Let us define the complex numbers as follows:

Let $z_1 = a + ib$, $z_2 = c + id$, $z_3 = e + if$, and $z_4 = g + ih$.

The given equation can be rewritten in terms of these complex numbers as:

$z_1 z_2 z_3 z_4 = A + iB$

We are required to prove that

$(a^2 + b^2) (c^2 + d^2) (e^2 + f^2) (g^2 + h^2) = A^2 + B^2.$

Step 1: Express the LHS using the modulus property.

From the algebraic identities and properties demonstrated in the context (specifically Example 8 where $x^2 + y^2 = (x + iy)(x - iy)$), we know that for any complex number $z = x + iy$, $x^2 + y^2 = z \bar{z}$, where $\bar{z}$ is the conjugate.

Applying this to each term on the Left Hand Side (LHS):

$a^2 + b^2 = (a + ib)(a - ib) = z_1 \bar{z}_1$

$c^2 + d^2 = (c + id)(c - id) = z_2 \bar{z}_2$

$e^2 + f^2 = (e + if)(e - if) = z_3 \bar{z}_3$

$g^2 + h^2 = (g + ih)(g - ih) = z_4 \bar{z}_4$

Substituting these into the LHS of the required equation:

$\text{LHS} = (z_1 \bar{z}_1)(z_2 \bar{z}_2)(z_3 \bar{z}_3)(z_4 \bar{z}_4)$

Step 2: Express the RHS using the given relation.

Consider the Right Hand Side (RHS), $A^2 + B^2$.

Using the same property,

$A^2 + B^2 = (A + iB)(A - iB).$

We are given that $A + iB = z_1 z_2 z_3 z_4$.

To find $A - iB$, we take the conjugate of both sides of the given equation.

$\overline{A + iB} = \overline{z_1 z_2 z_3 z_4}$

Using the property that the conjugate of a product is the product of conjugates:

$A - iB = \bar{z}_1 \bar{z}_2 \bar{z}_3 \bar{z}_4$

Step 3: Equate LHS and RHS.

Now, substitute the expressions for $A + iB$ and $A - iB$ into the RHS:

$\text{RHS} = (A + iB)(A - iB)$

$= (z_1 z_2 z_3 z_4)(\bar{z}_1 \bar{z}_2 \bar{z}_3 \bar{z}_4)$

By rearranging the terms (using the commutative law of multiplication mentioned in the context):

$\text{RHS} = (z_1 \bar{z}_1)(z_2 \bar{z}_2)(z_3 \bar{z}_3)(z_4 \bar{z}_4)$

Comparing the results from Step 1 and Step 3:

$\text{LHS} = (z_1 \bar{z}_1)(z_2 \bar{z}_2)(z_3 \bar{z}_3)(z_4 \bar{z}_4)$

$\text{RHS} = (z_1 \bar{z}_1)(z_2 \bar{z}_2)(z_3 \bar{z}_3)(z_4 \bar{z}_4)$

Thus, LHS = RHS.

Hence, $(a^2 + b^2) (c^2 + d^2) (e^2 + f^2) (g^2 + h^2) = A^2 + B^2$ is proved. ∴