Prove that the points A(9,1), B(4,13), C(-8,8), and D(-3,-4) form a square.

Let the complex numbers be:
$z_A = 9+j$
$z_B = 4+j13$
$z_C = -8+j8$
$z_D = -3-j4$

Vector $\vec{AB} = z_B - z_A = (4+j13) - (9+j) = (4-9) + j(13-1) = -5 + j12$
Vector $\vec{BC} = z_C - z_B = (-8+j8) - (4+j13) = (-8-4) + j(8-13) = -12 - j5$
Vector $\vec{CD} = z_D - z_C = (-3-j4) - (-8+j8) = (-3-(-8)) + j(-4-8) = 5 - j12$
Vector $\vec{DA} = z_A - z_D = (9+j) - (-3-j4) = (9-(-3)) + j(1-(-4)) = 12 + j5$

The magnitude of a complex number $x+jy$ is $\sqrt{x^2+y^2}$.
Length of AB:

Length of BC:

Length of CD:

Length of DA:

Since all four sides have equal length (13 units), ABCD is either a rhombus or a square.

Two vectors $(x_1, y_1)$ and $(x_2, y_2)$ are perpendicular if their dot product $x_1 x_2 + y_1 y_2 = 0$.
Let's check vectors $\vec{AB} = -5 + j12$ (corresponding to $(-5, 12)$) and $\vec{BC} = -12 - j5$ (corresponding to $(-12, -5)$).
Dot product of $\vec{AB}$ and $\vec{BC}$:

Since the dot product is 0, $\vec{AB}$ is perpendicular to $\vec{BC}$. This means that angle ABC is $90^\circ$.

Since all four sides of the quadrilateral ABCD are equal in length and at least one interior angle is $90^\circ$, the quadrilateral ABCD is a square.