# Homework Help: Caucy-Riemann in polar coordinates question

1. Nov 27, 2006

### redrzewski

I'm well out of school, and brushing up on math/physics as a hobby.

My question is if there is an intuitive explanation for the exp(I*theta) factor?

From Shankar, dz = (dr+ir*d(theta))*exp(I*theta)

Here's a good derivation of CR equations in polar:

http://planetmath.org/encyclopedia/ProofToCauchyRiemannEquationsPolarCoordinates.html [Broken]

Where the exp(-I*theta) cancels.

Shankar's question implies that there is an intuitive meaning for this term, which looks like a phase factor.

thanks
brian

Last edited by a moderator: May 2, 2017
2. Nov 28, 2006

### quasar987

"intuitive meaning" is a broad term. I would say that the two following fact fit in that category:

1) If you have a complez number z and multiply it by exp{ix}, the effect is to rotate t by an angle x counterclockwise.

2) If you have a sin wave represented by a complex exponential, then multiplying the wave by exp{ix} adds a phase factor of x to the wave.

*the wave could be truely complex, such as could be the case for wave function in QM, or it could be a real wave such as an E-M wave and in this case, we write the the wave is exp[kr-wt] but we mean that it is the real part of that

3. Nov 28, 2006

### marcusl

I hope I'm not missing the point of your question; the exponential comes from the definition of polar coordinates. A point z={x,iy} is written in polar coordinates as z=r*exp(i*theta), where r=sqrt(x^2 + y^2) and theta = arctan(y/x). The formula you wrote above for dz comes from the chain rule. The "intuitive" meaning of theta comes from looking at the point z in the complex plane. Think of the unit circle abs(z)=r=1, then theta is the angle around the circle.

4. Nov 28, 2006

### redrzewski

Thanks. Let me try to clarify.

I understand the derivation of dz, theta in the complex plane, etc, and why the exp(I*theta) term arises from the chain rule.

What I'm trying to understand (geometrically) is what the exp(I*theta) means in dz.

I.e. why does the differential contain an additional rotation?

5. Nov 28, 2006

### marcusl

It does the same thing it does to r*exp(i*theta), that is rotate it CCW from real axis by angle theta, as quasar said. Look at your expression:
1. dr is infinitesimal pointing along real axis
2. r*dtheta is classic infinitesimal of polar coordinates. i*r*dtheta makes it point upward along imag axis so it's orthogonal to 1.
3. Multiplying sum (1+2) rotates the orthogonal pair CCW from real axis so its components point along/perpendicular to the line r*exp(i*theta)

6. Nov 28, 2006

### redrzewski

Let's try this.

In normal (non-complex) coords: we have 4 points:
(0,0) - (2, 0) and
(0,Pi) - (2,Pi)

dx = 2 - 0
dy = Pi - 0

In rectangular coords: area is dx * dy = 2 * Pi
In polar (treating the points above as (r,theta)), the area is 2 * (2*Pi) because of the standard r*d(theta).
dr = 2. dtheta = Pi.
Since we want them equal, we invoke the usual 1/r scale factor.

Now do the same thing in complex.
Rect = dx * Idy = 2I*Pi. (dz1 = 2, dz2 = I*PI, dz1*dz2 = 2I*pi)

Polar = 2*2*I*Pi * exp(I*(theta2 - theta1))

What's up with the additonal exp factor?

It'll go away if I compute (dz)^2 as you multiply by the complex conjugate, which'll cancel the exp factor.

Why can't I equate the areas from complex rectangular to complex polar?

Edit:
Also, I have specific numbers above for dr and dtheta, yet no clue what value theta takes in exp(I*theta). So it appears that the differential has a totally arbitrary rotation?

Last edited: Nov 28, 2006
7. Nov 28, 2006

### marcusl

What? You can't take a differential of a point, and in any case your dy expression makes no sense. Are you working with a book? You need to back up and review both calculus and polar coordinates. I always liked Thomas, Calculus and Analytic Geometry. Any edition is fine (I have one from the 70's, but I've seen the ninth edition as well). You also need a book on complex variables, at least the intro chapter or two--you can't go wrong with Churchill, or at least look through Schaum's outline.

Come back later if you still have questions!

8. Nov 29, 2006

### redrzewski

Ok. I'll keep digging at this.

Note that I pulled this area calculation method from the chapter on the jacobian in my calc book (Stein). I'm thinking that I didn't take the jacobian correctly in complex, which may cancel the exp.

As for the differential, I was taking the difference of points, not a single point.

P1: (0,0)
P2: (2, 0)
P3: (0,Pi)
P4: (2,Pi)

So I too dx as P2 - P1 and dy as P4 - P2.

Last edited: Nov 29, 2006
9. Nov 29, 2006

### HallsofIvy

Then those would better be written as $\Delta x$ and $Delta y$
Without some functions, you are not going to be able to get "dx" and "dy".