Sketch complex curve $z(t) = e^{-1t+it}$, $0 \le t \le b$ for some $b>0$ Sketch complex curve $z(t) = e^{-1t+it}$, $0 \le t \le b$ for some $b>0$ 
I tried plotting this using mathematica, but I get two curves.
Also, how do I find its length, is it just the integral?
This equation doesn't converge right?
Edit: I forgot the $t$ in front of the $1$ so it's not a circle of radius $e$

Since
$$
e^{-t+it}=e^{t(-1+i)}=e^{-t}(\cos(t)+i\sin(t))
$$
the image you're trying to find is an arc of a spiral with center on the origin, where the angular range is $[0,b]$.
