simply that it "isn't a real angle." To examine what a complex angle means, one can plug in a complex number in where the argument goes (i.e., in place of "theta" in e^i*theta) and take it from there. So I believe a complex angle of a + ib is equivalent to a multiplication factor of e^i(a + bi), or e(-b + ai).

This is my understanding:
A real angle acts on the direction of a complex number’s vector; an imaginary angle acts the vector’s length. An angle of ib is equivalent to multiplication by e^(-b), or division by e^b. A complex angle thus acts on both the direction and the length of the vector.

Last edited: Apr 14, 2009
12. Apr 15, 2009

### jwhipple

Another careless error: That should have been "or e^(-b +ai)." Sorry.

13. Jan 2, 2012