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I Intuition behind Cauchy–Riemann equations

  1. Dec 12, 2017 #1
    I know that if a complex function is analytic , it means that i can reach the neighborhood of every complex point using a certain "stretch and rotation".
    In which way this fact conducts us to the "Cauchy Riemann equations" ? What's the intuition behind them ?
    Thanks
     
  2. jcsd
  3. Dec 12, 2017 #2

    FactChecker

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    The derivative of an analytic function at a point must be a single complex number, regardless of the direction that a path approaches that point. So coming in from the x direction must be the same as coming in from the iy direction.

    That leads to consider ∂x = ux + ivx and ∂iy = -i∂y.
    f(z) = u(x,y) + iv(x,y),
    ∂f/∂x = ux + ivx and
    ∂f/∂iy = -iuy +i(-ivy) = -iuy + vy.
    Since we need ∂f/∂x = ∂f/∂iy, setting the real and imaginary parts equal gives the Cauchy Riemann equations.
     
  4. Dec 12, 2017 #3
    Thanks so much.
    The last doubt : geometrically, why this partial derivatives have this form ? In particular why in ∂f/∂iy there is a i in the denominator ? It's the first time i see a partial derivative with the imaginary number i
     
  5. Dec 12, 2017 #4

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    I'm not sure how to describe it geometrically. The derivative of a vector is the derivative of the individual coordinates.
    So ∂f/∂x = ∂u/∂x + i ∂v/∂x.
    And because d/dy (iy) = i, we have ∂iy = i ∂y. So ∂f/∂iy = ∂u/∂iy + i ∂v/∂iy = ∂u/i∂y + i ( ∂v/i∂y)) = -i(∂u/∂y) + (∂v/∂y) = ∂v/∂y - i ∂u/∂y
     
  6. Dec 12, 2017 #5

    fresh_42

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    I've tried to give a bit of a different explanation, better different representation, since the math is the same in here:
    https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

    Basically the Cauchy Riemann equations arise from the fact that ##\mathbb{C}## is a field, and as such more than simply ##\mathbb{R}^2##. However, we want to calculate and draw pictures with real and imaginary parts, which means this discrepancy between the field and the real vector space has to show up somewhere. ##i \cdot i## doesn't take place in the imaginary dimension, it leaves it and ends up in the real dimension - a property we don't have in ##\mathbb{R}^2##. Therefore there is an inherent mixture which results in the Cauchy Riemann equations.
     
  7. Dec 13, 2017 #6
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  8. Dec 14, 2017 #7

    lavinia

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    A function ##f(x,y)=(u(x,y),v(x,y))## is conformal if its Jacobian ##J(f)## is multiplication by a complex number. That is: ##J(f)## is a rotation multiplied by a scale factor. ##J(f)## is then of the form

    ##\begin{pmatrix}∂u/∂x&∂u/∂y\\
    ∂v/∂x&∂v/∂y\end{pmatrix} =r\begin{pmatrix}cos(θ)&-sin(θ)\\
    sin(θ)&cos(θ)\end{pmatrix}## from which one has ##∂u/∂x=∂v/∂y## and ##∂u/∂y=-∂v/∂x##.

    If ##f## is twice continuously differentiable then the partial derivatives with respect to ##x## and ##y## commute and ##u## and ##v## are both harmonic.

    If one thinks of ##u(x,y)## as a velocity potential then its gradient can be thought of as the velocity of a fluid in the plane. Since the Laplacian is zero, the flow has zero divergence and is incompressible. Since the flow has a potential, its curl is zero so it is also irrotational. The curves ##u=##constant are the lines of constant potential and the curves ##v=##constant are the stream lines of the flow.
     
    Last edited: Dec 14, 2017
  9. Dec 14, 2017 #8

    WWGD

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    But, don't we have something similar with irrational numbers in that, e.g. ##\sqrt2 \times \sqrt 2=2 ## is not Irrational? Why don't we then have an analog in Real-differentiability? EDIT: True that Irrationals do not occupy a different dimension than Rationals; actually EDT Irrationals are "most" of the Reals by many measures, so then one can argue that they "go into nowhere" or something similar when they disappear this way?
     
    Last edited: Dec 14, 2017
  10. Dec 14, 2017 #9

    fresh_42

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    Yes, we don't consider ##\mathbb{R}## or even only the algebraic numbers (Btw, do they have a letter? ##\mathbb{A}## perhaps?) as ##\mathbb{Q}-## vector space, at least not in analysis. I know you would like to!
    I found @lavinia's explanation good, to emphasize that the derivative is a ##\mathbb{C}-##linear map. This is the point: it has to be complex linear and not just real linear.
     
  11. Dec 14, 2017 #10

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    This statement surprises me, since the rationals have Lebesgue measure zero and are countably infinite -- far fewer than the irrationals.
     
  12. Dec 14, 2017 #11

    WWGD

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    Yes, right, I meant the Irrationals. Let me edit.
     
  13. Dec 15, 2017 #12

    lavinia

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    Dividing by i rotates the vector clockwise by 90 degrees.

    For every directional derivative one gets a rotation. For instance derivative in the negative x-direction is rotated by 180 degrees(multiplied by -1). What does this tell you about the complex derivative?
     
    Last edited: Dec 15, 2017
  14. Dec 15, 2017 #13

    WWGD

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    Indeed: The Complex Jacobian/Differential of an Analytic function , with entries M:= ( a b -b a ) acts like multiplication by a+ib , i.e., when you left-multiply (c+id) by M, you get the same as multiplying (a+ib)(c+id), i.e., you scale by the modulus and rotate by the argument .
     
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