#### Topics

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/ +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 θr 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

Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!
Paid Users Only!