Let z = x + i y and w = u + i v be two complex numbers, such that z = w = 1 and z 2 + w 2 = 1 . Then, the number of ordered pairs z , w is equal to (where, x , y , u , v ∈ R and i 2 = - 1 )

Question

Let z = x + i y and w = u + i v be two complex numbers, such that z = w = 1 and z 2 + w 2 = 1 . Then, the number of ordered pairs z , w is equal to (where, x , y , u , v ∈ R and i 2 = - 1 ),

MARKS App · Accepted Answer

Let, z = e i α and w = e i β α , β ∈ - π , π z 2 + w 2 = 1 ⇒ e i 2 α + e i 2 β = 1 ⇒ cos ⁡ 2 α + cos ⁡ 2 β = 1 and sin ⁡ 2 α + sin ⁡ 2 β = 0 ⇒ 2 c o s α + β c o s α - β = 1 and 2 s i n α + β c o s α - β = 0 ⇒ s i n α + β = 0 ⇒ α + β = n π For α + β = - π , we have cos ⁡ 2 α = 1 2 ⇒ α = - 5 π 6 , - π 6 ⇒ 2 pairs of α , β For α + β = 0 , we have cos ⁡ 2 α = 1 2 ⇒ 4 pairs of α , β For α + β = π , we have cos ⁡ 2 α = 1 2 ⇒ 2 pairs of α , β

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