Calculate the least value of arg z for points on this locus. I answered a question about a complex number's locus on an Argand diagram, and I got most of it correct. This last question, I can't figure out (I'm sure it's pretty straightforward, but it might be because I'm very tired that I can't figure this one out).
Calculate the least value of arg z for points on this locus.
Here's the diagram. It has something to do with tangent and normal, I assume, but I can't figure out how to find the common point at which the line would intersect the circle. I only know one point on the tangent, the origin, and no slope.

 A: The least value of arg $z$ for on this locus happens when the purple line touches the origin $(0,0)$ and the tangent point of the circle (as drawn in your sketch).
You can find the gradient of the purple line and then proceed to find arg $z$ since the gradient $m \ = \ \tan \theta$ and thus arg $z \ = \ \tan^{-1} \theta$.
Using distance from a point to a line we can find the gradient of the purple line and thus the least value of arg $z$.
The formula for the distance, $d$ of a point $(x_{1}, y_{1})$ from a line $Ax+By+C = 0 \ $ is
$$ d = \frac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$$
The distance, $d$ is the radius of the circle, $d = 1$
The purple line equation is $ \ y = mx \ $ or $ \ mx \ - \ y \  = 0$
The point is the centre of the circle $(2, 3)$
\begin{align*}
 1   \ &=  \ \frac{|2m \ - \ 3|}{\sqrt{m^2+(-1)^{2}}} \\[3ex]
 \sqrt{m^2+(-1)^{2}}  \ &= \ |2m \ - \ 3| \\[1ex]
m^{2} + 1 \ &= \ (2m-3)^{2} \\[2ex]
3m^2 \ - \ 12m + 8 \ &= \ 0 \\[2ex]
m \ &= \ 2 \pm \frac{2}{3}\sqrt{3}
\end{align*}
Use the smaller value of $m$,
\begin{align*}
   m \ &= \ 2 \ - \ \frac{2}{3}\sqrt{3}  \\[1ex]
\tan \theta \ &= \ 2 \ - \ \frac{2}{3}\sqrt{3} \\[1ex]
\theta \ &= \ \tan^{-1} \Big( 2 \ - \ \frac{2}{3}\sqrt{3}  \Big)\\[1ex]
\theta \ &= \ 0.702...
\end{align*}
Therefore, the least of arg $z \ = \ 0.702...$
Cheers,
Mr Will
A: Let the slope of the tangent line be $m$. Consider the point $P = (3/m, 3)$ which lies on the line and call $Q = (2, 3)$. Computing the area of $\Delta OPQ$ two different ways:
$$\frac{1}{2} (3/m - 2) \cdot 3 = \frac{1}{2} \sqrt{(3/m)^2 + 3^2} \cdot 1$$
$$ \Rightarrow 9(3/m - 2)^2 = 9/m^2 + 9$$
$$ \Rightarrow 9/m^2 - 12/m + 4 = 1/m^2 + 1$$
$$\Rightarrow 8 - 12m + 3m^2 = 0 \tag{$m \ne 0$}$$
$$\Rightarrow m = \frac{12 ± \sqrt{48}}{6}$$
and we are looking for the smaller value of $m$ from the diagram, hence $m = \tan \theta = 2 - \sqrt{4/3}$.
