If 3 + i 100 = 2 99 a + i b , then show that a 2 + b 2 = 4 .

Given, ( 3 + i ) 100 = 2 99 ( a + i b ) We know that, z n = | z | n ⇒ ( 3 + i ) 100 = | 3 + i | 100 = 2 99 | a + i b | ⇒ 4 100 = 2 99 a 2 + b 2 ⇒ 2 100 = 2 99 a 2 + b 2 ⇒ 2 = a 2 + b 2 ∴ a 2 + b 2 = 4 Hence, proved.

If the amplitude of ( z - 1 ) is π 2 , then find the locus of z .

Let, z = x + i y ⇒ z - 1 = x + i y - 1 ⇒ z - 1 = x - 1 + i y We know that, amplitude/argument of complex number x + i y = tan - 1 y x From given condition, ⇒ tan - 1 y x - 1 = π 2 ⇒ y x - 1 = tan π 2 ⇒ y x - 1 = 1 0 ⇒ x - 1 = 0 Therefore, locus of z is x = 1 , y ≠ 0 .

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