Area of Rectangle Formed by Roots of Complex Equation

To find the area of the rectangle formed by the roots of the equation, we begin by simplifying the given equation:
$z \bar{z}^3 + \bar{z} z^3 = 350$

Factoring out $z\bar{z}$, we get:
$z\bar{z}\left(\bar{z}^2 + z^2\right) = 350$

Given $z = x + iy$ where $x, y \in \mathbb{Z}$, we have:
$z\bar{z} = |z|^2 = x^2 + y^2$
$z^2 + \bar{z}^2 = (x + iy)^2 + (x - iy)^2 = 2(x^2 - y^2)$

Substituting these expressions back into the equation:
$(x^2 + y^2) \cdot 2(x^2 - y^2) = 350$
$2(x^4 - y^4) = 350$
$x^4 - y^4 = 175$

Factoring the difference of squares:
$(x^2 - y^2)(x^2 + y^2) = 175$

Since $x$ and $y$ are integers, $x^2$ and $y^2$ are non-negative integers. Therefore, $x^2 + y^2 > 0$ and $x^2 + y^2 \ge x^2 - y^2$.
The positive integer factor pairs of $175$ are $(1, 175)$, $(5, 35)$, and $(7, 25)$.

We test each pair:

1. If $x^2 - y^2 = 1$ and $x^2 + y^2 = 175$:
   $2x^2 = 176 \implies x^2 = 88 \quad (\text{not a perfect square})$

2. If $x^2 - y^2 = 5$ and $x^2 + y^2 = 35$:
   $2x^2 = 40 \implies x^2 = 20 \quad (\text{not a perfect square})$

3. If $x^2 - y^2 = 7$ and $x^2 + y^2 = 25$:
   $2x^2 = 32 \implies x^2 = 16 \implies x = \pm 4$
   $2y^2 = 18 \implies y^2 = 9 \implies y = \pm 3$
   Both $x$ and $y$ are integers.

Thus, the four vertices of the rectangle in the Argand plane are:
$(4, 3), \quad (4, -3), \quad (-4, 3), \quad (-4, -3)$

The dimensions of the rectangle are:

• Length along the real axis: $4 - (-4) = 8$
• Breadth along the imaginary axis: $3 - (-3) = 6$

Therefore, the area of the rectangle is:
$\text{Area} = 8 \times 6 = 48$

Hence, the correct option is A.