WRITE THE COMPLEX NUMBER IN POLAR FORM

Polar form or trigonometric form :

The complex number a + bi is written in polar form as,

z  =  r(cos θ + i sin θ)

(where a  =  r cos θ, and b  =  r sin θ)

The value of r is called the modulus of a complex number z. i.e, r  =  |z|

To find the r, we use the formula r  =  √(a² + b²).

The angle θ is called the argument of the complex number z. i.e, θ  =  arg(z). 

To find the arg(z), we will use α  =  tan^-1|b/a|.

To find the principal value θ of a complex number, we may use the following methods.

Find the trigonometric form of the complex number where the argument satisfies 0 ≤ θ < 2π :

Example 1 :

3i

Solution :

Given, z  =  0 + 3i

The polar form of the complex number z is

r(cos θ + i sin θ)

0 + 3i  =  r(cos θ + i sin θ) -----(1)

Since the complex number 0 + 3i is positive, z lies in the first quadrant.

So, the principal value θ  =  π/2

By applying the value of r and θ in equation (1), we get

0 + 3i  =  3(cos π/2 + i sin π/2)

So, the polar form of z is 3(cos π/2 + i sin π/2)

Example 2 :

-2i

Solution :

Given, z  =  0 - 2i

0 - 2i  =  r(cos θ + i sin θ) -----(1)

Since the complex number 0 - 2i is positive and negative, z lies in the fourth quadrant.

So, the principal value θ  =  - π/2

By applying the value of r and θ in equation (1), we get

0 – 2i  =  2(cos (-π/2) + i sin (-π/2))

So, the polar form of z is 2(cos (-π/2) + i sin (-π/2))

Example 3 :

2 + 2i

Solution :

Given, z  =  2 + 2i

2 + 2i  =  r(cos θ + i sin θ) -----(1)

Since the complex number 2 + 2i is positive, z lies in the first quadrant.

So, the principal value θ  =  π/4

By applying the value of r and θ in equation (1), we get

2 + 2i  =  2√2(cos π/4 + i sin π/4)

So, the polar form of the complex number z is

2√2(cos π/4 + i sin π/4)

Example 4 :

√3 + i

Solution :

Given, z  =  √3 + i

√3 + i  =  r(cos θ + i sin θ) -----(1)

Since the complex number √3 + i has a positive, z lies in the first quadrant.

So, the principal value θ  =  π/6

By applying the value of r and θ in equation (1), we get

√3 + i  =  2(cos π/6 + i sin π/6)

So, the polar form of z is 2(cos π/6 + i sin π/6)

Example 5 :

-2 + 2i√3

Solution :

Given, z  =  -2 + 2i√3

-2 + 2i√3  =  r(cos θ + i sin θ) -----(1)

Since the complex number -2 + 2i√3 is negative and positive, z lies in the second quadrant.

So, the principal value θ  =  π - π/3

θ  =  2π/3

By applying the value of r and θ in equation (1), we get

-2 + 2i√3  =  4(cos 2π/3 + i sin 2π/3)

So, the polar form of z is 4(cos 2π/3 + i sin 2π/3)

Example 6 :

3 - 3i

Solution :

Given, z  =  3 - 3i

3 – 3i  =  r(cos θ + i sin θ) -----(1)

Since the complex number 3 - 3i has a positive and negative, z lies in the fourth quadrant.

So, the principal value θ  =  – π/4

By applying the value of r and θ in equation (1), we get

3 - 3i  =  4(cos (-π/4) + i sin (-π/4))

So, the polar form of z is 4(cos (-π/4) + i sin (-π/4))

Related pages

• Product of complex numbers in polar form
  
• Division of complex numbers in polar form

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