Let r and θ be polar coordinates of the point P(x, y) that corresponds to a non zero complex number z = x + iy.
The real number θ represents the angle, measured in radians, that z makes with the positive real axis when z is interpreted as a radius vector.
The angle θ has an infinitely many possible values, including negative ones that differ by integral multiples of 2π.
Those values can be determined from the equation
tan θ = y/x.
θ = π-α
θ = -π+α
θ = -α
Find the polar coordinates of the point with the given rectangular coordinates
Problem 1 :
(-5, 12)
Solution :
Rectangular form (x, y) ==> (-5, 12)
Polar form (r, θ) ==> ?
Finding r :
r = √(x^{2} + y^{2})
r = √(-5)^{2} + 12^{2}
r = √(25 + 144)
r = √169
r = 13
Finding θ :
θ = tan^{-1}(y/x)
x = -5 and y = 12
θ lies in the 2^{nd} quadrant
θ = π - α
α = tan^{-1}(12/5)
θ = π - tan^{-1}(12/5)
θ = 180 - 67.38
θ = 112.62
So, the required polar coordinate is ( 13, 112.62 ).
Problem 2 :
(3, -3)
Solution :
Rectangular form (x, y) ==> (-3, 3)
Polar form (r, θ) ==> ?
Finding r :
r = √(x^{2} + y^{2})
r = √3^{2} + (-3)^{2}
r = √(9+9)
r = √18
r = 3√2
Finding θ :
α = tan^{-1}(y/x)
x = 3 and y = -3
θ lies in the 4th quadrant.
θ = -α
θ = -tan^{-1}(3/3)
θ = -tan^{-1}(1)
θ = -π/4
So, the required polar coordinate is (3√2, -π/4).
Problem 3 :
(-2, -2√3)
Solution :
Rectangular form (x, y) ==> (-2, -2√3)
Polar form (r, θ) ==> ?
Finding r :
r = √(x^{2} + y^{2})
r = √(-2)^{2} + (-2√3)^{2}
r = √(4+12)
r = √16
r = 4
Finding θ :
θ = tan^{-1}(y/x)
x = -2 and y = -2√3
θ lies in 3rd quadrant.
α = tan^{-1}(2√3/2)
α = tan^{-1}(√3)
α = π/3
θ = (π/3) - π
θ = (-2π/3)
So, the required polar coordinate is (4, (-2π/3)).
Problem 4 :
(0, -3)
Solution :
Rectangular form (x, y) ==> (0, -3)
Polar form (r, θ) ==> ?
Finding r :
r = √(x^{2} + y^{2})
r = √0^{2} + (-3)^{2}
r = √9
r = 3
Finding θ :
θ = tan^{-1}(y/x)
x = 0 and y = -3
α = tan^{-1}(0/3
α = tan^{-1}(0)
α = 0
θ = 0 - π
θ = -π
So, the required polar coordinate is (3, -π).
Problem 5 :
(5, 5)
Solution :
Rectangular form (x, y) ==> (5, 5)
Polar form (r, θ) ==> ?
Finding r :
r = √(x^{2} + y^{2})
r = √5^{2} + 5^{2}
r = 5√2
Finding θ :
θ = tan^{-1}(y/x)
x = 5 and y = 5
θ lies in the first quadrant.
α = tan^{-1}(5/5)
α = tan^{-1}(1)
α = π/4
θ = π/4
So, the required polar coordinate is (5√2, π/4).
Problem 6 :
(-1, √3)
Solution :
Rectangular form (x, y) ==> (-1, √3)
Polar form (r, θ) ==> ?
Finding r :
r = √(x^{2} + y^{2})
r = √(-1)^{2} + (√3)^{2}
r = √(1 + 3)
r = 2
Finding θ :
θ = tan^{-1}(y/x)
x = -1 and y = √3
θ lies in the second quadrant.
α = tan^{-1}(√3/1)
α = tan^{-1}(√3)
α = π/3
θ = π - (π/3)
θ = 2π/3
So, the required polar coordinate is (2, 2π/3).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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