NCERT Solutions
Class 11 Maths
Complex Numbers and Quadratic Equations
Ex.Misc.Q.5
Convert the following in the polar form:
(1) (1 + 7i) ÷ (2 - i)2
(2) (1 + 3i) ÷ (1 – 2i)
(1) Let z = (1 + 7i) ÷ (2 - i)2
= (1 + 7i) ÷ (4 + i2 – 4i)
= (1 + 7i) ÷ (4 - 1 – 4i)
= (1 + 7i) ÷ (3 – 4i)
= {(1 + 7i) ÷ (3 – 4i)} × {(3 + 4i) ÷ (3 + 4i)}
= (3 + 4i + 21i + 28i2) ÷ {32 – (4i)2}
= (3 + 4i + 21i - 28) ÷ (9 + 16)
= (-25 + 25i) ÷ 25
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2(cos2 θ + sin2 θ) = 1 + 1
=> r2 = 2
[Since cos2 θ + sin2 θ = 1]
=> r = √2
[Since r > 0]
Now, √2 cos θ = -1 and √2 sin θ = 1
=> cos θ = -
and sin θ =
=> θ = π – 
= 
Now, z = r cos θ + i r sin θ
= √2(cos + i sin )
This is the required polar form.
(2) Let z = (1 + 3i) ÷ (1 – 2i)
= {(1 + 3i) ÷ (1 – 2i)} × {(1 + 2i) ÷ (1 + 2i)}
= (1 + 2i + 3i + 6i2) ÷ {12 – (2i)2}
= (1 + 2i + 3i - 6) ÷ (1 + 4)
= (-5 + 5i) ÷ 5
= -1 + i
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2(cos2 θ + sin2 θ) = 1 + 1
=> r2 = 2
[Since cos2 θ + sin2 θ = 1]
=> r = √2
[Since r > 0]
Now, √2 cos θ = -1 and √2 sin θ = 1
=> cos θ = - and sin θ =
=> θ = π – =
Now, z = r cos θ + i r sin θ
= √2(cos + i sin )
This is the required polar form.