# Caucy-Riemann in polar coordinates question

• redrzewski
In summary: But if you have a function z(r,theta) and you have a rectangle in r,theta space, then you can use the Jacobian to get the area in z space.The exp factor in dz is the result of the chain rule, as mentioned earlier. It represents the rotation of dz as it moves from the real to the imaginary axis. It is arbitrary in the sense that its value depends on the angle theta, but it is not arbitrary in the sense that it is necessary for the derivation of dz from the chain rule. Without it, the derivation would not hold true.
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

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

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"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 truly 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

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.

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?

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)

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?

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redrzewski said:
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
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!

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.

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redrzewski said:
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.

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".

## 1. What is the Cauchy-Riemann equation in polar coordinates?

The Cauchy-Riemann equation in polar coordinates is a set of two partial differential equations that describe the relationship between the real and imaginary parts of a complex-valued function. It is expressed as:
$\frac{\partial&space;u}{\partial&space;r}&space;=\frac{1}{r}\frac{\partial&space;v}{\partial&space;\theta}&space;\quad&space;\text{and}&space;\quad&space;\frac{\partial&space;v}{\partial&space;r}&space;=-\frac{1}{r}\frac{\partial&space;u}{\partial&space;\theta}$

## 2. Why is the Cauchy-Riemann equation important?

The Cauchy-Riemann equation is important because it is a necessary and sufficient condition for a function to be analytic. This means that if a function satisfies the Cauchy-Riemann equation, it is guaranteed to have a derivative at every point in its domain. Analytic functions are used extensively in mathematics and physics, and the Cauchy-Riemann equation provides a useful tool for identifying and analyzing them.

## 3. How is the Cauchy-Riemann equation derived?

The Cauchy-Riemann equation is derived from the Cauchy-Riemann conditions, which are a set of necessary and sufficient conditions for a complex-valued function to be analytic. These conditions are derived from the Cauchy-Riemann equations in Cartesian coordinates, and they can be expressed in polar coordinates by using the chain rule and the definition of the complex conjugate.

## 4. What are some applications of the Cauchy-Riemann equation?

The Cauchy-Riemann equation has many applications in mathematics and physics. For example, it is used in the study of conformal mappings, which are transformations that preserve angles between curves. It is also used in the derivation of the Navier-Stokes equations, which describe the motion of fluid flows. In addition, the Cauchy-Riemann equation is used in the study of complex analysis, which has numerous applications in fields such as engineering, economics, and statistics.

## 5. Are there any limitations to the Cauchy-Riemann equation?

While the Cauchy-Riemann equation is a powerful tool for analyzing analytic functions, it does have some limitations. For example, it only applies to functions that are differentiable at a point, which means that it cannot be used to analyze functions with discontinuities or sharp corners. Additionally, the Cauchy-Riemann equation only applies to functions that are defined on a simply connected domain, which means that they cannot have any holes or gaps. Finally, the Cauchy-Riemann equation only applies to functions of a single complex variable, so it cannot be used to analyze functions of multiple variables.

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