Question

Prove that the least value of $r$ for which the two curves $\arg \left(z\right)=\frac{\pi }{6}$ and $|z-2\sqrt{3i}|=r$ intersect is $3$.

Answer 1

Change the two equations to Cartesian form
$\arg z=\frac{\pi }{6}$ $\Rightarrow$ $\frac{y}{x}=\tan {30}^o$ $\Rightarrow$ $x-\sqrt{3y}=0$
$|z-2\sqrt{3i}|=r$ represents a circle centred at $(0,2\sqrt{3})$ and radius $r$
$x^2+{(y-2\sqrt{3})}^2=r^2$
If the two curves touch, then $p=r$ and in case they intersect, then $p\le r$ where $p$ is perpendicular from centre $(0,2\sqrt{3})$ to line
$\therefore$ $\left|\frac{0-2\sqrt{3}.\sqrt{3}}{2}\right|\le r$ or $3\le r$ $\therefore r\ge 3$
Hence the least value of $r$ is 3 for the curves to intersect. (Touching is also intersection at two coincident points).