Arithmetic Operations on Complex Numbers

Addition of two complex numbers and is carried out thus:   .

The real part of is therefore and the imaginary part is .

In a similar way,   .

The real part of is therefore and the imaginary part .

## 1.1  Examples

If and what is

i) ii) Solutions:

i) .

ii) .

# 2  Multiplication

Multiplication of two complex numbers and is carried out thus: .

The real part of is therefore and the imaginary part is .

## 2.1  Examples

If and what is

i) ii) Solutions:

i)  .

ii) .

# 3  Division

Division is achieved through use of the complex congugate, , i.e. with the sign of the imaginary part reversed. It will be noted that i.e. a real, positive number.

Division of a complex number by another complex number is facilitated by multiplying both numerator and denominator by giving .

## 3.1  Examples

Convert the fraction to the form .

Solution: .

# 4  Powers of complex numbers

Powers of pure imaginary numbers of the form are found by calculating and separately and multiplying the two results together.

## 4.1  Example

Calculate where .

Solution: . . .

Therefore, .

Powers of complex numbers of the form can be calculated by expanding . However, for large values of this calculation may be tedious, or, indeed, impractical.

An easier method is to use de Moivre's formula .

## 4.2  Example

Calculate where .

Solution:

The modulus of is found by .

The argument of is found by (noting that is in the first quadrant). can therefore be re-written as Using de Moivre's formula,  .

# 5  Roots of complex numbers

Roots of complex numbers are also found using de Moivre's formula, this time in the form .

When dealing with real numbers, we are used to there being a maximum of two solutions when calculating roots i.e. two solutions, , for even roots, and one solution, , for odd roots. However, when calculating complex roots, there are different solutions for roots.

## 5.1  Examples

i) Find the complex 4th roots of .

ii) Find the complex 3rd (cube) roots of .

i) Solution:

The modulus of is clearly given by . The argument of is zero, since lies on the positive real axis. The complex 4th roots of are therefore given by i.e. the roots are . The results can easily be verified by calculation.

ii) Solution: The modulus of is given by .

The argument of is , since lies in the 2nd quadrant of the Argand diagram.

The complex cube roots of are therefore given by   