Geometric interpretation of analyticity? 
Suppose the real valued functions $u(x,y)$ and $v(x,y)$ are continuous and have continuous first order partial derivatives in a domain $D$. If $u$ and $v$ satisfy the Cauchy Riemann equations at all points of $D$, then the complex function $f(z)= u+iv$ is analytic in $D$.

Could someone please give me a geometric interpretation of the theorem above? 
The Cauchy Riemann equations can be interpreted as saying the the gradient of $u$ and $v$ must be perpendicular for the function to be differentiable.
 A: A more elegant statement of the same theorem would be to say:

Let $f:\mathbb{C}\rightarrow \mathbb C$ be a function. Then $f$ is holomorphic if any only if the best linear approximation for $f(z)$ about a point $z_0$ can be written as, for some $c\in\mathbb C$:
  $$f(z_0)+c(z-z_0).$$

Note that when we say "linear approximation" we mean "linear over $\mathbb R$" - for instance a function like
$$f(x+iy)=x+iy^2$$
is not holomorphic because near $z_0=0$, we can compute the best linear approximation of $f(x+iy)$ as $x$, since the derivative of $y^2$ with respect to $y$ (or $x$) is $0$ there. However, there is no $c$ such that $x=c(x+iy)$, contradicting hypothesis. This means that, near $0$, $f$ is acting like the projection of $(x+iy)$ onto the real axis.
However, a function like
$$f(x+iy)=(x+iy)^2=x^2-y^2+2xyi$$
has, near any point $z_0=x_0+iy_0$ that we can approximate
$$f(x+iy)\approx f(z_0) + 2x_0(x-x_0)-2y_0(y-y_0)+(2x_0(y-y_0)+2y_0(x-x_0))i$$
where we can rewrite the left-hand side, after grouping terms and thinking a bit as
$$f(x+iy)\approx f(z_0) + (2x_0+2iy_0)((x+iy)-(x_0+iy_0))$$
$$f(z)\approx f(z_0) + 2z_0(z-z_0)$$
which is of the desired form. However, a function like $f(z_0)+c(z-z_0)$ has a nice interpretation: It is a spiral similarity centered at $z_0$ which rotates by $\arg(c)$ and scales by $|c|$. This excludes the possibility $f$ could act like a projection, as it did in the non-holomorphic example. (Notice that $c=f'(z_0)$, so this gives a precise geometric meaning to the argument and modulus of the derivative at a point)
So, a really simple geometric interpretation is:

$f$ is holomorphic if and only if the best linear approximation for $f$ about any point is a spiral similarity.

We can get a bit deeper than this though. Notice that any spiral similarity preserves angles. Therefore, so long as $f'(z_0)$ is not zero at a point, near the point $z_0$, the function $f$ will act like a particular spiral similarity and will therefore preserve any angle with vertex at $z_0$. If $f'(z_0)$ is nowhere zero, then this means that $f$ is angle preserving - that is, if we drew any figure on $\mathbb C$, then looked at its image, the angles of the figure in the image would be equal to those in the original. We call a map "conformal" if it has this property of preserving angles which gives the interpretation:

If $f$ is a conformal map $\mathbb C\rightarrow \mathbb C$, then it is holomorphic.

and, more generally, since $f$ might not be conformal where $f'(z_0)=0$ - for instance, drawing a right angle at $0$ on the complex plane by drawing a path along the points $i$ then $0$ then $1$ then applying the holomorphic map $z\mapsto z^2$ gives the image as a path from $-1$ to $0$ to $1$, with an angle of $\pi$ - twice the original angle. (This is related to why Cauchy's integral theorem works).
