If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²

Solution:

It is given that 

= (a + ib) (c + id) (e + if) (g + ih) = A + iB

Taking modulus on both sides,

|(a + ib) (c + id) (e + if) (g + ih)| =|A + iB|

For any two complex numbers z and z₁, we have |zz₁| = |z| · |z₁|. Applying this to the left side part of the above equation,

|(a + ib)| |(c + id)| |(e + if)| |(g + ih)|| = |A + iB|

√(a² + b²) √(c² + d²) √(e² + f²) √(g² + h²) = √(A² + B²)

Squaring on both sides,

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

Hence proved

NCERT Solutions Class 11 Maths Chapter 5 Exercise ME Question 19

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²

Summary:

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then we have shown that (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²