Why do u,v components in Cauchy-Riemann conditions are irrotational? It's very strange to me! When we decompose a complex function to a real part and an immaginary part, we have
$f(z) = u(x,y) + j v(x,y)$
following the conditions of analyticity we can derive the Cauchy-Riemann conditions where


*

*$u_x$ = $v_y$

*$v_x$ = -$u_y$


I see that my text book says that these components are irrotational while I see that it depends. When I calculate the curl I see:
Curl = 0 + 0 + ($v_x$ - $u_y$) = 2 * $v_x$
I see that if we consider u the second and v the first component it will be irrotational but in the default order the thing that I'd expected to work the curl doesn't appear to be 0. What is my error?
 A: The vector field $\vec f(x,y)$ that corresponds to the complex function $f(z)$ has a sign change on the $y$-component.  That is,
$$\vec f(x,y) = u(x,y) \hat x - v(x,y) \hat y$$
The curl and divergence of this vector field are zero as a consequence of the Cauchy-Riemann conditions.  In vector fields that are divergenceless and curlless* are fully determined by their values on some bounding surface (or a hypersurface beyond 3d), just as 2d holomorphic functions are fully determined by their values on some closed curve.
*Curl doesn't exist beyond 3d, but there are ways to generalize it, and it's not a crime to call these generalizations "curl" also.

Edit: now why should there be a sign change?  Well, you could identify it by inspection, but that's a pretty unsatisfying answer.  A better (though more complicated) one is to construct a correspondence between complex functions and vector fields.  You can do that if they're both algebraic elements of the same algebra--a "clifford" algebra.
Clifford algebra works with a noncommutative product called the geometric product.  A 2d clifford algebra can be built from four basis elements:  call them $1, e_1, e_2, e_1 e_2$.  Basically, the multiplication works like so:  $e_1 e_1 = e_2 e_2 = 1$ (so it captures the dot product), and $e_1 e_2 = -e_2 e_1$ (so for orthogonal vectors, it anticommutes like the cross product), and $e_1 e_1 e_2 = (e_1 e_1) e_2 = e_2$, so it's associative.
Now, how does this all connect to complex numbers?  As it turns out, the object $e_1 e_2$ has a special property:
$$(e_1 e_2)^2 = e_1 e_2 e_1 e_2 = -e_1 e_1 e_2 e_2 = -1$$
So algebraically, $e_1 e_2$ behaves exactly like the imaginary unit, $j$!  The algebra of linear combinations $a + b e_1 e_2$ for real numbers $a, b$ is exactly the algebra of complex numbers.
But!  We have the extra benefit of still having $e_1, e_2$ around as basis vectors.  Vectors coexist alongside these complex numbers.  Indeed, you can take a position vector $x e_1 + y e_2$ and convert it to a complex number by multiplying on the left by $e_1$:
$$e_1 (x e_1 + y e_2) = x + y e_1 e_2$$
So, this demands the question why we can't do this for vector fields.  The answer comes from differentiation, from the use of $\nabla$.  Consider a function $f(x,y) = u(x,y) + e_1 e_2 v(x,y)$, like your complex function, and let's take a derivative with $\nabla$:
$$\nabla f = e_1 (\partial_x u - \partial_y v) + e_2 (\partial_y u + \partial_x v)$$
The components are those expressions in the Cauchy-Riemann condition, so if $\nabla f = 0$, then the C-R condition holds.
How can we build up a vector field from $f$, though?  We can't do it by multiplying by $e_1$ on the left:  $\nabla$ is in the way, and inserting an $e_1$ between $\nabla$ and $f$ isn't something you can just do whenever you like.
But you can multiply by $1 = e_1 e_1$ on the right. This gives us the expression $[\nabla (f e_1)] e_1$.  That $f e_1$ evaluates to
$$f e_1 = u e_1 + e_1 e_2 e_1 v = u e_1 - v e_2$$
I have, perhaps, gone a bit too far in trying to give evidence for this mathematical curiosity, but I hope it's piqued your interest.
