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Let $z$ be a complex number such that $|z|+|z-2019|=2019$. Note that $$|z+(2019-z)|=2019=|z|+|z-2019|=|z|+|2019-z|$$ This equality occurs when $0,z,2019-z$ are collinear.

But, how to show that z is a real number from that?

Note. By using the definition of modulus, i can show that $\text{Im}(z)=0$ from the equation $|z|+|z-2019| = 2019$. But, i wonder if i can get the same result with the previous way. Thanks.

Agung Izzul Haq
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  • If we approach in the way you you did, we can get $(x, y):= x + iy$ for $x, y \in \mathbb{R}$. Then we have that as $\mathbf{0} \equiv (0, 0), (x, y) $ and $(2019-x, y)$ these points are co-linear, then the slope is $0$ of the line joining the points. Therefore $$ 0 = \frac{y}{x} \implies y = 0$$Hence, $y = \text{Im}(z) = 0$ – Ralph Clausen Aug 07 '20 at 06:40

5 Answers5

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$|a+b|=|a|+|b|$ iff $a =t b$ for some $t \geq 0$ (or $b =t a$ for some $t \geq 0$). Here we get $z=t(z-2019)$ (which implies $t \neq 1$). So $z= \frac{(2019) t} {t-1}$ which is real.

Kavi Rama Murthy
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  • I got it. But, can you provide me the source? I want to read more about it. – Agung Izzul Haq Aug 07 '20 at 06:51
  • @AgungIzzulHaq $|a+b|=|a|+|b|$ iff $|a|^{2}+|b|^{2}+2 \Re (a\overline b) =|a|^{2}+|b|^{2}+2|a||b|$ iff $\Re a\overline b =|a||b|$. Now just take $t=\frac {|a|} {|b|}$ and verify by direct calculation that $|a-tb|^{2}=0$ which gives $a=tb$. – Kavi Rama Murthy Aug 07 '20 at 07:35
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As mentioned in comment too, If we approach in the way you you did, we can get $(x, y):= x + iy$ for $x, y \in \mathbb{R}$. Then we have that as $\mathbf{0} \equiv (0, 0), (x, y) $ and $(2019-x, - y)$ these points are co-linear, then the slope is $0$ of the line joining the points. Therefore $$ 0 = \frac{y}{x} \implies y = 0$$Hence, $y = \text{Im}(z) = 0$

Ralph Clausen
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Equations of form $$|z-a| + |z-b|=c$$ describe

  1. An ellipse (when $c> |a-b|$, with focus at $a,b$)
  2. A straight line (when $c=|a-b|$)
  3. Nothing (for the remainder)

on an Argand plane. For someone new to complex numbers or someone who is interested to see a visual I present you a graph made for you which can be found here.

In this case $|a-b|$ is $2019$ and $c$ is also $2019$. Hence, a straight line on the real axis with endpoints at $0$ and $2019$.

Anindya Prithvi
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  • Can you explain why did it a straight line, if $c=|a-b|$? I guess it's related to distance between a, b, and z. – Agung Izzul Haq Aug 07 '20 at 07:02
  • @AgungIzzulHaq see, $|z- t|$ denotes the distance between an arbitrary z and a fixed complex number $t$, so the primary equation in your question basically tells that, the sum of distance of an arbitrary point from (0+0i) and (2019+0i) should be a fixed quantity which is = 2019. Can you see now? For an analogy with the real world, fix two points on a cardboard and wrap a single string connecting the two points (keep it a bit loose), if you take a pencil and make the string taught, the pencil will trace an ellipse, if the string is previously taught, it will be perfectly straight – Anindya Prithvi Aug 07 '20 at 07:40
  • @AgungIzzulHaq I have added an interactive graph for your visual comprehension as well – Anindya Prithvi Aug 15 '20 at 19:54
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Let $w=\frac{z}{2019}$

$$|z|+|z-2019|=2019 \iff |w|+|w-1|=1\iff |w|+|1-w|=1$$

If $w$ and $1-w$ are collinear and in the same direction we obtain

  • $w=|w|e^{i\theta}$
  • $1-w=|w-1|e^{i\theta}$

and therefore

$$|w|e^{i\theta}+|1-w|e^{i\theta}=1 \implies e^{i\theta}=1 \implies\theta=2\pi k,\: k=\mathbb Z$$

if $w$ and $w-1$ are collinear and in the same direction we obtain

  • $w=|w|e^{i\theta}$
  • $w-1=|w-1|e^{i\theta}$

and therefore

$$|w|e^{i\theta}+|w-1|e^{i\theta}=2w-1 \implies e^{i\theta}=2w-1 \implies e^{i\theta}=\frac1{2|w|-1}$$

which also implies that $w$ is real.

user
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If $z$ is not real, then $0,2019,z,z-2019$ are vertices of a parallelogram, and consequently they satisfy the strict triangle inequality $$|z|+|z-2019|>2019.$$ Because the strict inequality doesn't hold, the four points are collinear. This proves that $z$ is real.

(Naturally, $z$ lies on the segment between $0$ and $2019,$ but this is more than was asked.)

user376343
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