Find the greatest value of $\arg z$ achieved on a circle in $\mathbb{C}$ Consider the circle $$|z-6i|=3;$$ its centre is $(0,6)$ and its radius is $3$.
I want to find the greatest value of $\arg z$ achieved on this circle. My idea is that the tangents to the circle from the origin create right angles with the radius. Then use the Pythagorean Theorem and trigonometry to find the angle $\theta$ between the tangent and the imaginaery axis.
Then the formula for finding the two values of $\arg z$ in the form $p\pi$ where $-1\lt{p}\le1$ are:
$$\frac{\pi-2\theta}{2}=argz$$
$$\frac{\pi-2\theta}{2}+2\theta=argz$$
 A: You're on the right track. Here I'll assume we're using the branch of the argument function that takes values in $(-\pi, \pi]$, but this argument will work just as well for any branch that is constant on open rays from the origin, takes the value $\frac{\pi}{2}$ on the positive imaginary axis, and is continuous on the given circle $C$. (In fact, with just a little tweaking we can drop the first hypothesis.)
The argument function $\arg z$ is constant on open rays centered at the origin $0$. Since it increases with anticlockwise angle from the positive real axis (at least up to an angle $\pi$), $\arg z$ is maximized among points on $C$ when that angle is as large as possible, and in particular for that angle the corresponding ray from $0$ must be tangent to $C$. By definition, the (unsigned) angle $\alpha$ such a tangent ray makes with the ray from $0$ through the center $Q$ of the circle satisfies
$$\sin \alpha = \frac{R}{d},$$ where $d := |\overline{0Q}|$ and $R$ is the radius of $C$.

In our case, $C$ is centered at $6 i$, so $d = |6 i| = 6$ and $R = 3$. Substituting gives $$\sin \alpha = \frac{(3)}{(6)} = \frac{1}{2},$$ and hence $$\alpha = \frac{\pi}{6}.$$  Since the ray from $\overrightarrow{0Q}$ has (constant) argument $\frac{\pi}{2}$, the rays from $0$ tangent to $C$ have arguments $$\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \quad \text{and} \quad \frac{\pi}{2} + \frac{\pi}{6} = \frac{2 \pi}{3}.$$ The larger of these is $${\boxed{\dfrac{2 \pi}{3}}}.$$

A similar argument works for any circle that does not contain the origin by the way, that is, whenever $R < d$, provided that one works with an argument function continuous on $C$.
A: $z=x+iy\implies x^2+(y-6)^2=3$
We can write $y-6=\sqrt3\sin\theta,x=\sqrt3\cos\theta$
arg$(z)=\arctan\left(\dfrac{y-6}x\right)=\arctan(\tan\theta)$ 
Using atan2, arg$(z)\le\pi$ and
the equality occurs if $x=\sqrt3\cos\pi=-\sqrt3$ and $y-6=\cdots$
