CONVERTING COMPLEX NUMBERS TO POLAR FORM

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

1^st quadrant

2^nd quadrant

3^rd quadrant

4^th 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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CONVERTING COMPLEX NUMBERS TO POLAR FORM

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