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 $e$

(You have received answers for the rest, so let me focus on the length.)
The length of the curve is given by
$$\int_0^b |z'(t)|\,dt = \int_0^b |(-1+i)e^{(-1+i)t}|\,dt = \int_0^b \sqrt 2 e^{-t}\,dt.$$
I think you can work out what happens when $b\to\infty$ now.
