Finding the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$

Question

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?

Answer 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 $\sec(\pi+u)=-\sec u$

Answer 2

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

$$\sqrt{\frac{1}{4} \times \left[36 + 18\sqrt{3}\right]} ~= ~\frac{1}{2} \left[3 + 3\sqrt{3}\right].$$

Answer 3

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 $z = 2i + re^{\frac{\pi}{6}i},\quad r>0$.

Answer 4

As you remarked, $ \ |z - (3 - i)| \ $ describes the set of points at some constant distance from $ \ z_1 \ = \ 3 - i \ \ . \ $ The smallest such circle that has points on the ray given by $ \ \arg ( z - 2i) $ $ = \ \arg ( z - z_2) \ = \frac{\pi}{6} \ $ will be the circle to which this ray is just tangent (at $ \ z_3 \ ) \ ; $ the radius from $ \ z_1 \ $ to $ \ z_3 \ $ is then perpendicular to that ray, making its length the perpendicular distance from $ \ z_1 \ $ to the ray.

We would solve this problem vectorially then by calculating the scalar projection of $ \ \overrightarrow{z_1 z_2} \ $ onto $ \ \overrightarrow{z_1 z_3} \ \ , \ $ which has the desired value $ \ \Vert \overrightarrow{z_1 z_2} \Vert · \cos \theta \ \ , \ $ where $ \ \theta \ $ is the included angle between the two vectors. We learn that $$ \cos \theta \ \ = \ \ \frac{\overrightarrow{z_1 z_2} \ \centerdot \ \overrightarrow{z_1 z_3}}{\Vert \overrightarrow{z_1 z_2} \Vert \ \Vert \overrightarrow{z_1 z_3} \Vert} \ \ , $$ making the scalar projection of $ \ \overrightarrow{z_1 z_2} \ $ onto $ \ \overrightarrow{z_1 z_3} \ $ $$ s \ \ = \ \ \frac{\overrightarrow{z_1 z_2} \ \centerdot \ \overrightarrow{z_1 z_3}}{ \Vert \overrightarrow{z_1 z_3} \Vert} \ \ . $$

We are given the coordinates of the "endpoints" of $ \ \overrightarrow{z_1 z_2} \ \ , \ $ so we have $ \ \overrightarrow{z_1 z_2} \ = \ \langle \ (0 - 3) \ , \ (2 - (-1) \ \rangle $ $ = \ \langle \ -3 \ , \ 3 \ \rangle \ $ and $ \ \Vert \overrightarrow{z_1 z_2} \Vert \ = \ \sqrt{ \ (-3)^2 + 3^2 } \ = \ 3·\sqrt2 \ \ . \ $ Since the ray extending from $ \ z_2 \ $ makes an angle $ \ \frac{\pi}{6} \ $ to the positive-real axis, it has a slope $ \ \tan \frac{\pi}{6} \ = \ \frac{1}{\sqrt3} \ $ or a "direction vector $ \ \langle \ \sqrt3 \ , \ 1 \ \rangle \ \ . \ $ From what we know about slopes of perpendicular lines, we can thus represent $ \ \overrightarrow{z_1 z_3} \ $ by its direction vector, $ \ \langle \ 1 \ , \ -\sqrt3 \ \rangle \ \ $ (alternatively, we see that the dot product of these two direction vectors is zero). We can make $ \ \overrightarrow{z_1 z_3} \ $ an arbitrary scalar multiple of its direction vector, with $ \ \Vert \overrightarrow{z_1 z_3} \Vert \ = \ k·\sqrt{ \ 1^2 \ + \ (-\sqrt3)^2} \ = \ k·2 \ \ . \ $ So the minimum value of $ \ |z - (3 - i)| \ $ is the absolute-value of $$ s \ \ = \ \ \frac{\langle \ -3 \ , \ 3 \ \rangle \ \centerdot \ \langle \ k·1 \ , \ k·(-\sqrt3) \ \rangle}{ k·2} \ \ = \ \ \frac{ (-3)·k \ + \ 3·(-\sqrt3)·k }{ k·2} \ \ = \ \ \frac{ (-3) \ - \ 3·\sqrt3 }{ 2} \ \ , $$ or $ \ \boxed{ \ |s| \ = \ \frac32·(1 \ + \ \sqrt3 ) \ } \ \approx \ 4.098 \ . $

We can re-cast this calculation using complex numbers. Using the vectors we determined above, we designate $ \ u \ = \ z_2 - z_1 \ = \ (2i) - (3 - i) \ = \ -3 + 3i \ $ and $ \ v \ = \ z_3 - z_1 \ = \ k·(1 - \sqrt3·i) \ \ . \ $ If we represent two vectors $ \ \vec{u} \ = \ \langle \ a \ , \ b \ \rangle \ \ , \ \ \vec{v} \ = \ \langle \ c \ , \ d \ \rangle \ $ by the complex numbers $ \ u \ = \ a + bi \ \ , $ $ v \ = \ c + di \ \ , \ $ then $ \ u \overline{v} \ = \ (ac + bd) - (ad - bc)i \ \ ; \ $ the dot product of the vectors is then represented by $ \ \vec{u}·\vec{v} \ = \ \mathfrak{Re}(u \overline{v}) \ \ . \ $ The "scalar projection" we calculated above would now be found from $$ s \ \ = \ \ \frac{\mathfrak{Re}(u \overline{v})}{|v|} \ \ = \ \ \frac{\mathfrak{Re}[ \ (-3 + 3i) \ ( k·(1 + \sqrt3·i) \ ]}{k·\sqrt{ \ 1^2 \ + \ (-\sqrt3)^2}} \ \ = \ \ \frac{ (-3)·k·1 \ + \ 3·k· \sqrt3·i^2}{k·2} $$ $$ = \ \ \frac{ k·(-3 \ - \ 3 \sqrt3)}{k·2} \ \ = \ \ \frac{ -3 \ - \ 3 \sqrt3}{2} \ \ , $$

as before.

[Note that we have not needed the value of $ \ z_3 \ $ in order to complete the calculation, although we can now easily determine that

$$ z_3 \ \ = \ \ z_1 \ + \ s·\frac{v}{|v|} \ \ = \ \ (3 \ - \ i) \ \ + \ \ [ \ -\frac32·(1 \ + \ \sqrt3 ) \ ]·\frac{ (1 \ - \ \sqrt3·i)}{2} $$ $$ = \ \ \left(\frac{9 \ - \ 3 \sqrt3}{4} \right) \ + \ \left(\frac{5 \ + \ 3 \sqrt3}{4} \right)·i \ \ \approx \ \ 0.951 \ + \ 2.549·i \ \ . \ $$

We can check that $$ z_3 \ - \ z_2 \ \ = \ \ \left[ \ \left(\frac{9 \ - \ 3 \sqrt3}{4} \right) \ + \ \left(\frac{5 \ + \ 3 \sqrt3}{4} \right)·i \ \right] \ - \ 2i $$ $$ = \ \ \left(\frac{9 \ - \ 3 \sqrt3}{4} \right) \ + \ \left(\frac{-3 \ + \ 3 \sqrt3}{4} \right)·i $$ $$ \Rightarrow \ \ \arg(z_3 \ - \ z_2) \ \ = \ \ \arctan \ \left(\frac{\frac{-3 \ + \ 3 \sqrt3}{4} }{\frac{9 \ - \ 3 \sqrt3}{4}} \right) \ \ = \ \ \arctan \ \left(\frac{-3 \ + \ 3 \sqrt3 }{9 \ - \ 3 \sqrt3} \right) \ \ $$ $$ = \ \ \arctan \ \left(\frac{(27 - 9)· \sqrt3 }{81 \ - \ 27} \right) \ \ = \ \ \arctan \ \left(\frac{18· \sqrt3 }{54} \right) \ \ = \ \ \arctan \ \frac{\sqrt3}{3} \ \ = \ \ \frac{\pi}{6} \ \ , $$

the argument for the specified ray.]