Given a curved line, and any point on that line is a radial vector with only a known radius $r$ and known tangent angle $a$, how does one express the point $(r,a)$ as vector coordinates $(x, y)$?
I have a spiral defined by its changing tangent angle and radius. I want to know the angle from any given point $(r,a)$ to the $x$ axis (the $\arg(x+iy)$ on the complex plane for instance) or the vector coordinates $(x,y)$ that corresponds to point $(r,a)$, as either will provide the other.
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The usual method for finding the tangent angle $a$ of a radial vector follows
$$\tan(a)=r(dr/d \arg(z))^{-1}.$$ However, on the complex plane for this curve/spiral, the argument of $z$ cancels out entirely (in how $r$ is defined), such that the antiderivative of $d\arg(z)$,
$$\int jt^{-1} dt=\arg(z),$$
is multivalued (in multiples of the inverse of the real part of $z$), as the integral $$-\int_{1}^{e^k} jt^{-1} dt=\arg(z): k=e^{\arg(z)/j} $$ is indeterminant: $$j \ln(e^{arg(z)/j}) = j, e^{\arg(z)/j}>=0.$$
Yet, there is a single argument of $z$, it's just that I do not know how to solve for it even given the radius and the tangent angle. This is not an Archimedean spiral in that the angle changes per a constant value and it is not quite a logarithmic spiral either (somewhere between the two). So if the spiral could be thought of as a particle curling in a field toward zero, it would be accelerating...not having a constant radial speed.
Any help would be greatly appreciated.
Consider a particle starting from the center at t=0. The angular position of the particle is given by $\theta=\omega t$, where $\omega$ is a constant. We also have a change in radial distance (radial speed) $\dot{r}=u$ where $u$ is a constant. Now the position in the xy-plane at time t is given by
$$\mathbf{r}(t)=(x(t),y(t))=(ut\cos(\omega t), ut\sin(\omega t))$$
Spiral t=0 to 9, $u=1$, $\omega=2$
Edit: It seems you edited your question while I was answering, hence no mention of the complex plane, but I think you can still use the information to figure out what to do :)
The $(r,a)$ point z is
$$z=x+iy=r(\cos a, i\sin a)$$