Provide parametisations for the triangle with vertices $1, 2i$ and $-1$

Question

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 $z_3(t)$?

Answer 1

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.