Finding the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$
Solution 1:
Instead of $(0,3)$ the straight line passes through many arbitrary points, right?
Also the distance has to be from some point on the circle to some other point on the straight line.
WLOG if $z=x+iy$
let $x-3+i (y+1)=r(\cos t+i\sin t)$ where $t$ is real and $r\ge0$
$$\tan\dfrac\pi6=\dfrac{y-2}x=\dfrac{r\sin t-3}{r\cos t+3}$$
$$\sqrt3(r\sin t-3)=r\cos t+3$$
$$\iff2r\cos (t-\pi/3)=-3(\sqrt3+1)$$
$$r=\cdots=\dfrac{3(\sqrt3+1)\sec(t+2\pi/3)}2\ge3(\sqrt3+1)/2$$ as $\sec(\pi+u)=-\sec u$