Complex Amplitudes and Exp'l Increase/Decrease
In the figure a the function being represented must be of the type \${\bf S}(t)=Ae^{jωt}\$, as it is described. But in the figure b the function must be 
\${\bf S}(t)=Ae^{(jω+\sigma)t}\$ to have the trajectory depicted.  


  Question1: I'm not able think about a complex number whose derivative is not i times something, since any complex number can be represented by \$e^{jt}\$, their derivative is given by \$je^{jt}\$ which means the velocity vector is perpendicular to the current position. But in this "general case" the velocity vector is pointing at an arbitrary angle, how?


Think of \$e^{(j\omega +a)t}\$, the derivative is \$(j\omega +a)e^{(j\omega +a)t}\$, do observe that the \$j\omega e^{(j\omega +a)t}\$ is the part perpendicular to \$e^{(j\omega +a)t}\$, the angular velocity, and \$ae^{(j\omega +a)t}\$ is parallel to \$e^{(j\omega +a)t}\$, the velocity going away from the center. Hence you can an arbitrary angle.


  Question 2: If the velocity vector is pointing in an arbitrary direction, then it's stated that the amplitude increases or decays "exponentially" Why is this true? Why not some other rate of decrease or increase?


It has the behavior you describe if the function is of the form \$e^{(j\omega +a)t}\$, the \$a\$ will determine the exponential growth/decay. But you could have a function such as \$t \cdot e^{j\omega t}\$, that will not grow exponentially. The reason complex functions of the form \$e^{(j\omega +a)t}\$ are more used/studied is that they are solutions to many ODEs, being used a lot in engineering.

obs: if the function is of the type \$2^{at} \cdot e^{jωt} = e^{ln(2)at} \cdot e^{jωt}\$, and \$ln(2)a\$ can be sen as a new number, like, call it \$c=ln(2)a\$, so 

\$2^{at} \cdot e^{jωt} = e^{(j\omega +c)t}\$