Question

-16/1+i√3 convert in to polar form.. Plz plz

Answer 1

Answer:

-16/1+i√3

=-16+√3

Step-by-step explanation:

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Answer 2

Solution:

∴ z = - 16/(1 + i√3)

= - 16 (1 - i√3) / {(1 + i√3) (1 - i√3)}

= - 16 (1 - i√3) / (1 - 3i²)

= - 16 (1 - i√3) / (1 + 3) [ ∵ i² = - 1 ]

= - 16 (1 - i√3) / 4

= - 4 (1 - i√3)

= - 4 + i 4√3

Let, - 4 + i 4√3 = r (cosθ + i sinθ)

Then, r cosθ = - 4 & r sinθ = 4√3

We have

r² cos²θ + r² sin²θ = (- 4)² + (4√3)²

or, r² (cos²θ + sin²θ) = 16 + 48

or, r² = 64

or, r = 8

∴ cosθ = - 1/2 & sinθ = √3/2

and this determinates θ = 2π/3

∴ mod z = 8 and arg z = 2π/3

∴ - 4 + i 4√3 = 8 {cos(2π/3) + i sin(2π/3)}

i.e., - 16/(1 + i√3) = 8 {cos(2π/3) + i sin(2π/3)} ,

which is the required polar form.