Find the modulus and argument of the complex number 1 - i 1 + i and hence express in the polar form.

Given, 1 - i 1 + i Rationalising the denominator, 1 - i 1 + i × 1 - i 1 - i = 1 + i 2 - 2 i 1 - i 2 Let, z = - i Now, r = z = 0 + 1 = 1 Since point ( 0 , - 1 ) lies on the negative direction of the imaginary axis, the argument of z is given by 3 π 2 . Polar form = r ( cos θ + i sin θ ) = cos 3 π 2 + i sin 3 π 2 = cos 2 π - π 2 + i sin 2 π - π 2 = cos π 2 - i sin π 2

Find the modulus and argument of the complex number 1 - i 1 + i and hence express in the polar form.

Given, 1 - i 1 + i Rationalising the denominator, 1 - i 1 + i × 1 - i 1 - i = 1 + i 2 - 2 i 1 - i 2 Let, z = - i Now, r = z = 0 + 1 = 1 Since point ( 0 , - 1 ) lies on the negative direction of the imaginary axis, the argument of z is given by 3 π 2 . Polar form = r ( cos θ + i sin θ ) = cos 3 π 2 + i sin 3 π 2 = cos 2 π - π 2 + i sin 2 π - π 2 = cos π 2 - i sin π 2

Find the modulus and argument of the complex number 1 1 + i and hence express in the polar form.

Given, 1 1 + i Rationalising the denominator: 1 1 + i × 1 - i 1 - i = 1 - i 1 - i 2 = 1 - i 2 ∵ i 2 = - 1 Let, z = 1 2 - i 2 r = | z | = 1 4 + 1 4 = 1 2 Let tan α = lm ( z ) Re ( z ) ∴ tan α = 1 2 - 1 2 = 1 ⇒ α = π 4 Since point 1 2 , - 1 2 lies in the fourth quadrant, the argument is given by θ = - a = - π 4 . Polar form = r ( cos θ + i sin θ ) = 1 2 cos - π 4 + i sin - π 4 = 1 2 cos π 4 - i sin π 4

Find the modulus and argument of the complex number 1 + 2 i 1 - 3 i and hence express in the polar form.

Given, 1 + 2 i 1 - 3 i Rationalising the denominator: 1 + 2 i 1 - 3 i × 1 + 3 i 1 + 3 i = 1 + 3 i + 2 i + 6 i 2 1 - 9 i 2 Let, z = - 1 2 + i 2 r = | z | = 1 4 + 1 4 = 1 2 Let, tan α = lm ( z ) Re ( z ) Then, tan α = 1 2 - 1 2 = 1 ⇒ a = π 4 Since point - 1 2 , 1 2 lies in the second quadrant, the argument is given by θ = π - α = 3 π 4 Polar form = r ( cos θ + i sin θ ) = 1 2 cos 3 π 4 + i sin 3 π 4

Find the modulus and argument of the complex number - 16 1 + i 3 and hence express in the polar form.

Given, - 16 1 + i 3 Rationalising the denominator: - 16 1 + 3 × 1 - i 3 1 - 3 = - 16 + 16 3 i 1 - 3 i 2 = - 16 + 16 3 i 4 ∵ i 2 = - 1 Let, z = - 4 + 4 3 i r = | z | = 16 + 48 = 8 Let tan a = lm ( z ) Re ( z ) then, tan α = 4 3 - 4 = 3 ⇒ α = π 3 Since the point ( - 4 , 4 3 ) lies in the third quadrant, the argument is given by θ = x - α = 2 π 3 Polar form = r ( cos θ + i sin θ ) = 8 cos 2 π 3 + i sin 2 π 3