TOPICS
Exercise - 5.2

Question-1 :- Find the modulus and the arguments of the complex numbers :
z = -1 - i√3

Solution :-
  We have, z = -1 - i√3
  Now, -1 = r cos θ, -√3 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = (-1)2 + (-√3)2
  r2 (cos2 θ + sin2 θ) = 1 + 3
  r2 (cos2 θ + sin2 θ) = 4
  r x 1 = √4
  r = 2
  Modulus = 2

  Therefore, -1 = r cos θ and -√3 = r sin θ
  cos θ = -1/2 and sin θ = -√3/2, 
  Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in III quadrant,
  So, which gives θ = -(π - π/3) = -2π/3
  Arguments = -2π/3
   

Question-2 :-  Find the modulus and the arguments of the complex numbers :
z = -√3 + i

Solution :-
  We have, z = -√3 + i
  Now, -√3 = r cos θ, 1 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = (-√3)2 + 12
  r2 (cos2 θ + sin2 θ) = 3 + 1
  r2 (cos2 θ + sin2 θ) = 4
  r x 1 = √4
  r = 2
  Modulus = 2

  Therefore, -√3 = r cos θ and 1 = r sin θ
  cos θ = -√3/2 and sin θ = 1/2, 
  Since cos θ are negative and cosθ are negative in II quadrant,
  So, which gives θ = (π - π/6) = 5π/6
  Arguments = 5π/6
   

Question-3 :-  Convert the complex numbers in the polar form:
1 - i

Solution :-
  We have, z = 1 - i
  Now, 1 = r cos θ, -1 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = 12 + (-1)2
  r2 (cos2 θ + sin2 θ) = 1 + 1
  r2 (cos2 θ + sin2 θ) = 2
  r x 1 = √2
  r = √2

  Therefore, 1 = r cos θ and -1 = r sin θ
  cos θ = 1/√2 and sin θ = -1/√2, which gives θ = -π/4   [lies on IV Quadrant]
  Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(-π/4) + i sin(-π/4)]
    

Question-4 :-  Convert the complex numbers in the polar form:
-1 + i

Solution :-
  We have, z = -1 + i
  Now, -1 = r cos θ, 1 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = (-1)2 + 12
  r2 (cos2 θ + sin2 θ) = 1 + 1
  r2 (cos2 θ + sin2 θ) = 2
  r x 1 = √2
  r = √2

  Therefore, -1 = r cos θ and 1 = r sin θ
  cos θ = -1/√2 and sin θ = 1/√2, which gives θ = π - π/4 = 3π/4   [lies on II Quadrant]
  Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(3π/4) + i sin(3π/4)]
   

Question-5 :-  Convert the complex numbers in the polar form:
-1 - i

Solution :-
  We have, z = -1 - i
  Now, -1 = r cos θ, -1 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = (-1)2 + (-1)2
  r2 (cos2 θ + sin2 θ) = 1 + 1
  r2 (cos2 θ + sin2 θ) = 2
  r x 1 = √2
  r = √2

  Therefore, -1 = r cos θ and -1 = r sin θ
  cos θ = -1/√2 and sin θ = -1/√2, which gives θ = -(π - π/4) = -3π/4   [lies on III Quadrant]
  Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(-3π/4) + i sin(-3π/4)]
   

Question-6 :-  Convert the complex numbers in the polar form:
-3

Solution :-
  We have, z = -3 + 0.i
  Now, -3 = r cos θ, 0 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = (-3)2 + 0
  r2 (cos2 θ + sin2 θ) = 9 + 0
  r2 (cos2 θ + sin2 θ) = 9
  r x 1 = 9
  r = 3

  Therefore, -3 = r cos θ and 0 = r sin θ
  cos θ = -3/3 = -1 and sin θ = 0, which gives θ = π
  Therefore, required polar form is z = r[cosθ + i sinθ] = 3[cos(π) + i sin(π)]
    

Question-7 :-  Convert the complex numbers in the polar form:
√3 + i

Solution :-
  We have, z = √3 + i√3
  Now, √3 = r cos θ, 1 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = (√3)2 + 12
  r2 (cos2 θ + sin2 θ) = 3 + 1
  r2 (cos2 θ + sin2 θ) = 4
  r x 1 = √4
  r = 2

  Therefore, √3 = r cos θ and 1 = r sin θ
  cos θ = √3/2 and sin θ = 1/2, which gives θ = π/6     [lies on I quadrant]
  Therefore, required polar form is z = r[cosθ + i sinθ] = 2[cos(π/6) + i sin(π/6)]
   

Question-8 :-  Convert the complex numbers in the polar form:
i

Solution :-
  We have, z = 0 + 1.i
  Now, 0 = r cos θ, 1 = r sin θ
  By squaring and adding, we get 
  r2 cos2 θ + r2 sin2 θ = 0 + 12
  r2 (cos2 θ + sin2 θ) = 0 + 1
  r2 (cos2 θ + sin2 θ) = 1
  r x 1 = √1
  r = 1

  Therefore, 0 = r cos θ and 1 = r sin θ
  cos θ = 0 and sin θ = 1, which gives θ = π/2
  Therefore, required polar form is z = r[cosθ + i sinθ] = [cos(π/2) + i sin(π/3)]
   
CLASSES

Connect with us:

Copyright © 2015-17 by a1classes. All Rights Reserved.

www.000webhost.com