If a and b are complex numbers, show that the points a-b and…
If $a$ and $b$ are complex numbers, show that the points $a-b$ and $b$ are symmetric with respect to the point $\dfrac{a}{2}$.
Explanation of Method
Two points $z_1,z_2$ are symmetric about a point $c$ exactly when $c$ is their midpoint, i.e. $c=\frac{z_1+z_2}2$. It suffices to compute the midpoint of $a-b$ and $b$ directly.
Step 1 · Compute the midpoint
$$ \frac{(a-b)+b}{2} = \frac a2 $$
which is exactly the claimed center of symmetry.
Step 2 · Equivalent symmetry condition
Symmetry about $c=\frac a2$ means $(a-b)+b=2c$:
$$ (a-b)+b=a=2\cdot\frac a2 \quad \checkmark $$
Step 3 · Equal distances, opposite directions
$$ (a-b)-\frac a2=\frac a2-b, \qquad b-\frac a2 = -\left(\frac a2-b\right) $$
so the vectors from $\frac a2$ to each point are negatives of one another — equal in length, opposite in direction, confirming $\frac a2$ is their midpoint.
$\frac a2$ is the midpoint of the segment joining $b$ and $a-b$.
Step 4 · Numerical example
Let $a=6+4i$, $b=2+i$: then $a-b=4+3i$, and the midpoint of $(4,3)$ and $(2,1)$ is $\left(\frac{4+2}2,\frac{3+1}2\right)=(3,2)$, which equals $\frac a2=\frac{6+4i}2=3+2i$. ✓
Result
So the points $a-b$ and $b$ are symmetric about $\dfrac a2$ for every choice of complex $a,b$.