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Hey,
(a + ib)(c + id)(e + if)(g + ih) = A + iB
take modulus both sides,
|(a + ib)(c + id)(e + if)(g + ih)| = |A+iB|
we know,
|z1.z2.z3.......zn| = |z1|.|z2|........|zn| use this here,
|a+ib|.|c+id|.|e+if|.|g+ih| = |A+iB|
we also know,
|x + iy| = √(x²+y²) use it here,
√(a²+b²).√(c²+d²).√(e²+f²).√(g²+h²) = √(A²+B²)
squaring both sides,
(a²+b²).(c²+d²).(e²+f²).(g²+h²) = A²+B²
hence, proved//
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