1. Simply 3 (1 – 2i ) + ( -4 – 5i ) + ( -8 + 3i )
Ans: 3 ( 1 – 2i ) + ( -4 - 5i ) + ( -8 + 3 I ) = 3 – 6 i – 4 - 5i – 8 + 3i
                                                               = - 9 - 8i

2, Simplify ( 2 + 3 i ) ( 4 – 5i )
Ans: ( 2 + 3i) ( 4 – 5 i ) = 8 – 10 i + 12 i - 15 i²
                                      = 8 + 2 i + 15 = 23 + 2i

3, If ( 1 + i ) z = ( 1 – i ) , then find the value of z.
Ans: Let z = x + i y
   ( 1 + i ) z = ( 1 –i )  
∴  ( 1 + i ) ( x+ iy )  = ( 1 - i )  ( x - iy )

⇒(x-y)+i(x+y) = (x-y) - i(x+y)
⇒    x + y = 0
  ∴   z = x - ix = x ( 1 - i ), ∀ x  ∈ R

4, Express ( 5 i ) ( - ³/₅ i ) in the form a + ib.
Ans: ( 5i) ( -³/₅ i ) = -3i² = -3 x -1 = 3.

5.  Express i⁹ + i ¹⁹ in the form a + ib.
Ans: i⁹ + i ¹⁹ = ( i⁴ ) ² . i + ( i ⁴ )⁴ . i³
                      = i + i³ = i + i². i = i - i = 0

6, Find the multiplicative inverse of 4 - 3 i
Ans: Let z = 4 -3i

7,

Ans:

8, Evaluate : 
       a) i ⁷ b) i⁵¹ c) i ⁵⁰⁰
Ans: 
a) i ⁷ = i⁴, i³ = i³ = -i
b) i ⁵¹ = i ⁴⁸. i ³ = i³ = -i
c) i⁵⁰⁰ = ( i⁴) ¹²⁵ = 1

 9, Find the  value of : 
  a) √-25  ×  √-81     b) 4√-4   + 5 √-9  + 3 √-16

a) √-25  ×  √-81 = 5i x 9i
                         = 45 i² = -45.
b)  4√-4   + 5 √-9  + 3 √-16
              = 4 x 2 i + 5 x 3i + 3 x 4i
              = 8i + 15i + 12 i
              = 35i

10,Prove that i ¹⁰⁷ + i ¹¹² + i¹¹⁷ + i ¹²² = 0.
               LHS = i ¹⁰⁷ + i ¹¹² + i¹¹⁷ + i ¹²²
                       = i³ + i⁰ + i¹ + i²
                       = -1 + 1 + i - i = 0 = RHS

11. 

Ans:

12, Find a complex  number when multiplied by 5 + 3i gives 3 - 4 i
Ans:  Let z be the requried complex number  
     
 

13, Find a least postiive integer value for n  
  So that  
Ans: 

14, Find the value of θ so that    is  purely real. 
Ans:  
   
 For a complex number to be purely real, then imaginary part = 0.
       

15, Find the polar form of 2 + i 2 √3. 
Ans:  Let  2 + i 2 √3  = 4 ( cos θ + 1 sin  θ )
 Equating real and imaginary parts
     r cos  θ  = 2   →   ( 1 )
     r sin   θ    = 2 √3   → ( 2 )
Squaring and adding ( 1 ) and ( 2 )
r²  ( sin ² θ   + cos²  θ )  = 4 + 12.
  ⇒ r² = 16       ⇒  r =  4. 
  ^{r sin θ}/ _{r cos θ}  = ^2 √3/₂   
 tan θ  =  √3
 ⇒   θ   = tan ^-1 ( √3 ) 
          = ^π/₃
  ∴ Z = 4 ( cos  ^π/₃  +i sin  ^π/₃  )
         = 4 ( ¹/₂ + i ³/₂ ). 

16, Find the  polar form of 5 + 12 i
Ans:  Let z = 5 + 12 i  = r ( cos θ + isin θ )
Equating real and imaginary parts.
r cos θ  = 5    →  ( 1 )
 r sin θ  = 12   → ( 2 )
Squaring and adding ( 1 )  & ( 2 ).
r² ( cos ² θ + sin ² θ )  = 25 + 144
    ⇒ r² = 169       ⇒  r =  13. 
    ^{r sin θ}/ _{r cos θ}  = ¹²/₅  
    ⇒ tan  θ      ⇒  θ  =  tan ^{-1  12}/₅.
  ∴ Z = 13  [ cos ( tan ^{-1  12}/₅)  + i  sin ( tan ^{-1  12}/₅)]

17, Find the conjugate and the modulus of 7 + 24 i ?
Ans :  Let Z = 7 + 24 i
   

18, x + iy = ^{a + ib}/ _{a - ib}   prove that x² + y² = 1. 
   

19, 
Ans:
 

20,
Ans:

21, 
Ans:
  

22. If ( 1 + i ) ( 1 + 2i ) ( 1 + 3 i )  - - - - -  ( 1 + n i ) = x + iy  Prove that 2 x 5 x 10 x . . . . . . ( 1 + n² ) = x² + y²
 Ans:  Given that ( 1 + i )  ( 1 + 2i ) ( 1 + 3 i )  - - - ( 1 + n i ) = x + iy  
Taking modulus,  
   |  (1 + i )  ( 1 + 2i ) ( 1 + 3 i )  - - - ( 1 + n i )| =   | x + iy |
     ⇒ |1 + i |  | 1 + 2i | | 1 + 3 i |  - - - | 1 + n i | =   | x + iy |
 
Squaring  on  both sides,
 2 .5.10   - - - - - -  ( 1 + n² )  = x²+y².

23, Solve x² - 7 ix - 12 = 0
 Ans: a = 1,  b = - 7 i,  c = -12
  

24, Solve √5 x²  +  x + √5  = 0.
Ans:   a = √5,   b = 1,  c = √5 
 

25. Solve x² + 3x + 5 = 0.
    a = 1,  b = 3,  c = 5