WRITE THE COMPLEX NUMBER IN POLAR FORM

Plot each complex number. Then write the complex number in polar form. You may express the argument in degrees or radians.

Problem 1 :

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)

Problem 2 :

1 + i√3

Solution:

Given, z = 1 + i√3

1 + i√3 = r(cos θ + i sin θ) ---(1)

Since the complex number 1 + i√3 is positive, z lies in the first quadrant.

So, the principal value θ = π/3

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

1 + i√3 = 2(cos π/3 + i sin π/3)

Problem 3 :

-1 - i

Solution:

Given, z = -1 - i

-1 - i = r(cos θ + i sin θ) ---(1)

Since the complex number -1 - i lies in the third quadrant, has the principal value θ = - π + π/4.

So, the principal value θ = -3π/4

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

Problem 4 :

2 - 2i

Solution:

Given, z = 2 - 2i

2 - 2i = r(cos θ + i sin θ) ---(1)

Since the complex number 2 - 2i 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

Problem 5 :

-4i

Solution:

Given, z = 0 - 4i

0 - 4i = r(cos θ + i sin θ) ---(1)

Since the complex number 0 - 4i has a 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

Problem 6 :

-3i

Solution:

Given, z = 0 - 3i

0 - 3i = r(cos θ + i sin θ) ---(1)

Since the complex number 0 - 3i has a 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

Problem 7 :

2√3 - 2i

Solution:

Given, z = 2√3 - 2i

2√3 - 2i = r(cos θ + i sin θ) ---(1)

Since the complex number 2√3 - 2i has a positive and negative, z lies in the fourth quadrant.

So, the principal value θ = -π/6

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

Problem 8 :

-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 has a negative and positive, z lies in the second quadrant.

So, the principal value θ = -π/3

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

Problem 9 :

-3

Solution:

Given, z = -3 + 0i

-3 + 0i = r(cos θ + i sin θ) ---(1)

Since the complex number -3 + 0i has a negative and positive, z lies in the second quadrant.

So, the principal value θ = π  

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

-3 + 0i = 3(cos π + i sin π)

Problem 10 :

-4

Solution:

Given, z = -4 + 0i

-4 + 0i = r(cos θ + i sin θ) ---(1)

Since the complex number -4 + 0i has a negative and positive, z lies in the second quadrant.

So, the principal value θ = π  

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

-4 + 0i = 4(cos π + i sin π)

Problem 11 :

3√2 - 3i√2 

Solution:

Given, z = 3√2 - 3i√2 

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

Since the complex number 3√2 - 3i√2 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

Problem 12 :

-3 + 4i

Solution:

Given, z = -3 + 4i

-3 + 4i  = r(cos θ + i sin θ) ---(1)

Since the complex number -3 + 4i has a negative and positive, z lies in the second quadrant.

θ = tan^-1(4/-3) + π

= π - tan^-1(4/3)

So, the principal value θ = π - tan-1(4/3)

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

Problem 13 :

-2 + 3i

Solution:

Given, z = -2 + 3i

-2 + 3i  = r(cos θ + i sin θ) ---(1)

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

θ = tan^-1(3/-2) + π

= π - tan^-1(3/2)

So, the principal value θ = π - tan^-1(3/2)

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

Problem 14 :

2 - i√3

Solution:

Given, z = 2 - i√3

2 - i√3 = r(cos θ + i sin θ) ---(1)

Since the complex number 2 - √3i has a positive and negative, z lies in the fourth quadrant.

θ = -tan^-1(√3/2) 

So, the principal value θ = -tan^-1(√3/2) 

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

Problem 15 :

1 - i√5

Solution:

Given, z = 1 - i√5

1 - i√5 = r(cos θ + i sin θ) ---(1)

Since the complex number 1 - √5i has a positive and negative, z lies in the fourth quadrant.

θ = -tan^-1(√5) 

So, the principal value θ = -tan^-1(√5) 

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