Let $\gamma: [a, b] \to \mathbb{C}$, $\gamma(t) = r(t)e^{i\phi(t)}$ be a continuously differentiable curve, given in it's polar coordinates, where $r, \phi: [a, b] \to \mathbb{R}$ are continuously differentiable.
Why is the arclength of $\gamma$ then given by the formula:
$$L(\gamma) = \int_a^b \sqrt{r(t)^2 \dot\phi(t)^2 + \dot r(t)^2}dt$$
My attempt: differentiating $r(t)e^{i\phi(t)}$ would give the derivative $\dot r(t) e^{i\phi(t)} + i\cdot r(t) \dot \phi(t) e^{i \phi(t)}$, and inserting that into $L(\gamma) = \int_a^b ||\dot \gamma(t)||dt$ results in:
$$\int_a^b ||\dot r(t) e^{i\phi(t)} + i\cdot r(t) \dot \phi(t) e^{i \phi(t)}||dt$$
But how would I continue from there?
The two parts of your sum are orthogonal so the norm is computed applying Pythagorean theorem.