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Caucy-Riemann in polar coordinates question

  1. Nov 27, 2006 #1
    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.

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


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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 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
  4. Nov 28, 2006 #3


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    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.
  5. Nov 28, 2006 #4
    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?
  6. Nov 28, 2006 #5


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    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)
  7. Nov 28, 2006 #6
    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?

    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
  8. Nov 28, 2006 #7


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    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!
  9. Nov 29, 2006 #8
    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
  10. Nov 29, 2006 #9


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