FINDING MODULUS AND ARGUMENT OF A COMPLEX NUMBER

Find the complex conjugate and the modulus of the number.

Problem 1 :

12 - 5i

Solution :

Given, 12 - 5i

Real part is 12.

Complex part = -5i

To find the complex conjugate, we have to change the sign of complex part. Therefore, the complex conjugate is 12 + 5i.

|12 - 5i| is the modulus of the complex number.

Modulus of a complex number a + bi is given by √(a² + b²).

|12 - 5i| = √(12)² + (-5)²

= √(144 + 25)

= √169

= 13

Problem 2 :

-1 + 2√2 i

Solution :

Given, -1 + 2√2 i

Real part is -1.

Complex part = 2√2 i

To find the complex conjugate, we have to change the sign of complex part. Therefore, the complex conjugate is -1 - 2√2 i.

|-1 + 2√2 i| is the modulus of the complex number.

Modulus of a complex number a + bi is given by √(a² + b²).

|-1 + 2√2 i| = √[(-1)² + 2² (√2)²]

= √(1 + 8)

= √9

= 3

Problem 3 :

 -4i

Solution :

Given, -4i

Real part is 0.

Complex part = -4i

To find the complex conjugate, we have to change the sign of complex part. Therefore, the complex conjugate is +4i.

|4i| is the modulus of the complex number.

Modulus of a complex number a + bi is given by √(a² + b²).

|4i| = √(0² + 4²)

= √16

= 4

Write the number in polar form with argument between 0 and 2π.

Problem 4 :

 -3 + 3i

Solution :

Given,  z = -3 + 3i

a + bi = r(cos θ + i sin θ)

-3 + 3i =  r(cos θ + i sin θ) --- (1)

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

So, the principal value θ = π - π/4

θ = (4π - π)/4

θ = 3π/4

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

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

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

Problem 5 :

 1 - √3 i

Solution :

Given,  z = 1 - √3 i

a + bi = r(cos θ + i sin θ)

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

Since, the complex number  1 - √3i has a  positive and negative, z lies in the fourth quadrant.

So, the principal value θ =  π/3

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

1 - √3i  = 2(cos π/3 + i sin π/3)

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

Problem 6 :

 3 + 4i

Solution :

Given,  z = 3 + 4i

a + bi = r(cos θ + i sin θ)

3 + 4i =  r(cos θ + i sin θ) --- (1)

Since, the complex number 3 + 4i has a positive, z lies in the first quadrant.

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

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

3 + 4i =  5(cos tan^-1(4/3) + i sin tan^-1(4/3))

So, the polar form of z is 5[cos tan^-1(4/3) + i sin tan^-1(4/3)].

Problem 7 :

8i

Solution :

Given,  z = 0 + 8i

a + bi = r(cos θ + i sin θ)

0 + 8i =  r(cos θ + i sin θ) --- (1)

Since, the complex number 0 + 8i has a 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 + 8i =  8(cos π/2 + i sin π/2)

So, the polar form of z is 8(cos π/2 + i sin π/2).