Parametrization of a half circle in a different interval other than from 0 to $\pi$. If I have a half circle parametrization such that $e^{it}$, $0 \leq t \leq \pi$, to get the parameterization of the same circle on interval $0 \leq t \leq 1$, is it just $e^{it\pi}, \quad 0\leq t \leq 1$? The contour is an upper half circle centred at the origin connected to a straight line, to be parametrized from $0 \leq t \leq 2$. That's why I thought of having the interval from $0 \leq t \leq 1$. If anyone has better suggestions for parametrizing both the circle and line segment from 0 to 2 then that would be great. This is on the complex plane.

Why not consider $z=e^{i\theta}$ for $2\pi\leq\theta\leq 3\pi$? Or, if you want something "less trivial," then we can consider the nondegenerate interval $[a,b]$. Since we can map this interval onto the interval $[0,\pi]$ via $\pi\left(\frac{t-a}{b-a}\right)$, we can have any interval $[a,b]$ parametrize the top half of the unit circle by $$ \exp{i\left(\pi\frac{t-a}{b-a}\right)}.$$

In your particular case, if you're paramtrizing this circle for $0\leq t\leq 1$, then we have $$\exp\left(i(\pi\cdot t)\right).$$ Now to join the segment along the real axis, we have the parametrization $2t-3$. Thus, let's suppose you're integrating $f(z)$ over this contour. Then we have from the definition: $$\int_\Gamma f(z)\,dz=\int_0^1 f\big(e^{i\pi t}\big)\cdot i\pi e^{i\pi t}\,dt+\int_1^2f(2t-3)\cdot 2\,dt.$$I hope this helps!