Calculate path length of solution to differential equation Suppose I have a parameterized path $\gamma:[0,\infty)\rightarrow\mathbb C$ with initial condition $\gamma(0)=1$ and which satisfies the following:
$$\gamma'(t)=u\Big(i\gamma(t)-\gamma(t)\Big)=\exp\Big(\frac{3\pi}{4}i\Big)\frac{\gamma(t)}{|\gamma(t)|}$$
where $u:\mathbb C^*\rightarrow\mathbb D$ denotes the unit vector function. In other words, $\gamma(t)$ is always moving at constant unit speed towards its rotation in the plane by $\pi/2$. I found computationally that
$$\lim_{t\rightarrow\infty}\gamma(t)=0$$

Questions:


*

*Is there a formula for $\gamma(t)$?

*Does $\gamma$ reach the origin in finite time? In other words, does there exist $T<\infty$ such that $\lim_{t\rightarrow T}\gamma(t)=0$?
Edit: Originally we had $\gamma'(t)=s(i-1)\frac{\gamma(y)}{|\gamma(t)|}$ but without loss of generality set $s=1/\sqrt{2}$ so that $\gamma'(t)$ has unit magnitude. Also I realize that the path length $\int_0^\infty|\gamma'(t)|dt$ must be infinite since $|\gamma'(t)|$ is constant.
 A: It appears that this equation can be well-addressed via a transformation to polar coordinates; in polars, we write
$\gamma(t) = r(t)e^{i\theta(t)}; \tag 1$
then
$\dot \gamma = \dot r e^{i\theta} + r i \dot \theta e^{i\theta}; \tag 2$
also,
$\vert \gamma \vert = r, \tag 3$
whence
$\dfrac{\gamma}{\vert \gamma \vert} = e^{i\theta}; \tag 4$
assembling all this into the given equation
$\dot \gamma = s(i - 1) \dfrac{\gamma}{\vert \gamma \vert} \tag 5$
yields
$\dot r e^{i\theta} + r i \dot \theta e^{i\theta} = s(i - 1) e^{i\theta}; \tag 6$
we divide out $e^{i\theta}$:
$\dot r + ri\dot \theta = s(i - 1) = -s + is; \tag 7$
equating real and imaginary parts,
$\dot r = -s, \tag 8$
$ri\dot \theta = is; \tag 9$
$r\dot \theta = s; \tag{10}$
we may easily integrate (8) 'twixt $t_0$ and $t$, assuming $r(t_0) = r_0$:
$r - r_0 = -s(t - t_0), \tag{11}$
$r = r_0 - s(t - t_0); \tag{12}$
we observe that the coordinate constraint $r > 0$, along with the condition $s > 0$, implies that $t$ must remain less than $r_0/s + t_0$; we shall soon find that the polar solution circumvents this limit on $t$; inserting (12) into (10):
$(r_0 - s(t - t_0))\dot \theta = s; \tag{13}$
$\dot \theta = \dfrac{s}{r_0 - s(t - t_0)}, \tag{14}$
which is also easily integrated 'twixt $t_0$ and $t$:
$\theta - \theta_0 = -\ln(r_0 - s(t - t_0)) + \ln r_0; \tag{15}$
$\theta = \theta_0 - \ln(r_0 - s(t - t_0)) + \ln r_0; \tag{16}$
using (12), we may express $r$ in term of $\theta$:
$\theta = \theta_0 - \ln r + \ln r_0; \tag{17}$
$e^\theta = e^{\theta_0} \dfrac{r_0}{r}; \tag{18}$
$r = r_0 e^{\theta_0 - \theta}; \tag{19}$
the above presents the general solution for $t = t_0$, $r(t_0) = r_0$, $\theta(t_0) = \theta_0$; we specialize to the case at hand:
$t_0 = 0, \; \gamma(0) = 1 + i; \tag{20}$
$r_0 = \vert \gamma(0) \vert = \sqrt 2; \tag{21}$
$\theta_0 = \dfrac{\pi}{4}; \tag{22}$
from (19):
$r = \sqrt 2 e^{\pi/4 - \theta}; \tag{23}$
from (12) and (16):
$r(t) = \sqrt 2 - st; \tag{24}$
$\theta(t) = \dfrac{\pi}{4} - \ln(\sqrt 2 - s(t - t_0)) + \ln \sqrt 2. \tag{25}$
The path length is perhaps easiest found by returning to the expressions for $r(t)$ and $\theta(t)$, in terms of which
$\vert \dot \gamma \vert = \sqrt{ (\dot r)^2 + r^2 (\dot \theta)^2} = \sqrt{s^2 + s^2} = \sqrt 2 s; \tag{26}$
since $t$ is restricted to $[0, r_0/s]$, the length of the curve is
$\displaystyle \int_0^{r_0/s} \vert \dot \gamma \vert \; dt = \displaystyle \int_0^{r_0/s} \sqrt 2 s \; dt = \sqrt 2 r_0. \tag{27}$
As concerns our OP M. Nestor's final remarks:  
The curve $\gamma(t)$ (1), expressed as a function of the independent variable $\theta$ is, in polar coordiates, given by (23); it is easy to see this curve is a spiral which approaches the origin as $\theta \to \infty$. 
Now, as for item (1), there is indeed a formula for $\gamma(t)$; in polar coordinates the equations are given in (12) and (16); from these we may derive
cartesian equations for $x(t)$, $y(t)$ by setting
$x(t) = r(t) \cos \theta (t), \; y(t) = r(t) \sin \theta(t); \tag{28}$ 
and as for item (2), we have shown above ca. (12) that
$\displaystyle \lim_{t \to t_0 + r_0/s} r(t) = 0, \tag{29}$
where $\gamma(t_0) = (r_0, \theta_0)$; but it should be noted that $\gamma(t)$ never "reaches the origin", though it does become arbitrarily close as $t \to (t_0 + r_0 / s)^-$.
Finally, in response to M. Nestor's final edit, we remark that the integral
$\displaystyle \int_{t_0}^\infty \vert \gamma'(t) \vert \; dt \tag{30}$
is not defined, since $\gamma(t)$ can't be extended past $t = t_0 + r_0/s$; the length may instead be found via (27).
