<s>[INST] 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?
 [/INST] 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
$$\sqrt{\frac{1}{4} \times \left[36 + 18\sqrt{3}\right]}
~= ~\frac{1}{2} \left[3 + 3\sqrt{3}\right].$$
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