Undergrad Intuition behind Cauchy–Riemann equations

[themagiciant95]

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

 

[FactChecker]

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 = u_x + iv_x and ∂_iy = -i∂_y.
f(z) = u(x,y) + iv(x,y),
∂f/∂x = u_x + iv_x and
∂f/∂iy = -iu_y +i(-iv_y) = -iu_y + v_y.
Since we need ∂f/∂x = ∂f/∂iy, setting the real and imaginary parts equal gives the Cauchy Riemann equations.

 

[themagiciant95]

∂f/∂x = u_x + iv_x and
∂f/∂iy = -iu_y +i(-iv_y) = -iu_y + v_y.

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

 

[FactChecker]

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

 

[fresh_42]

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.

 

[themagiciant95]

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

up

 

[lavinia]

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.

 

[WWGD]

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.

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?

 

[fresh_42]

True that Irrationals do not occupy a different dimension than Rationals

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.

 

[FactChecker]

actually Rationals are "most" of the Reals by many measures

This statement surprises me, since the rationals have Lebesgue measure zero and are countably infinite -- far fewer than the irrationals.

 

[WWGD]

This statement surprises me, since the rationals have Lebesgue measure zero and are countably infinite -- far fewer than the irrationals.

Yes, right, I meant the Irrationals. Let me edit.

 

[lavinia]

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

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?

 

[WWGD]

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 .