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.
 A: 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!
A: Instead of $\exp(it)$, you can use $\exp (ikt)$ over $[0,\frac \pi k]$  This should indicate many options-any continuous invertible function over $[0,\pi]$ can transform the interval to another one.  Following the comments, you can use $\exp(it-\pi )$ to get $[\pi, 2\pi)$
A: I'm assuming here that you want the semi-circle and the line to be each parameterised over $[0,2]$.
The semi-circle is:
$$ S(t) = \cos\tfrac{t\pi}{2} + i  \sin\tfrac{t\pi}{2}  = \exp(\tfrac{1}{2}t\pi) \quad (0 \le t \le 2)  $$
The line is 
$$L(t) = (1 - \tfrac{1}{2}t)P +   \tfrac{1}{2}tQ \quad (0 \le t \le 2)$$
where the points $P$ and $Q$ are given by $P=(1,0)$, $Q=(-1,0)$.
A: On the other hand, if you want the entire contour (formed by stringing the two curves together) to be continously parameterised over $[0,2]$, then:
The semi-circle is:
$$ S(t) = \cos t\pi + i  \sin t\pi = e^{it\pi} \quad (0 \le t \le 1)  $$
The line is 
$$L(t) = (2 - t)P +   (t-1)Q \quad (1 \le t \le 2)$$
where the points $P$ and $Q$ are given by $P=(-1,0)$, $Q=(1,0)$.
Or, if you want a purely complex-number-ish description, $P = -1 +0i$ and $Q = 1 + 0i$, so $L(t) = 2t - 3$, as the comment indicated.
