Provide parametisations for the triangle with vertices $1, 2i$ and $-1$ 
Sketch these paths and provide parametisations for them:

*

*The straight line from $z=0$ to $z=2$.


My idea:
From the logic that a line from $p$ to $q$ in $\mathbb C$ can be parametised as $z(t)=(1-t)p+tq$ for $0\leq t \leq 1$
$z(t)=2t$ for $0\leq t\leq 1$
Real answer:
$z(t)=t$ with $0 \leq t \leq 2$.


*

*The triangle with vertices $1, 2i$ and $-1$ traced anticlockwise.


My idea:
Matches with $z_1(t)$ and $z_2(t)$ but $z_3(t)=-1+2t$ for $0 \leq t \leq 1$
Real answer:
$z_1(t)=(1-t)\cdot1+2it=1-(1-2i)t$ for $0\leq t \leq 1$.
$z_2(t)=(1-t)\cdot2i+t\cdot(-1)=2i-(1+2i)t$ for $0 \leq t \leq 1$.
$z_3(t)=t$ for $-1\leq t \leq 1$
Can anyone show me how these parameters have been adjusted in the cases of $z(t)$ and $z_3(t)$?
 A: First of all your logic is sound and your calculations are correct. From my point of view they are also preferrable, since they always use the same interval $[0,1]$ for parametrisation which makes reading easier.

First case: 
\begin{align*}
z(t)=2t&\qquad 0\leq t\leq 1\\
z(u)=u&\qquad 0\leq u \leq 2
\end{align*}
In both cases we have as codomain of $z$ the closed interval $[0,2]$. The mapping between the two representations is given via $u(t)=2t$. We could somewhat sloppily say $t$ moves twice as fast as the $u$ from $0$ to $2$.
Second case:
\begin{align*}
z_3(t)&=-1+2t\qquad 0\leq t\leq 1\\
z_3(u)&=u\ \ \qquad\quad -1\leq u \leq 1
\end{align*}
In both cases we have as codomain of $z$ the closed interval $[-1,1]$. The mapping between the two representations is given via $u(t)=-1+2t$. Again we could say $t$ moves twice as fast as the $u$ from $-1$ to $1$.

A reason for the specific choice of the parametrisation interval for $z$ and $z_3$ might be convenience by simply using the identity function. Nevertheless I prefer the more concise way presented by OP.
