Question

If (a+i)^2 / (2a-i) = p + iq Show that: p^2 + q^2 = (a^2 + i)^2 / ( 4a^2 + 1 )

Answer 1

Given :

• ( a + i )² / ( 2 a - i ) = p + i q ___equation (1)

To prove :

• p² + q² = ( a² + i )² / ( 4 a² + 1 )

Proof :

Given that,

→ ( a + i )² / ( 2 a - i ) = p + i q

Taking conjugate both sides

we will get,

→ ( a - i )² / ( 2 a + i ) = p - i q  _____equation (2)

Multiplying equation (1) and (2)

→ [ (a + i)²/(2 a - i) ] × [ ( a - i )²/(2 a + i) ] = ( p + i q ) ( p - i q )

using algebraic identity (x+y) (x-y) = x² - y² in RHS and denominator of LHS

→ [ (a + i)² (a - i)² ] / [ (2 a)² - ( i )² ] = ( p )² - ( i q )²

→ [{(a + i)(a - i )}²] / ( 4 a² - i² ) = p² - i² q²

→ ( a² - i² )² / ( 4 a² - i² ) = p² - i² q²

putting i² = -1 in the equation

→  ( a² - ( - 1 ) )² / ( 4 a² - ( - 1 ) ) = p² - ( - 1 ) q²

→ ( a² + 1 )² / ( 4 a² + 1 ) = p² + q²

PROVED .