Tangent to Parametric Polar Curve

Question

If we have some $$\gamma(t)=r(t)e^{i\theta(t)}$$

Where $\gamma(t)$ is some complex parametric curve; how would one express the tangent vector to that curve, without just converting straight to rectangular cordinates?

Answer 1

Suppose you are looking for the tangent in $t_0$. There is two cases :

1. If $r(t_0) = 0$ (it means that we pass by the pole) then the tangente is the line with equation $t = t_0$ (line with angle $t_0$ to the abscisse axe).
2. If $r(t_0) \neq 0$ (we're not in the pole) the the tangent is directed by the vector $r'(t_0) \vec{u} + r(t_0) \vec{v}$ where : $$u = \vec{i} \cos t_0 + \vec{j} \sin t_0$$ and : $$v = - \vec{i} \sin t_0 + \vec{j} \cos t_0$$ $\vec{u}$ is the unit vector passing by the pole and the point $\gamma(t_0)$ where we are looking for the tangent and $\vec{v}$ orthogonal to $\vec{u}$.