The endpoints of a line segment are z_1 and z_2 . Find its…
The endpoints of a line segment are $z_1$ and $z_2$. Find its midpoint.
Explanation of Method
Each complex number $z=x+iy$ corresponds to the point $(x,y)$, so operations on complex numbers mirror the corresponding vector operations in the plane. The midpoint of two points is their coordinate-wise average, which translates directly into the arithmetic mean of the two complex numbers.
Step 1 · Set up coordinates
Let $z_1 = x_1+iy_1 \leftrightarrow (x_1,y_1)$ and $z_2 = x_2+iy_2 \leftrightarrow (x_2,y_2)$.
Step 2 · Cartesian midpoint formula
The midpoint of the segment between $(x_1,y_1)$ and $(x_2,y_2)$ is:
$$ M = \left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right) $$
Step 3 · Translate back to a complex number
$$ z_M = \frac{x_1+x_2}{2} + i\,\frac{y_1+y_2}{2} = \frac{1}{2}\big[(x_1+iy_1)+(x_2+iy_2)\big] = \frac{z_1+z_2}{2} $$
Step 4 · Vector cross-check
Equivalently, starting at $z_1$ and moving halfway toward $z_2$:
$$ z_M = z_1 + \frac{1}{2}(z_2-z_1) = \frac{1}{2}z_1 + \frac{1}{2}z_2 = \frac{z_1+z_2}{2} $$
Step 5 · Verify the defining properties
Equidistance:
$$ z_M - z_1 = \frac{z_2-z_1}{2}, \qquad z_M - z_2 = \frac{z_1-z_2}{2} $$
$$ |z_M-z_1| = \frac{|z_2-z_1|}{2} = \frac{|z_1-z_2|}{2} = |z_M-z_2| \quad \checkmark $$
Collinearity: since $z_M - z_1 = \tfrac12(z_2-z_1)$ is a real scalar multiple of $z_2-z_1$, the points $z_1, z_M, z_2$ lie on the same line. ✓
Step 6 · Examples
Example 1: $z_1=0$, $z_2=2+2i$: $$ z_M = \frac{0+(2+2i)}{2} = 1+i $$ matching the Cartesian midpoint of $(0,0)$ and $(2,2)$, which is $(1,1)$.
Example 2: $z_1=1+i$, $z_2=3+5i$: $$ z_M = \frac{(1+i)+(3+5i)}{2} = \frac{4+6i}{2} = 2+3i $$ matching the Cartesian midpoint of $(1,1)$ and $(3,5)$, which is $\left(\frac{1+3}{2},\frac{1+5}{2}\right)=(2,3)$. ✓
Step 7 · General division point
The formula generalizes to any point dividing the segment from $z_1$ to $z_2$ in ratio $t:(1-t)$, $0\le t\le 1$:
$$ z(t) = (1-t)z_1 + t z_2 = z_1 + t(z_2-z_1) $$
the midpoint is the special case $t = \tfrac12$.
$z_M$ lies exactly halfway along the segment from $z_1$ to $z_2$, equidistant from both.
Result
The midpoint of the segment with endpoints $z_1$ and $z_2$ is the arithmetic mean of the two complex numbers:
This matches the standard coordinate-geometry midpoint formula, satisfies $|z_M-z_1|=|z_M-z_2|$, keeps $z_1, z_M, z_2$ collinear, and is the $t=\tfrac12$ case of the general division-point formula $z(t)=(1-t)z_1+tz_2$.