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It would be really awesome if you guys could give me a hand on this question!

Question: Given that the equation of a line is $y=-\sqrt{3}x$. This line can be also expressed in $|z-1|=|z-k|$, where $k \in\Bbb C$. Find $k$ in cartesian form.

What I've done so far:

  • Since the perpendicular gradient is $\frac{1}{\sqrt3}$, I made the equation $\frac{y-0}{x+1}=\frac{1}{\sqrt3}$. This lead me to no-where
  • I've also considered using $cis\frac{-2\pi}{3}$, since it's perpendicular, but I'm confused on what to do from here

The answer is $-\frac{1}{2} - \frac{\sqrt3}{2}i$

Please correct me if I am wrong.

Shubham Johri
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Munchies
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    How did you make the equation $\frac{y-0}{x+1}=\frac1{\sqrt3}$? Shouldn't it be $y=\frac1{\sqrt3}(x\color{red}-1)$? – J. W. Tanner Nov 06 '20 at 05:34

4 Answers4

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The line is $y=-\sqrt3 x$,

so a line that passes through $(1,0)$ and is perpendicular to the line is $y=\dfrac x{\sqrt3}-\dfrac1{\sqrt3}.$

These lines intersect when $-\sqrt3 x=\dfrac x{\sqrt3}-\dfrac1{\sqrt3}$ ; i.e., $x=\dfrac14$ and $y=\dfrac{-\sqrt3}4$.

Now double the segment from $(1,0) $ to $(\frac14,\frac{-\sqrt3}4)$ to get $(\frac{-1}2,\frac{-\sqrt3}2).$

J. W. Tanner
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You can solve this in a purely algebraic way as follows:

$z=0$ lies on the line, hence

$$|0-1| = |0-k| \Rightarrow |k|=1$$

$z = 1-\sqrt 3 i$ lies also on the line. Hence, plugging this into the equation $|z-1| = |z-k|$ gives

$$\sqrt 3 = |1-\sqrt 3 i - k|$$

Squaring and setting $k = \cos t + i \sin t$ gives

$$3 = (1-\cos t)^2 + (-\sqrt 3 - \sin t)^2$$

Expanding and rearranging gives

$$-1 = -\cos t + \sqrt 3 \sin t = 2\cos \left(t-\frac{2\pi}{3}\right)$$

The solutions for $t \in [0,2\pi)$ are $t=0$ and $t=\frac{4\pi}{3}$. Hence, the looked for $k$ is

$$k = \cos\left(\frac{4\pi}{3}\right) + i \sin \left(\frac{4\pi}{3}\right) = -\frac 12 - \frac{\sqrt 3}{2}i$$

trancelocation
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I think drawing out your problem will help you understand the equation you seek

Now, the line given is $y = -\sqrt3 x$. This line passes through the origin, and has a slope of $\tan \theta = -\sqrt3$

Now, the equation $|z-1| = |z-k|$ describes the line that is the perpendicular bisector of the points $(1,0)$ and $(k_x, k_y)$

What is the slope of the line segment joining these two points?

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{k_y - 0}{k_x - 1}$$

Now, another equation we have for $m$ comes from the perpendicular

$$m\times(- \sqrt3) = -1$$

$$\implies m = \frac{1}{\sqrt3}$$

So now you have two equations to help find $k_x, k_y$ -

  1. $$\sqrt3 (k_y) = k_x - 1$$
  2. $$\frac{k_y}{2} = -\sqrt3 \left(\frac{k_x+1}{2}\right)$$
Dhanvi Sreenivasan
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Here is an intuitive way to solve the problem: $|z-1|=|z-k|$ is the set of all points equidistant from $(1,0)$ and $k$. This is the perpendicular bisector of the line segment joining $(1,0)$ and $k$. Thus the point $k$ is the reflection of $(1,0)$ with respect to the perpedicular bisector $y=-x\sqrt3$.

Say $k=(a,b)$, then mid-point of $k$ and $(1,0)$ lies on bisector, i.e. $\frac b2=-\sqrt3(1+a)/2\implies b=-(1+a)\sqrt3$ .

Further the slope of segment joining $k$ and $(1,0)$ is $1/\sqrt3$, so $\frac{b}{a-1}=1/\sqrt 3$. Solving these two equations gives $a=-1/2,b=-\sqrt3/2$.

Shubham Johri
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