` ( 2 + i ) /( 4i + ( 1 + i ) ^ 2 ) ` का संयुग्मी तथा कोणाक निकालिये |

` ( 2 + i ) /( 4i + ( 1 + i ) ^ 2 ) ` का संयुग्मी

1.	` ( 2 + i ) /( 4i + ( 1 + i ) ^ 2 ) ` का संयुग्मी तथा कोणाक निकालिये \|
Answer» माना कि ` z = ( 2 + i) /( 4i + ( 1 + i ) ^ 2) = ( 2 + i ) /( 4i + 1 + i ^ 2 + 2i ) = ( 2 + i ) /( 6i ) ` ` = ((2 + i ) /( 6i)) (( - 6i )/( -6i)) = ( 6 - 12 i ) /( 36 ) = ( 1 ) /( 6 ) - ( 1 ) / ( 3 ) i ` ` therefore bar z = ( 1 ) /( 6 ) + ( 1 ) / ( 3 ) i ` पुनः ` z = ( 1 ) /( 6 ) - ( 1 ) /( 3 ) i `, यहाँ ` x = ( 1 ) /( 6 ) , y = - ( 1 )/ ( 3 ) ` ` tan theta = \| ( y ) / ( x ) \|\| (- ( 1 ) / ( 3 )) / ( ( 1 ) / ( 6 )) \| = \| - 2 \| = 2 therefore theta = tan ^( -1) 2, 0 le theta le ( pi ) / ( 2 ) ` ` because x = ( 1 ) /( 6 ) gt 0 ` तथा ` y = - ( 1) / ( 3 ) lt 0 ` अतः z चतुर्थ पाद में होगा \| ` therefore arg z = 2pi - theta = 2pi - tan ^( - 1) 2 ` Note : Principal value of arg z = ` ( 2pi - tan ^ ( - 1 ) 2 ) - 2 pi = - tan ^( - 1) 2 `