If z 1 = 3 + 2 i and z 2 = 5 - 3 i , then show that z 1 z ¯ 2 and z ¯ 1 z 2 are conjugates of each other.

Given that z 1 = 3 + 2 i and z 2 = 5 - 3 i Also, z ¯ 1 = 3 - 2 i and z ¯ 2 = 5 + 3 i So, z 1 z ¯ 2 = 3 + 2 i 5 + 3 i = 15 + 9 i + 10 i + 6 i 2 = 9 + 19 i and z ¯ 1 z 2 = z 2 z ¯ 1 = 5 - 3 i 3 - 2 i = 15 - 10 i - 9 i + 6 i 2 = 9 - 19 i Since, z 1 z ¯ 2 = z ¯ 1 z 2 and z ¯ 1 z 2 = z 1 z ¯ 2 Hence, z 1 z ¯ 2 and z ¯ 1 z 2 are conjugates of each other.

Find all non-zero complex numbers z satisfying z ¯ = i z 2

Given z ¯ = i z 2 … ( i ) Let z = a + i b ; a , b ∈ R Then z ¯ = a - i b … ( i i ) & z 2 = a + i b 2 ⇒ z 2 = a 2 + i 2 b 2 + 2 a b i ⇒ z 2 = a 2 - b 2 + 2 a b i ; i 2 = - 1 … ( i i i ) Using equation ( i ) , ( i i ) & ( i i i ) we can write a - i b = i a 2 - b 2 + 2 a b i ⇒ a - i b = a 2 - b 2 i + 2 a b i 2 ⇒ a - i b = a 2 - b 2 i + 2 a b - 1 ⇒ a - i b = - 2 a b + a 2 - b 2 i For the equation a - i b = - 2 a b + a 2 - b 2 i comparing both sides (real and imaginary part), we can write a = - 2 a b … ( i v ) and - b = a 2 - b 2 … ( v ) Using equation i v a + 2 a b = 0 ⇒ a 1 + 2 b = 0 ⇒ a = 0 o r 1 + 2 b = 0 ⇒ a = 0 o r b = - 1 2 Using equation v a 2 - b 2 + b = 0 When a = 0 ⇒ b - b 2 = 0 ⇒ b 1 - b = 0 ⇒ b = 0 o r 1 - b = 0 ⇒ b = 0 o r b = 1 a , b = 0 , 0 o r 0 , 1 When b = - 1 2 ⇒ a 2 - - 1 2 2 + - 1 2 = 0 ⇒ a 2 - 1 4 - 1 2 = 0 ⇒ a 2 = 3 4 ⇒ a = ± 3 2 a , b = 3 2 , - 1 2 o r - 3 2 , - 1 2 So possible values of a , b are 0 , 0 , 0 , 1 , 3 2 , - 1 2 o r - 3 2 , - 1 2 Hence z = 0 + 0 i (rejected, z ≠ 0 ) z = 0 + i z = 3 2 - 1 2 i or z = - 3 2 - 1 2 i

Write ( i - 3 ) 13 in the polar form.

Given the complex number is i - 3 . Let r cos θ = - 3 and r sin θ = 1 Then, r 2 cos 2 θ + r 2 sin 2 θ = 3 + 1 = 4 ⇒ r 2 = 4 ⇒ r = 2 ⇒ cos θ = - 3 2 , sin θ = 1 2 ⇒ θ = 5 π 6 ⇒ i - 3 = 2 cos 5 π 6 + i sin 5 π 6 ⇒ i - 3 13 = 2 13 cos 5 π 6 + i sin 5 π 6 13 = 2 13 cos 65 π 6 + i sin 65 π 6 .