Maths Question Polar forms Help please!!~~?

Question

Write z1 and z2 in polar form. Find the product z1z2 and the quotients z1/z2 and 1/z1. Leave your answer in polar form.
a)

  z1=4(squareroot of 3) - 4i

   z2=8i

b)

    z1=3+4i

    z2=2-2i

Answer 1

a)

z1 rewrite the polar form

z1 = 4√3 - 4i = 4(√3 - i) = 8(√3 / 2 - i / 2)

                                         = 8(cosπ/6 + isinπ/6)

z2 rewrite the polar form

z2 = 8i = 0 + 8i = 8(0 + i) = 8(cosπ/2 + isinπ/2)

Answer 2

z1z2 = (4√3 - 4i)8i = 32i√3 - 32i² = 32(i√3 + 1)       ( Substitute i² = -1)

z1 / z2 = 4√3 - 4i / 8i = (√3 - i) / 2i = 1 / 2(-i√3 - 1) = -1 / 2(1 + i√3) (Substitute -i² = 1)

Comments

z₁ / z₂ = (4√3 - 4i) / (8i) = (√3 - i) / (2i) = (√3 - i)i / (2i)i = (i√3 - i²) / (2i²) = (i√3 + 1) / (- 2) =  - 1/2 - i√3/2.

Answer 3

1 / z1 = 1 / (4√3 - 4i)

           = 1 / 4((√3 - i))

Multiply numerator and denominator by (√3 + i)

          = (1 / 4)((√3 + i) / (√3 - i)(√3 + i)

          = (1 / 4)((√3 + i) / (3 - i²)

          = (1 / 4)((√3 + i) / (3 + 1)          (Substitute i² = -1)

          = (1 / 16)((√3 + i)

          = (√3 + i) / 16.

Answer 4

b)

z2 = 2- 2i

It can rewrite the polar form

z2 = 2 - 2i

|z2| = r = √2² + (-2)² = √4 + 4 = √8 = 2√2

z2 = 2√2(2 - 2i) / 2√2 = 2√2(1 / √2 - i / √2) = 2√2(cosπ/4 - isinπ/4).

Answer 5

The polar form of a complex number z = a + bi is z = r (cos θ + i sin θ), where r = | z | = √(a² + b²), a = r cos θ, and b = r sin θ, and θ = tan^{- 1}(b / a) for a > 0 or θ = tan^{- 1}(b / a) + π or θ = tan^{- 1}(b / a) + 180^o for a < 0.

The complex number is z₁ = 3 + 4i.

The polar form of a complex number z = a + bi is z = r (cos θ + i sin θ).

Here a = 3 > 0 and b = 4.

So, first find the absolute value of r .

r = | z | = √(a² + b²)

            = √[ (3)² + (4)² ]

            = √[ 9 + 16 ]

            = √[ 25 ]

            = 5.

Now find the argument θ.

Since a = 3 > 0, use the formula θ = tan^{- 1}(b / a).

θ = tan^{- 1}(4/3)^o

θ ≅ tan^{- 1}(1.333)^o

θ ≅ 53.13^o

Note that here θ is measured in degrees.

Therefore, the polar form of 3 + 4i is about 5[ cos(53.13^o) + i sin(53.13^o) ].

z₁ = 3 + 4i and z₂ = 2 - 2i.

z₁ · z₂ = (3 + 4i) · (2 - 2i) = 6 - 6i + 8i - 8i² = 3 + 2i + 8 = 14 + 2i.

z₁ / z₂ = (3 + 4i) / (2 - 2i)

          = (3 + 4i)(2 + 2i) / (2 - 2i)(2 + 2i)

          = (6 + 6i + 8i + 8i²) / [4 - 4i²]

          = (6 + 14i - 8) / (4 + 4)

          = (- 2 + 14i) / 8

          = - 1/4 + 7i/4.