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 polar form or trigonometric form of a complex number P is
z = r (cos θ + i sin θ)
The value "r" represents the absolute value or modulus of the complex number z .
The angle θ is called the argument or amplitude of the complex number z denoted by θ = arg(z).
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
To find the principal argument of a complex number, we may use the following methods
1st quadrant 2nd quadrant 3rd quadrant 4th quadrant |
θ = α θ = π - α θ = - π + α θ = -α |
The capital A is important here to distinguish the principal value from the general value. Evidently, in practice to find the principal angle θ, we usually compute α = tan−1 |y/x| and adjust for the quadrant problem by adding or subtracting α with π appropriately
arg z = Arg z + 2nπ , n ∈ z.
Write the following complex numbers in polar form.
Problem 1 :
2 + i 2√3
Solution :
2 + i 2√3 = r (cos θ + i sin θ)
r = √22 + (2√3)2 r = √(4+12) r = √16 r = 4 |
α = tan-1|y/x| α = tan-1 (2√3/2) α = tan-1 (√3) α = π/3 |
Since the complex number 2 + i 2√3 lies in the first quadrant, has the principal value θ = α.
So, the polar form of the given complex number is
Problem 2 :
3 - i √3
Solution :
3 - i √3 = r (cos θ + i sin θ)
r = √32 + (-√3)2 r = √(9+3) r = √12 r = 2√3 |
α = tan-1|y/x| α = tan-1 |-√3/3| α = tan-1 (1/√3) α = π/6 |
Since the complex number 3-i√3 lies in the fourth quadrant, has the principal value θ = -α.
θ = -π/6
3 - i √3 = 2√3 (cos (-π/6) + i sin (-π/6)
3 - i √3 = 2√3 (cos (π/6) - i sin (π/6))
So, the polar form of the given complex number is
Problem 3 :
-2 - i2
Solution :
−2 − i2 = r (cos θ + i sin θ)
r = √(-2)2 + (-2)2 r = √(4+4) r = 2√2 |
α = tan-1|y/x| α = tan-1 |2/2| α = tan-1 (1) α = π/4 |
Since the complex number −2 − i2 lies in the third quadrant, has the principal value θ = -π+α.
θ = -π + π/4
θ = (-4π+π)/4
θ = -3π/4
−2 − i2 = 2√3 (cos ( -3π/4) + i sin ( -3π/4))
So, the polar form of the given complex number is
Problem 4 :
(i - 1) / [cos (π/3) + i sin (π/3)]
Solution :
= (i - 1) / [cos (π/3) + i sin (π/3)]
= (i - 1) / [(1/2) + i (√3/2)]
= 2(i - 1) / (1 + i√3)
= (2i - 2) / (1 + i√3)
r = √[(4 + 2√3 + 4 - 2√3)/4]
r = √2
α = tan-1|(√3+1)/(√3-1)|
α = tan-1 (5π/12)
tan 75 = tan (30 + 45)
= (tan 30 + tan 45)/(1 - tan 30 tan 45)
= [(1/√3) + 1]/[1 - (1/√3)1]
= (1 + √3)/(√3 - 1)
= (√3 + 1)/(√3 - 1)
(i - 1) / [cos (π/3) + i sin (π/3)]
= √2 (cos (5π/12) + i sin (5π/12))
So, the polar form of the given complex number is
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