The following are quotes from a textbook,
If S(t) is represented as a rotating phasor, the angular frequency of the phasor can be thought of as velocity at the end of the phasor. In particular the velocity $\omega$ is always at right angles to the phasor. $$\mathbf S(t) = Ae^{i\omega t}$$
However when we consider the general case when the velocity vector is inclined at an arbitrary angle $\psi$. In this case is a velocity is given by the symbol $s$ which now is composed of a component $\omega$ at right angle to the phasor as well as a component sigma which is parallel to $s$
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 $e^{it}$, their derivative is given by $i e^{it}$ 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?
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?
For question 1, you have $S(t)=Ae^{i\omega t}$. You are thinking that $S'(t)=i\omega Ae^{i\omega t}$. That's true only if $A$ is a constant in time. If $A$ changes, you have $$S'(t)=i\omega Ae^{i\omega t}+\frac{dA}{dt}e^{i\omega t}$$
If the angle $\psi$ is constant, then we have the ratio between the two components (radial and tangential) is constant. $$\frac{\frac{dA}{dt}}{\omega A}=c$$ or $$\frac{dA}{dt}=c\omega A$$ The solution of this equation is an exponential $A(t)=A_0e^{c\omega t}$.