Given four points A, B, C, and D, represented by complex numbers, prove whether the quadrilateral ABCD is a square.

Let $z_A = 9+j$
Let $z_B = 4+j13$
Let $z_C = -8+j8$
Let $z_D = -3-j4$

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: $|AB| = |-5+j12| = \sqrt{(-5)^2 + (12)^2} = \sqrt{25+144} = \sqrt{169} = 13$
Length of BC: $|BC| = |-12-j5| = \sqrt{(-12)^2 + (-5)^2} = \sqrt{144+25} = \sqrt{169} = 13$
Length of CD: $|CD| = |5-j12| = \sqrt{(5)^2 + (-12)^2} = \sqrt{25+144} = \sqrt{169} = 13$
Length of DA: $|DA| = |12+j5| = \sqrt{(12)^2 + (5)^2} = \sqrt{144+25} = \sqrt{169} = 13$
Since all four sides have equal length (13 units), the quadrilateral ABCD is a rhombus.

We can check if the angle at vertex B is a right angle. This means checking if vector $\vec{BA}$ is perpendicular to vector $\vec{BC}$.
Vector $\vec{BA} = z_A - z_B = (9+j) - (4+j13) = (9-4) + j(1-13) = 5 - j12$.
Vector $\vec{BC} = -12 - j5$.
To check for perpendicularity, we examine the ratio $\frac{\vec{BC}}{\vec{BA}}$. If this ratio is purely imaginary, the vectors are perpendicular.

Multiply the numerator and denominator by the conjugate of the denominator, which is $5+j12$:

Since $j^2 = -1$:

Since the ratio $\frac{\vec{BC}}{\vec{BA}}$ is $-j$, which is a purely imaginary number, the vectors $\vec{BA}$ and $\vec{BC}$ are perpendicular. This means that the angle $\angle ABC$ is $90^\circ$.

A quadrilateral with four equal sides (a rhombus) and at least one right angle is a square.
Therefore, ABCD is a square.