Simplify the complex number ( 2 + 4 i ) ( - 1 + 2 i ) ( - 1 - i ) ( 3 - i ) and find their modulus.

Let, z = ( 2 + 4 i ) ( - 1 + 2 i ) ( - 1 - i ) ( 3 - i ) = - 2 + 4 i - 4 i + 8 i 2 - 3 + i - 3 i + i 2 = - 10 - 4 - 2 i ∴ i 2 = - 1 = 5 2 + i = 5 ( 2 - i ) ( 2 + i ) ( 2 - i ) = 5 ( 2 - i ) 4 + 1 ⇒ z = 2 - i Modulus of z is | z | = x 2 + y 2 = 4 + 1 ∴ z = 5

Simplify the complex number ( 2 + 4 i ) ( - 1 + 2 i ) ( - 1 - i ) ( 3 - i ) and find their modulus.

Let, z = ( 2 + 4 i ) ( - 1 + 2 i ) ( - 1 - i ) ( 3 - i ) = - 2 + 4 i - 4 i + 8 i 2 - 3 + i - 3 i + i 2 = - 10 - 4 - 2 i ∴ i 2 = - 1 = 5 2 + i = 5 ( 2 - i ) ( 2 + i ) ( 2 - i ) = 5 ( 2 - i ) 4 + 1 ⇒ z = 2 - i Modulus of z is | z | = x 2 + y 2 = 4 + 1 ∴ z = 5

Simplify the complex number ( 1 + i ) 3 ( 2 + i ) ( 1 + 2 i ) and find their modulus.

Let, z = ( 1 + i ) 3 ( 2 + i ) ( 1 + 2 i ) = 1 3 + i 3 + 3 1 2 i + 3 1 i 2 2 + 4 i + i + 2 i 2 = 1 - i + 3 i - 3 2 + 5 i - 2 = 2 i - 2 5 i = 2 i - 2 5 i × - i - i = - 2 i 2 + 2 i - 5 i 2 ⇒ z = 2 i + 2 5 Modulus of z is z = x 2 + y 2 = 2 5 2 + 2 5 2 = 4 + 4 25 = 8 25 ∴ z = 2 2 5

If ( 1 - i ) ( 2 - i ) ( 3 - i ) … . ( n - i ) = x - i y , then prove that 2 · 5 · 10 … . ( 1 + n ) 2 = x 2 + y 2 .

Given, ( 1 - i ) ( 2 - i ) ( 3 - i ) … ( n - i ) = x - i y We know that, z 1 z 2 ⋯ z n = z 1 z 2 ⋯ z n . Taking modulus on both sides, ⇒ | ( 1 - 1 ) ‖ ( 2 - i ) | 3 - i ⋯ | n - i | = | x - i y | ⇒ 2 · 5 · 10 ⋯ 1 + n 2 = x 2 + y 2 Squaring on both sides, we get the required term to be proved. ∴ 2 · 5 · 10 ⋯ 1 + n 2 = x 2 + y 2 Hence, proved.

If the real part of z + 1 z + i is 1 , then find the locus of z .

Given, z + 1 z + i Let, z = x + i y ⇒ z + 1 z + i = ( x + 1 ) + i y ( x ) + ( y + 1 ) i = [ ( x + 1 ) + i y ] [ x - ( y + 1 ) i ] [ x + ( y + 1 ) i ] [ x - ( y + 1 ) i ] = x ( x + 1 ) + y ( y + 1 ) + i ( x y - ( x + 1 ) ( y + 1 ) ) x 2 + ( y + 1 ) 2 = x ( x + 1 ) + y ( y + 1 ) x 2 + ( y + 1 ) 2 + i ( x y - ( x + 1 ) ( y + 1 ) ) x 2 + ( y + 1 ) 2 Given, real part = 1 ⇒ x ( x + 1 ) + y ( y + 1 ) x 2 + ( y + 1 ) 2 = 1 ⇒ x 2 + y 2 + x + y = x 2 + ( y + 1 ) 2 ⇒ x 2 + y 2 + x + y = x 2 + y 2 + 2 y + 1 ⇒ x - y = 1 ∴ Locus of z is x - y = 1 .