Showing distance from $z_0$ to the line parametrized by $z(t)=w_0+te^{i \theta}$ is Suppose $w_0$ and $z_0$ are in $\mathbb{C}$ and $\theta$ is a fixed angle with $0 \le \theta \le 2 \pi$. Show that the distance from the point $z_0$ to the line parametrized by $z(t)=w_0+te^{i \theta}$, $t \in \mathbb{R}$ is $$|Im((z_0-w_0)e^{-i \theta})|$$. 
How do I take the length of $|z(t)|$ to show that this is equal to $|w_0sin \theta-z_0 sin \theta|$? Thanks in advanced. 
 A: Write $z(t) = (x(t), y(t)) = w_0 + t\cdot e^{i\theta} = (x_0 + t\cdot cos\theta, y_0 + t\cdot sin\theta)$ with $w_0 = (x_0, y_0)$ and $z_0 = (a_0, b_0)$. Consider $f(t) = (x_0 - a_0 + t\cdot cos\theta)^2 + (y_0 - b_0 + t\cdot sin\theta)^2 = t^2 + (x_0 - a_0)^2 + 2(x_0 - a_0)\cdot t\cdot cos\theta + (y_0 - b_0)^2 + 2(y_0 - b_0)\cdot t\cdot sin\theta$. Taking derivative with respect to $t$:
$f'(t) = 2t + 2(x_0 - a_0)\cdot cos\theta + 2(y_0 - b_0)\cdot sin\theta = 0$ gives:
$t = -(x_0 - a_0)\cdot cos\theta - (y_0 - b_0)\cdot sin\theta$. This is the critical value  $t_0$. So $d = \sqrt{f_{min}} = \sqrt{f(t_0)} = \sqrt{((x_0 - a_0)\cdot sin\theta - (y_0 - b_0)\cdot cos\theta)^2} = |(x_0 - a_0)\cdot sin\theta - (y_0 - b_0)\cdot cos\theta| = |Im((z_0 - w_0)e^{-i\theta})|$
A: Note:  I'm a taking $\theta \in [0, 2\pi)$, i.e. $0 \le \theta < 2\pi$.  This doesn't restrict $z(t)$ in any way, and by eliminating a possible ambiguity in $\theta$, makes things a little more clear.  To my mind, in any event.
Write $z_0 - w_0 = re^{i\phi}$ for some $\phi \in [0, 2\pi)$.  Note that $\vert z_0 - w_0 \vert = r$, the ordinary Euclidean length of the line segment $\overline{z_0 w_0}$ joining the points $z_0$ and $w_0$ in the complex plane.  Furthermore, $\phi - \theta$ is the angle between the segment $\overline{z_0 w_0}$ and the line $w_0 + te^{i\theta}$.  We have
$(z_0 - w_0)e^{-i\theta} = re^{i\phi}e^{-i\theta} = re^{i(\phi - \theta)}$
$= r\cos(\phi - \theta) + ir \sin(\phi - \theta), \tag{1}$
and it is evident from (1) that
$\vert \Im ((z_0 - w_0)e^{-i\theta}) \vert = r \vert \sin(\phi - \theta) \vert \tag{2}$
since $r \ge 0$.  If $z_0$ is not on the line $z(t)$, let $L$ be the line perpendicular to $z(t)$ through $z_0$.   Let $y \in \Bbb C$ be the point of intersection of line $L$ and $z(t)$.  If $y = w_0$, then $\vert z_0 - y \vert = \vert z_0 - w_0\vert = r$ is the distance between $z_0$ and $z(t)$ since it is then the length of the perpendicular segment $\overline{z_0 y}$ joining $z_0$ and $z(t)$; in this case we have at least one of $\phi = \theta \pm \pi /2$ holds, depending on the relative positions of $\phi, \theta \in [0, 2\pi)$, so at least one of $\phi - \theta = \pm \pi / 2$ holds, whence at least one of $\sin(\phi - \theta) = \pm 1$ holds as well, so that (2) becomes
$\vert \Im ((z_0 - w_0)e^{-i\theta}) \vert = r, \tag{3}$
showing the hypothesized formula holds in the case $y = w_0$.  If $y \ne w_0$, consider the triangle $w_0 z_0 y$.  Since $\overline{z_0 y} \bot z(t)$, $\triangle w_0 z_0 y$ is a right triangle with right angle $\angle z_0 y w_0$, hypoteneuse $\overline{w_0 z_0}$ of length $\vert z_0 - w_0 \vert = r$, and $\angle z_0 w_0 y = \phi - \theta$ opposite side $\overline{z_0 y}$.  It follows that the length of side $\overline{z_0 y}$, which is $\vert z_0 - y \vert$, the perpendicular distance between $z_0$ and $z(t)$, is $\vert r \sin (\phi - \theta) \vert =  r\vert \sin (\phi - \theta) \vert = \vert \Im ((z_0 - w_0)e^{-i\theta}) \vert$.  Finally, in the event that $z_0 \in z(t)$, we must have $\phi = \theta$, $\phi = \theta + \pi$ or $\phi = \theta - \pi$; in each of these cases, $\sin (\theta - \phi) = 0$, so the formula applies when $z_0 \in z(t)$ as well.  QED.
I find this argument interesting insofar as it combines techniques from both plane geometry and complex analysis in one unified setting.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
