Finding the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$ I need to find the the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$. My issue is that just looking at the graph of $\arg{(z-2i)}=\frac{\pi}{6}$ (which is a ray from $(0,2)$ on the Argand diagram) and $|z-3+i|$ (a circle with centre at $(3,-1)$) wouldn't the lowest value just be the distance between $(0,2)$ and $(3,-1)$? However the question is worth $4$ marks and this isn't $4$ marks of working so I feel like I'm seriously overlooking something. Could someone clarify what I've missed or is the question that simple?
 A: 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 $\sec(\pi+u)=-\sec u$
A: The condition $\arg (z - 2i) = \frac{\pi}{6}$ is equivalent to $z = 2i + re^{\frac{\pi}{6}i},\quad r>0$ which does describe a ray.
For the quantity $|z - 3 + 2i|$ to be minimized, it is necessary and sufficient for $z$ to be on the smallest circle that shares a point with the aforementioned ray. i.e on the circle centered at $3 - 2i$ which is tangent to our ray.
Finding the radius of this circle is equivalent to finding the distance between $3 - 2i$ and the ray $z = 2i + re^{\frac{\pi}{6}i},\quad r>0$.
A: Alternative approach:
$\displaystyle \tan(\pi/6) = \frac{1}{\sqrt{3}}$.
Therefore, with $z = x + iy$, two constraints must be satisfied:

*

*$\displaystyle \frac{y-2}{x} = \frac{1}{\sqrt{3}}.$

*Both $(x)$ and $(y-2)$ must be positive (i.e. in the 1st quadrant), rather than both being negative.

The above constraints imply that $0 < x = (y-2)\sqrt{3}$.
You can minimize the absolute value of a complex number by minimizing the square of the absolute value.
Therefore, you want to minimize 
$|z - 3 + i|^2 = (x - 3)^2 + (y + 1)^2$ 
$ = D(y) = [(y - 2)\sqrt{3} - 3]^2 + (y + 1)^2.$
Examining derivatives:
$D'(y) = 2[(y - 2)\sqrt{3} - 3]\sqrt{3} + 2(y + 1)$.
This simplifies to $(y)[8] + [(-12) + (-6\sqrt{3}) + (2)]
= 8y - 10 - 6\sqrt{3}$.
So, $~\displaystyle D'(y) = 0 \iff y = \frac{5 + 3\sqrt{3}}{4} \implies (y - 2) > 0.$
Further, $D''(y) = 8 > 0.$
Therefore, $D$ is minimized at
$\displaystyle y = \frac{5 + 3\sqrt{3}}{4}.$
At that value for $y$, you have that
$\displaystyle D(y) = \left[\frac{3\sqrt{3} - 3}{4}\sqrt{3} - 3\right]^2 + \left[\frac{9 + 3\sqrt{3}}{4}\right]^2$
$\displaystyle = \left[\frac{- 3 - 3\sqrt{3}}{4}\right]^2 + \left[\frac{9 + 3\sqrt{3}}{4}\right]^2$
$\displaystyle = \frac{1}{16} ~~\times 
~~\left\{ ~\left[9 + 27 + 18\sqrt{3}\right]
~+ \left[81 + 27 + 54\sqrt{3}\right] ~\right\}$
$\displaystyle = \frac{1}{16} ~~\times 
~~\left\{ ~\left[144 + 72\sqrt{3}\right]
 ~\right\} = \frac{1}{4} ~~\times ~~\left[36 + 18\sqrt{3}\right].$
In order to compute $\sqrt{D(y)}$, you must find $a,b \in \Bbb{R}$ such that

*

*$\left(a + b\sqrt{3}\right) > 0$

*$a^2 + 3b^2 = 36$

*$(2ab) = 18$.

Note that the $3$rd constraint above implies that $a$ and $b$ are both positive or both negative.  Then, the first constraint above implies that $a,b$ are both positive.
You can guess that $a = b = 3$, or (more formally)
Setting $\displaystyle a = \frac{9}{b}$ leads to
$$\frac{81}{b^2} + 3b^2 = 36 \implies 
3b^4 - 36b^2 + 81 = 0.\tag1 $$
Regarding (1) above as a quadratic in $b^2$ gives:
$\displaystyle b^2 = \frac{1}{6} \left[36 \pm \sqrt{1296 - (972)}\right] = \frac{1}{6} \left[36 \pm 18\right].$
The convenient try is $\displaystyle b^2 = \frac{36 + 18}{6} = 9 \implies b = 3.$
Therefore, the minimum value for $\sqrt{D(y)}$ is
$$\sqrt{\frac{1}{4} \times \left[36 + 18\sqrt{3}\right]}
~= ~\frac{1}{2} \left[3 + 3\sqrt{3}\right].$$
