Why are the Cauchy-Riemann equations in polar form 'obvious'? In my book on complex analysis I'm asked to prove the Cauchy-Riemann equations in polar form, which I did. However, at the end of the question the author asks why these relations are 'almost obvious'. 
Now I get the derivation using chain rules and also the idea of approach along a circle and along a radial line and then equating. But the fact that the author asks this question to me suggests that there is an even simpler way of seeing this. That or maybe the author just regards one of these approaches to be 'almost obvious'.
So I'm looking for a more intuitive (so it also does not need to be 100% rigorous) way of thinking about the Cauchy-Riemann relations in polar form. These  relations are 
$$u_r=\frac{1}{r}v_\theta, \quad \frac{1}{r}u_\theta = -v_r$$
 A: A function $u+iv$ is complex differentiable  if the derivative of $u$ in any direction $h$ is equal to the derivative of $v$ in direction obtained by rotating $h$ by $90$ degrees counterclockwise. This should make sense if you think of imaginary axis as real axis rotated by $90$ degrees counterclockwise.
The Cauchy-Riemann criterion for differentiability  is the observation that it's enough to check the above for horizontal and vertical $h$. (Because these directions span the plane.) This is expressed by 
$$u_x = v_y\quad\text{ and }\quad u_y = - v_x$$
where $-v_x$ is the derivative of $v$ in the direction of negative $x$-axis (the result of rotating the $y$ axis counterclockwise.) 
In polar coordinates, we use the radial and tangential directions instead. The rate of change in the tangential direction is $\frac{1}{r}\frac{\partial}{\partial \theta}$ because of angle-distance conversion. Since rotating the positive $r$-direction gives positive $\theta$-direction, 
$$
u_r = \frac{1}{r}v_\theta
$$
And rotating positive $\theta$-direction counterclockwise gives negative $r$-direction, so 
$$
\frac{1}{r}u_\theta = -v_r
$$
