Q.Show that the points in the Argand plane represented by the complex numbers $-2 + 7i$, $-\dfrac{3}{2} + \dfrac{1}{2}i$, $4 - 3i$, $\dfrac{7}{2}(1 + i)$ are the vertices of a rhombus.

Plot the four points as $(x,y)$ pairs, compute all four side lengths — they come out equal — and confirm the figure is a parallelogram by showing the diagonals share the same midpoint.

Treat each complex number $x+iy$ as the point $(x,y)$:

$A=(-2,7),\quad B=\left(-\tfrac32,\tfrac12\right),\quad C=(4,-3),\quad D=\left(\tfrac72,\tfrac72\right)$

Side lengths (using distance formula):

$AB^2 = \left(-\tfrac32+2\right)^2+\left(\tfrac12-7\right)^2 = \left(\tfrac12\right)^2+\left(-\tfrac{13}2\right)^2 = 0.25+42.25 = 42.5$

$BC^2 = \left(4+\tfrac32\right)^2+\left(-3-\tfrac12\right)^2 = \left(\tfrac{11}2\right)^2+\left(-\tfrac72\right)^2 = 30.25+12.25 = 42.5$

$CD^2 = \left(\tfrac72-4\right)^2+\left(\tfrac72+3\right)^2 = \left(-\tfrac12\right)^2+\left(\tfrac{13}2\right)^2 = 0.25+42.25=42.5$

$DA^2 = \left(-2-\tfrac72\right)^2+\left(7-\tfrac72\right)^2 = \left(-\tfrac{11}2\right)^2+\left(\tfrac72\right)^2 = 30.25+12.25=42.5$

So $AB=BC=CD=DA=\sqrt{42.5}$ — all four sides are equal.

Confirm it's a (non-degenerate) parallelogram — the diagonals $AC$ and $BD$ must bisect each other:

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