CONVERTING POLAR TO RECTANGULAR FORM COMPLEX NUMBERS

Properties of Polar Form

Property 1 :

If z = r (cos θ + i sin θ), then

z^-1  =  (1/r) (cos θ - i sin θ)

Property 2 :

If z₁ = r₁ (cos θ₁ + i sin θ₁) and z₂ = r₂ (cos θ₂ + i sin θ₂) then

 z₁ z₂  =  r₁r₂ (cos (θ₁ + θ₂) + i sin (θ₁ + θ₂))

Property 3 :

If z₁ = r₁ (cos θ₁ + i sin θ₁) and z₂ = r₂ (cos θ₂ + i sin θ₂) then

 z_1/ z₂  =  (r_1/r₂) (cos (θ₁ - θ₂) + i sin (θ₁ - θ₂))

Find the rectangular form of the following complex numbers :

Problem 1 :

[cos (π/6) + i sin (π/6)] [cos (π/12) + i sin (π/12)] 

Solution :

=  [cos (π/6) + i sin (π/6)] [cos (π/12) + i sin (π/12)] 

=  [cos ((π/6) + (π/12)) + i sin ((π/6) + (π/12))]

=  cos (2π + π)/12 + i sin (2π + π)/12

=  cos (3π/12) + i sin (3π/12)

=  cos (π/4) + i sin (π/4)

  =  (1/√2) + i (1/√2)

=  (1/√2)(1 + i)

Problem 2 :

[cos (π/6) - i sin (π/6)]/2 [cos (π/3) + i sin (π/3)] 

Solution :

  =  [cos (π/6) - i sin (π/6)cos (π/3) + i sin (π/3)] 

  =  (1/2) [cos (-π/6)+sin (-π/6)] [cos (-π/3)+i sin (-π/3)] 

  =  (1/2) [cos (-3π/6) + sin (-3π/6)]

   =  (1/2) [cos (-π/2) + sin (-π/2)]

   =  (1/2) [0 + (-1)]

   =  (-1/2) 

Problem 3 :

If (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) ................(x_n + iy_n)  =  a + ib show that 

(x₁² + y₁²) (x₂² + y₂²)............ (x_n² + y_n²)  =  a² + b²

Solution :

(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) ................(x_n + iy_n)  =  a + ib

Taking modulus on both sides, we get

|(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) ................(x_n + iy_n)|  =  |a + ib|

|(x₁ + iy₁)| |(x₂ + iy₂)| |(x₃ + iy₃)|............|(x_n + iy_n)|  =  |a + ib|

√(x₁² + y₁²) √(x₂² + y₂²)√(x₃² + y₃²) ...............√(x_n² + y²)  =  √(a² + b²)

√[(x₁² + y₁²) (x₂² + y₂²)(x₃² + y₃²) ..............(x_n² + y²)]  =  √(a² + b²)

Taking squares on both sides, we get 

(x₁²+y₁²) (x₂² + y₂²)(x₃² + y₃²) ..............(x_n² + y²) =  (a²+b²)

Problem 4 :

If (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) ................(x_n + iy_n)  =  a + ib show that 

∑_{r = 1 to n} tan^-1(y_r/x_r) = tan^-1(b/a)  + 2kπ, k ∈ z

Solution :

 (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) ................(x_n + iy_n)  =  a + ib

By taking argument on both sides, we get  

arg [(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) ................(x_n + iy_n)]  =  arg(a + ib)

 arg(x₁+ iy₁) + arg(x₂+ iy₂)+................+ arg(x₂+ iy₂)  =  arg(a + ib)

tan^-1(y₁/x₁) + tan^-1(y₂/x₂) + ..................+ tan^-1(y_n/x_n)  =  tan^-1(b/a) 

∑_{r = 1 to n} tan^-1(y_r/x_r) = tan^-1(b/a)  + 2kπ, k ∈ z

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