# Tangible meaning of complex exponential integral

Discussion in 'Math' started by blah2222, Jun 12, 2013.

1. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
Hi, looking through some old notes about complex exponentials and I was wondering what does the integral of a complex exponential equate to, within the complex plane?

For example:

$
\int_{0}^{\frac{\pi}{2}} e^{ix} dx = \int_{0}^{\frac{\pi}{2}} cos(x) + isin(x) dx = [sin(\frac{\pi}{2}) - icos(\frac{\pi}{2})] - [sin(0) - icos(0)] = [1 - 0] - [0 - i(1)] = 1 + i
$

What I am reading this result as is: the integral of a unit circle in the complex plane from angle 0 to pi/2 is equal to "1 + i".

What does this "1 + i" mean for the complex plane? Doesn't seem to be the area under the curve.

Any ideas? Thanks!

2. ### mikeleeson New Member

Aug 22, 2012
26
4
The complex plane is a way of displaying a complex number or a complex function. If you plot the value of your complex exponential as a function of x, it will be a line that looks like a quarter of a circle. The integral is performed along this line (and is called a line integral).

It is still finding "the area under the curve" but the y values of your curve are complex numbers and the x value is phase or angle.

It would be wrong (very wrong) to look at the complex plane and think "there is a y axis and an x axis, so the integral must be the area".

In your example, the function represents a vector of length 1 that rotates about the origin (hence it traces out a line which is part of a circle). The integral gives you the length of the line - which is still a complex number!

PS Did I mention just how wrong it would be to find the area on a complex plane? I just wanted to be clear!!!

3. ### WBahn Moderator

Mar 31, 2012
18,281
4,953

You say that

$
\int_{0}^{\frac{\pi}{2}} e^{ix} dx
$

Is somehow the area of the unit circle. Yet the limits have the angle going from 0 to pi/2. You are hardly integrating over the entire unit circle.

Then, to be anea, the integrand needs to be a differential area. Your integrand is scalar, not area. So your integral will yield a distance, not an area.

Think of just integrating one a normal x-y plane to get the area of a circle. You would have:

$
A = \int dA
dA = \frac{1}{2} \cdot R \cdot Rd\theta
$

So

$
A = \int_0^{2\pi} \frac{1}{2} \cdot R \cdot Rd\theta
\
A = \frac{R^2}{2} \int_0^{2\pi} d\theta
\
A = \frac{R^2}{2} \theta |_0^{2\pi}
\
A = \pi R^2
$

So if your integral isn't the area of something (be it a unit circle or whatever), then what is it?

Again, think of it in the normal 2D plane. You have a vector that is sweeping out a quarter-circular arc. Imagine that a tiny weight is placed at the tip of the vector at each step as it sweeps along. Or imagine that you have a continuous wire that is shaped like this quarter circle. What you are finding is related to the centroid of such a wire. The difference is that you are not then dividing by the area of the wire, which would be pi/2. Thus the centroid would be located at [(2/∏),i(2/∏)]

Note that the units are that of distance, not area. This is a bit hard to see because we are working with a pure mathematical plane that is dimensionless. But what you can do is imagine the x and y dimensions to have units of length and then track them through. You will see that they always add, never multiply.

Taking a different problem, what would the area be of a rectangular area of the complex plane bordered by the axes and the lines x=2 and y=2i? By inspection it would be A=(2)(2i)=4i.

Do it by integration:

$
A = \int_0^2 \int_0^{2i} dy dx
\
A = \int_0^2 y |_0^{2i} dx
\
A = \int_0^2 2i dx
\
A = 2i \int_0^2 dx
\
A = 2i x |_0^2
\
A = 2i 2
\
A = 4i
$

Nothing mysterious going on.

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4. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
Correct me if I am wrong, but wouldn't the length of "1 + i" be √2? Maybe I am blind but I am still not seeing a relation to a 90 degree rotation and a complex integral value of "1 + i".

Nowhere in my question did I say that this was the area of the unit circle. I said that this was "the integral of a unit circle in the complex plane from angle 0 to pi/2", meaning that if you had a unit circle drawn on the plane, you would only integrate from 0 to pi/2 (ie: a quarter circle).

5. ### WBahn Moderator

Mar 31, 2012
18,281
4,953
Okay, I'll accept that.

So what is the "area" that you are saying that this integral represents?

6. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
It was more of a sarcastic comment on my part and I should have been more clear.

As you know, in the usual plot of a dependent function on the y-axis and independent on the x-axis, an integral would bring the area under the curve created by that function and I was just pointing out how the complex plane is different.

I am just trying to make sense of this "1 + i" value for a 90 degree rotation on a quarter unit-circle.

7. ### mikeleeson New Member

Aug 22, 2012
26
4
I think that I have misled you by calling it the length. For each small change in x, the vector rotates and the end point moves by a small amount. If the change in x is δx, and the end point moves by a distance δz, then δz = R.δx. R=1 in this case.

The integral adds up all of the δz - but as each δz is a complex number, then the total (sum) is also a complex number. Hence 1 + i.

The last point is where I got it wrong: each value of δz is complex, whereas the length of the arc is found by adding up all the moduli |δz|. Sorry about that!

8. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
Okay, so by that logic shouldn't the integral be: -1 + i because you are moving from the point (1, 0) to (0, i)?

9. ### WBahn Moderator

Mar 31, 2012
18,281
4,953
Because you are talking about a complex number, it is basically impossible to put a "real" meaning to it. The best you can do is talk in abstract terms.

Read my post again and think abou this quarter circular arc as being a piece of wire. Now imagine wanting to find the center of gravity of that wire, which would be the place that you could balance it on a point, even though it might mean having to add a "weightless" element to connect the wire to the balance point.

To find the balance point you need to find the "first moment of area" which is what your integral is finding, and divide that by the total area (the zero moment of area). Now, technically your integral needs to have Rdθdr in order for the differential element to be an area, but in your case R=1 and is dimensionless. Then the integral over r can be left in integral form and allowed to cancel out the same integral that will appear in the denominator.

So that's the closest I can come to a "tangible" meaning for the integral -- that is is, in essence, the first moment of area of the first quadrant of the unit circle in the complex plane.

These area moments give you information about the distribution of the points in your function, just like the mean and the standard deviation give you information about the distribution of a set of scalar points. The metrics give you SOME information, but they don't tell you everything. But if you had ALL of the moments (the zeroth, the first, the second, the third, and on and on), then those, as a whole, DO tell you the EXACT distribution of points you are working with. In that regard, the moments are like the Fourier series -- each one tells you something and if you have enough of them you can reconstruct the data set to within an acceptably small deviation.

blah2222 likes this.
10. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Yes one complex plane will display one complex number, but you need two complex planes to display a complex function. Look up 'conformal mapping'

The integral has another meaning besides area, from which the symbol is derived. It is also a sum.

So the integral in this case is the sum of many small steps of the stated complex function.

11. ### mikeleeson New Member

Aug 22, 2012
26
4
I didn't actually calculate the integral - I just copied the value you had used! But when I did calculate the integral I also get 1 + j.

So I will have to go home, open a beer and think more about it...

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12. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
Thanks for your explanation. I think it will take some time to put all the pieces together. I might need to read this through in a bit to get a better jist.

13. ### mikeleeson New Member

Aug 22, 2012
26
4
Well I found my mistake...

I forgot that $\delta Z=R \cdot e^{\left(j\theta)} \delta\theta$ is still a complex number and hence a vector on the complex plane. It has a length Rδθ but it lies along the same direction as $R \cdot e^{\left(j\theta)}$. I was thinking that each δZ was lying along the arc - which was wrong.

Sorry to have misled you!

May 3, 2010
565
33