The Cauchy-Riemann equations may look complicated, but like most mathematical
equations they are really just a
restatement of the obvious using fancy symbols.
Let's play a game. I've drawn an arrow on the ground in the direction we'll both call "forward". I'm going to watch where you walk, and wherever you go I'll go twice as far in the same direction. Let's call
x and
y, respectively, the amount you walked in the forward and leftward directions, and
u and
v will be the amount that I walked in these same directions. So, for example, if you took one step forward and one step to the left, then
x=1 and
y=1, and since I said I would match your direction and twice your distance I would take two steps forward and two to the left, so
u=2 and
v=2.
So having set the game up in this way, it isn't particularly
deep to say that the number of steps I will walk forward,
u, when you walk
x steps forward is the same as the number I'll walk left,
v, when you walk left the same distance,
y. The first Cauchy-Riemann equation is just a fancy restatement of this
obvious fact:
du dv
-- = --
dx dy
That is, divide the number of steps I took forward by the number you took forward, and you'll find it equals the number of steps I took left divided by the number you took left. In both cases the amount will be
2, since that was the rule of the game that I chose.
Now let's play a
slight variation of this game. For every step you take, I will take one step in the direction
left of where you walked. So if you take one step forward, I'll take one step left. And if you take one step left, I'll go one step
backwards, since that's the direction that is left of left. Put another way, the amount of steps that I'll go left,
v, when you go forward
x steps is the
opposite of the number of steps I'll go forward,
u, when you go the same amount left,
y, since in the latter case I'll be walking backwards. This gets us the second Cauchy-Riemann equation,
dv du
-- = - --
dx dy
These equations themselves are
not very deep. What's deep is the fact that I was able to pick a rule for a game such that the amount I walked in was a fixed rotation of where you walked (that is, no rotation in the first game, 90 degrees of rotation in the second) times a constant value. So basically what I walked was just a multiple of what you walked --- if you're willing to expand your idea of multiplication to include rotations, which is
exactly what
complex numbers do! So in the first case I walked where you walked times 2, and in the second case I walked where you walked times
i, where
i means
left. These were two of the simplest possible cases; I could just have easily as said that for every step you take forward, I'll take a step 45 degrees left of where you walked --- or in complex number form, I will walk
(1+i)/sqrt(2) times where you walked. (That
sqrt(2) has to be there due to the
Pythagorean theorem -- walking one step forward and one step left means walking a total distance of
sqrt(1+1)=sqrt(2), so I had to divide that factor out if I wanted to keep my distance the same as yours.)
I don't have to play a game that takes this special form.
I could just as easily pick a rule that says, "for every step forward you take I'll walk two steps backwards, and for every step left you take I'll walk three steps left", and you will have a hard time reducing this to multiplication by a complex number. So the Cauchy-Riemann equations are not significant in and of themselves as much as they say something significant about the game --- that it can be reduced to multiplication by a complex number.
Of course, the equations aren't about
games, they are about
functions. But the same idea still basically holds; let
z be the spot where you are standing,
f be where I am standing,
dz the (
small) distance that you decide to walk, and
df the (
small) amount that
f changes in response. Thus,
df/dz is the rate at which I move with respect to how much you moved. This quantity may be different at different locations (that is, for different
z), and this is why I said that
df and
dz had to be small.
If df/dz can be given by a single complex number -- just like our games -- then
f necessarily satisfies the Cauchy-Riemann equations, since these equations are nothing more than a re-statement of what it means for
df/dz to be a single complex number, just as before the equations were nothing more than a re-statement of the rules of our games.
So like most things in mathematics, these equations really
just tell us what we already knew, just in a different form.