Picture and length of an exponential complex curve 
Problem 4:  Draw a picture of the curve $\gamma(t) - e^{-t}e^{it}$ for $0\le t\le b$ and calculate the length of this curve.  What happens if you let $b \to\infty$?  Does this infinite curve have finite length?

How do I do this one? I'm pretty sure the graph should look like a spiral.
 A: I'm taking the curve to be
$\gamma(t) = e^{-t}e^{it}; \tag{1}$
then further
$\gamma(t) =  e^{-t}e^{it} = e^{t(i - 1)}, \tag{2}$
so that
$\gamma'(t) = (i - 1)e^{t(i - 1)} = (i - 1)\gamma(t); \tag{3}$
thus from (2) and (3)
$\Vert \gamma'(t) \Vert^2 = (\overline{i - 1})(i - 1)\overline{\gamma(t)}\gamma(t)  =  (\overline{i - 1})(i - 1)(e^{-t}e^{it})(e^{-t}e^{-it}) =2e^{-2t}, \tag{4}$
so that
$\Vert \gamma'(t) \Vert = \sqrt {2} e^{-t}; \tag{5}$
the length of the curve between $\gamma(0)$ and $\gamma(b)$, $l(b)$, is thus
$l(b) = \int_0^b \Vert \gamma'(t) \Vert dt = \int_0^b \sqrt {2} e^{-t}dt = \sqrt{2}\int_0^b  e^{-t}dt = \sqrt{2}(1 - e^{-b}). \tag{6}$
As $b \to \infty$, we see that the length $l(b)$ approaches:
$\lim_{b \to \infty} l(b) = \lim_{b \to \infty}\sqrt{2}(1 - e^{-b}) = \sqrt{2}; \tag{7}$
apparently "infinite curve" has finite length $\sqrt{2}$.
I dont' have the grahics tools to draw this curve, but it's pretty easy to see what happens:  the point $\gamma(t)$ on the curve starts at $\gamma(0) = 1$ and rotates in a counter-clockwise direction, all the while approaching the origin $0$; the distance between $\gamma(t)$ and $0$ is in fact $e^{-t}$ since $\Vert \gamma(t) \Vert = e^{-t}$; by the time $t = 2\pi$ we are again back on the real axis with a value of $\gamma(2\pi) = e^{-2\pi}$.  After each revolution the radius has decreased by a factor of $e^{-2\pi}$ in geometric progression.  It's pretty easy to visualize.
It gets even better in polars.  Since $\gamma(t) = e^{-t}e^{it} = e^{-t}(\cos t+ i \sin t) = r(\cos \theta + i\sin \theta)$, we can see by inspection that $r = e^{-t}$; $\theta = t$.  Thus the polar form is $r = e^{-\theta}$; the spiral nature of this curve is clear.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
