# CONVERTING COMPLEX NUMBERS TO POLAR FORM

Converting Complex Numbers to Polar Form :

Here we are going to see some example problems based on converting complex numbers to polar form.

## 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

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.

## Converting Complex Numbers to Polar Form - Examples

Question 1 :

Write in polar form of the following complex numbers

(i) 2 + i 23

Solution :

2 + i 23  =  r (cos θ + i sin θ)

r = √22 + (23)2

r = √4 + 12

r = √16

r = 4

α  = tan-1|y/x|

α  =  tan-1 (23/2)

α  =  tan-1 (3)

α  =  tan-1 (π/3)

Since the complex number 2 + i 23 lies in the first quadrant, has the principal value θ  =  α.

Hence the polar form of the given complex number 2 + i 23 is (ii) 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  =  tan-1 (π/6)

Since the complex number 3-i3 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))

Hence the polar form of the given complex number 3 - i 3 is (iii) −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  =  tan-1 (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))

Hence the polar form of the given complex number −2 − i2 (iv) (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 + i3)

=  (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))

Hence the polar form of the given complex number (i - 1) / [cos (π/3) + i sin  (π/3)] is  After having gone through the stuff given above, we hope that the students would have understood, "Converting Complex Numbers to Polar Form".

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