Polar form of a complex number:

Let OP = r, then x = r cosθ , and y = r sinθ => z = x + iy = r cosθ + ir sinθ = r ( cosθ + i sinθ). That is called as Polar form(Trigonometric form) of a Complex Number. Here we could take the main value of θ.

For normal values of the argument

z = r [ cos( 2n∏ + θ) + i sin ( 2n∏ + θ)]       (where n is an integer)

Note: Occasionally cos q + i sin q is, in short, defined as cis(θ).

Euler's formula: cosθ + i sinθ = e^iθ  .

Note:

      When complex numbers are multiply their arguments get included i..e

      (cosθ₁ + i sinθ₁) (cos θ₂ + i sinθ₂) = cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)

      Generalizing,

      (cosθ₁+ i sinθ₁)(cosθ₂+ i sinθ₂)...(cosθ_n + i sinθ_n) = cos(θ₁+ θ₂+ ...+θ_n) + i sin(θ₁+ θ₂+...+θ_n)

      when complex numbers are divided, Arguments of complex number get subtracted

       = cos (θ₁ - θ₂) + i sin (θ₁ - θ₂)

Problem:           Calculate the modulus and the principal argument of the given numbers:

                              (i)   6(cos 310° - i sin 310°)         (ii)        

Solution:              (i)   6(cos 310° - i sin 310°) = 6[cos(360° - 50°) - i sin(360° - 50°)]

                                                                                          = 6(cos 50° + i sin 50°)

                                                                                          = 

                                    ∴ modulus = 6 and principal value of the given argument = 5Π/18

                              (ii)  

                                  ∴  Modulus = 1 and the principal value of the argument = - Π/2

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