Let C be the right half of the circle $\{ z \in \mathbb{C} : |z| = 2\}$. Consider the parameterizations $$ \sigma(t) = 2e^{i\theta}, \frac {-\pi}{2} < \theta < \frac {\pi}{2} $$ and $$ \delta(t)=\sqrt{4-t^2}+it, -2 < t < 2 $$ of C.
Verify the function $\phi: (-2, 2) \to (-\pi/2, \pi/2)$ given by $\phi(t)= \arctan(\frac {t}{\sqrt(4-t^2)})$ is a reparameterization from $\delta$ to $\sigma$.
I've been stuck on this problem for about a day now. I understand how each parameterization describes C in the complex plane, but I'm not really sure how to approach the problem. Any help would be appreciated, thank you!
$\renewcommand{\paren}[1]{ \Bigl( #1 \Bigr) }$ \begin{align} &\begin{aligned} \phi(t) = \arctan\paren{\frac{t}{\sqrt{4-t^2}}} &\implies \tan\phi = \frac{t}{\sqrt{4-t^2}} \\ &\implies 1 + \tan^2\phi = \frac4{4-t^2} \\ &\implies \cos\phi = \frac{ \sqrt{4 - t^2} }2~,~~\sin\phi = \frac{t}2 \\ \end{aligned} \\ &\implies \delta = \sqrt{4 - t^2} + it = 2(\cos\phi + i \sin\phi) = 2e^{i\phi} = \sigma \end{align}