Find a square root for the complex number - 47 + i 8 3 .

Let, z = - 47 + i 8 3 Let the square root of z be a + i b , ⇒ ( a + i b ) 2 = - 47 + i 8 3 ⇒ a 2 - b 2 + 2 a i b = - 47 + i 8 3 ⇒ a 2 - b 2 = - 47 ...(i), 2 a b = 8 3 Now, | a + i b | = - 47 + i 8 3 Squaring on both sides, ⇒ a 2 + b 2 = ( - 47 ) 2 + ( 8 3 ) 2 = 2209 + 192 = 2401 ⇒ a 2 + b 2 = 49 ...(ii) Adding (i) and (ii), we get 2 a 2 = 2 ⇒ a 2 = 1 ⇒ a = ± 1 Subtracting (i) and (ii), we get 2 b 2 = 96 ⇒ b 2 = 48 ⇒ b = ± 4 3 ∴ Square root of z is a + i b = ± ( 1 + 4 3 i ) .

Find a square root for the complex number - 47 + i 8 3 .

Let, z = - 47 + i 8 3 Let the square root of z be a + i b , ⇒ ( a + i b ) 2 = - 47 + i 8 3 ⇒ a 2 - b 2 + 2 a i b = - 47 + i 8 3 ⇒ a 2 - b 2 = - 47 ...(i), 2 a b = 8 3 Now, | a + i b | = - 47 + i 8 3 Squaring on both sides, ⇒ a 2 + b 2 = ( - 47 ) 2 + ( 8 3 ) 2 = 2209 + 192 = 2401 ⇒ a 2 + b 2 = 49 ...(ii) Adding (i) and (ii), we get 2 a 2 = 2 ⇒ a 2 = 1 ⇒ a = ± 1 Subtracting (i) and (ii), we get 2 b 2 = 96 ⇒ b 2 = 48 ⇒ b = ± 4 3 ∴ Square root of z is a + i b = ± ( 1 + 4 3 i ) .

If x + i y = 3 2 + cos θ + i sin θ , then show that x 2 + y 2 = 4 x - 3 .

Given, x + i y = 3 2 + cos θ + i sin θ Rationalize the denominator, ⇒ x + i y = 3 ( 2 + cos θ - i sin θ ) ( 2 + cos θ ) 2 - i 2 sin 2 θ ⇒ x + i y = 3 ( 2 + cos θ - i sin θ ) 4 + cos 2 θ + 4 cos θ + sin 2 θ ⇒ x + i y = 6 + 3 cos θ - 3 i sin θ 5 + 4 cos θ ∴ x = 6 + 3 cos θ 5 + 4 cos θ , y = - 3 sin θ 5 + 4 cos θ To prove, x 2 + y 2 = 4 x - 3 LHS = x 2 + y 2 = 6 + 3 cos θ 5 + 4 cos θ 2 + - 3 sin θ 5 + 4 cos θ 2 = 36 + 9 cos 2 θ + 36 cos θ + 9 sin 2 θ ( 5 + 4 cos θ ) 2 = 45 + 36 cos θ ( 5 + 4 cos θ ) 2 = 9 ( 5 + 4 cos θ ) ( 5 + 4 cos θ ) 2 ⇒ x 2 + y 2 = 9 5 + 4 cos θ RHS = 4 x - 3 = 4 ( 6 + 3 cos θ ) 5 + 4 cos θ - 3 = 24 + 12 cos θ - 15 - 12 cos θ 5 + 4 cos θ = 9 5 + 4 cos θ ∴ LHS = RHS Hence, proved.

If x + i y = 1 1 + cos θ + i sin θ , then show that 4 x 2 - 1 = 0 .

Given, x + i y = 1 1 + cos θ + i sin θ ⇒ x + i y = 1 + cos θ - i sin θ ( 1 + cos θ ) 2 - i 2 sin 2 θ ⇒ x + i y = 1 + cos θ - i sin θ 1 + 2 cos θ + cos 2 θ - i 2 sin 2 θ ⇒ x + i y = 1 + cos θ - i sin θ 1 + 2 cos θ + 1 ⇒ x + i y = 1 + cos θ 2 + 2 cos θ - - i sin θ 2 + 2 cos θ Then, x = 1 + cos θ 2 + 2 cos θ ⇒ x = 1 + cos θ 2 ( 1 + cos θ ) ⇒ x = 1 2 ⇒ 2 x = 1 ⇒ 4 x 2 = 1 ∴ 4 x 2 - 1 = 0 Hence, proved.

If u + i v = 2 + i z + 3 and z = x + i y , then find u , v .

Given, u + i v = 2 + i z + 3 and z = x + i y ⇒ u + i v = 2 + i ( x + 3 ) + i y ⇒ u + i v = ( 2 + i ) ( x + 3 - i y ) ( x + 3 ) 2 + y 2 ⇒ u + i v = ( x + 3 ) 2 + y ( x + 3 ) 2 + y 2 + i ( x + 3 - 2 y ) ( x + 3 ) 2 + y 2 ∴ u = 2 ( x + 3 ) + y ( x + 3 ) 2 + y 2 , v = x + 3 - 2 y ( x + 3 ) 2 + y 2