HOW TO SHOW THAT GIVEN COMPLEX NUMBER IS PURELY REAL AND IMAGINARY

Question :

Find the least value of the positive integer n for which (√3 + i)ⁿ (i) real (ii) purely imaginary.

Solution : 

(i)  Purely real 

If n = 1

   =  (√3 + i)¹

  =   (√3 + i)

If n = 2

   =  (√3 + i)²

  =   √3² + i² + i2√3

  =   2 + i 2√3

If n = 3

   =  (√3 + i)³

  =   √3³ + 9i + 3√3(-1) - i

  =   3√3 + 9i - 3√3 - i

=  8i (purely imaginary)

If n = 6

   =  (√3 + i)⁶

   =  {(√3 + i)³}²

=  (8i)²

=  64 i²

=  -64 (purely real)

Hence the least value of n is 6 to make (√3 + i)ⁿ as purely real, make it as purely imaginary, we have to put n = 3.

Question 2 :

Show that (i) (2 + i√3)¹⁰ - (2 - i√3)¹⁰ is purely imaginary

Solution :

Let z = (2 + i√3)¹⁰, then z bar = (2 - i√3)¹⁰

By using properties of conjugate, we get 

(z - z bar)/2  =  Im (z)

Hence the given is purely imaginary.

(ii)  [(19 - 7i)/(9 + i)]¹² + [(20 - 5i)/(7 - 6i)]¹² is real

Solution :

Let z = (19 - 7i)/(9 + i)

  =  [(19 - 7i)/(9 + i)][(9 - i)/(9 - i)]

  =  [(19 - 7i)(9 - i)/(9 + i)(9 - i)]

  =  (171 - 19i - 63i + 7i²) / (81 - (1))

  =  (164 - 82i) / 82

z  =  2 - i

(20 - 5i)/(7 - 6i)

  =  [(20 - 5i)/(7 - 6i)] [(7 + 6i)/(7 + 6i)]

  =  [(20 - 5i)(7 + 6i)/(7 - 6i) (7 + 6i)]

  =  (140 + 120i - 35i - 30i²)/(49  - 36(-1))

  =  (140 + 120i - 35i + 30)/(49  + 36)

  =  (170 + 85i)/85

z bar  =  2 + i

By adding a complex number and its conjugate, we get the real part of the complex number.

Hence the sum of above complex numbers is purely real.

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