If $z \in \mathbb C$ such that $|z|+|z-2019|=2019$ then $z \in \mathbb R$ 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.
 A: $|a+b|=|a|+|b|$ 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.
A: 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, $y = \text{Im}(z) = 0$
A: Equations of form $$|z-a| + |z-b|=c$$ describe

*

*An ellipse (when $c> |a-b|$, with focus at $a,b$)

*A straight line (when $c=|a-b|$)

*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 $2019$.
A: 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.
A: 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.)
