Midpoint between two complex numbers

Question

I have the following exercise:

Let $z_1$ and $z_2$ be complex numbers. Prove that the complex number $z_3 = \frac{z_1 + z_2}{2}$ is the midpoint between $z_1$ and $z_2$.

My attempt:

It is easy to prove that $|z_1 - z_3| - |z_2 - z_3| = 0$, which implies that the distance between $z_1$ and $z_2$ is the same that the distance between $z_2$ and $z_3$. Is it sufficient to prove that $|z_1 - z_3| + |z_2 - z_3| = |z_1 - z_2|$?

Thank you in advance.

Answer 1

I assume the definition of midpoint of $z_1$ and $z_2$ is a $w$ so that $|z_1 - w| = |z_2 -w| = \frac 12 |z_2 - z_1|$

Just do it.

It's easy to see that $w =\frac {z_1 + z_2}2$ fits the bill.

To prove it is the only one that fits the bill

(do you need to prove that or is it taken for granted that for any two points exactly one midpoint exist? If so, stop reading)

, let

$z_1 = x_1 + y_1 i$ and $z_2 = x_2 + y_2 i$ and $w = a + bi$ then

$\sqrt{(x_1 - a)^2 + (y_1 - b)^2} = \sqrt{(x_2 - a)^2 + (y_2 - b)^2}= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ so

1) $(x_1 -a)^2 + (y_1 -b)^2 = (x_2 - a)^2 + (y_2 - b)^2$

2) $(x_1 -a)^2 + (y_1 -b)^2 = \frac 14[(x_1-x_2)^2 + (y_1 - y_2)^2$

3) $(x_2 -a)^2 + (y_2 -b)^2 = \frac 14[(x_1-x_2)^2 + (y_1 - y_2)^2$

Solve for $a,b$ there will be only one solution.

Answer 2

Let $M_1,M_2$ and $M_3$ the images of $z_1,z_2$ and $z_3$. You have to prove that:

$\overrightarrow{M_1M_3}+\overrightarrow{M_2M_3}=\overrightarrow{0}$ which is easy since $(z_3-z_1)+(z_3-z_2)=2z_3-(z_1+z_2)=0$

Answer 3

One deifinition of mid-point is the following: $z_3$ is the mid point of $z_1$ and $z_2$ if it lies on the line joining $z_1$ and $z_2$ and is equidistant from these points. You have proved the second property. To prove the first property note that $(z_1-z_3)=(z_3-z_2)$. This implies that $z_3$ lies on the line joining $z_1$ and $z_2$.