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2. Basic Operations with Complex Numbers

by M. Bourne

Addition and subtraction of complex numbers works in a similar way to that of adding and subtracting surds. This is not surprising, since the imaginary number j is defined as `j=sqrt(-1)`.

Addition of Complex Numbers

Add real parts, add imaginary parts.

Subtraction of Complex Numbers

Subtract real parts, subtract imaginary parts.

Example 1- Addition & Subtraction

a. `(6 + 7j) + (3 − 5j) =`

`(6 + 3) + (7 − 5)j = 9 + 2j`

b. `(12 + 6j) − (4 + 5j) =`

`(12 − 4) + (6 − 5)j = 8 + j`

Continues below

Multiplication of Complex Numbers

Expand brackets as usual, but care with `j^2`!

Example 2 - Multiplication

Multiply the following.

a. `5(2 + 7j)`

Answer

`5(2 + 7j) = 10 + 35j`

b. `(6 − j)(5j)`

Answer

`(6 − j)(5j)`

`= 30j − 5j^2`

`= 30j − 5(−1) `

`= 5 + 30j`

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c. `(2 − j)(3 + j)`

Answer

`(2 − j)(3 + j)`

`= 6 − 3j + 2j − j^2`

`= 6 − j − (−1)`

`= 6 − j + 1 `

`= 7 − j`

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d. `(5 + 3j)^2`

Answer

We apply the algebraic expansion `(a+b)^2 = a^2 + 2ab + b^2` as follows:

`(5 + 3j)^2 = 25 + 2(5)(3)j + 9(j^2)`

`= 25 + 30j + 9(-1)`

`= 25 + 30j - 9`

`= 16 + 30j`

e. `(2sqrt(-9)-3)(3sqrt(-16)-1)`

Answer

`(2sqrt(-9)-3)(3sqrt(-16)-1)`

`=(2j(3)-3)(3j(4)-1)`

`=(6j-3)(12j-1)`

`=72(j^2)-36j-6j+3`

`=-69-42j`

f. `(3 + 2j)(3 − 2j)`

Answer

`(3 + 2j)(3 − 2j) `

`= (3)^2 − (2j)^2`

`= 9 − 4j^2`

`= 9 + 4`

`= 13`

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Multiplying by the conjugate

Example 2(f) is a special case.

`3 + 2j` is the conjugate of `3 − 2j`.

In general:

`x + yj` is the conjugate of `x − yj`

and

`x − yj` is the conjugate of `x + yj`.

Notice that when we multiply conjugates, our final answer is real only (it does not contain any imaginary terms.

We use the idea of conjugate when dividing complex numbers.

Division of Complex Numbers

Earlier, we learned how to rationalise the denominator of an expression like:

`5/(3-sqrt2)`

To simplify the expression, we multiplied numerator and denominator by the conjugate of the denominator, `3 + sqrt2` as follows:

`5/(3-sqrt2)xx(3+sqrt2)/(3+sqrt2)`

`=(15+5sqrt2)/(9-2)`

`=(15+5sqrt2)/7`

We did this so that we would be left with no radical (square root) in the denominator.

Dividing by a complex number is a similar process to the above - we multiply top and bottom of the fraction by the conjugate of the bottom.

Example 3 - Division

a. Express

`(3-j)/(4-2j)`

in the form x + yj.

Answer

The conjugate of `4 − 2j` is `4 + 2j`. We multiply the top and bottom of the fraction by this conjugate.

`(3-j)/(4-2j) xx (4+2j)/(4+2j)`

`=(12+6j-4j-2j^2)/(16-4j^2)`

`=(12+2+6j-4j)/(16+4)`

`=(14+2j)/20`

`=(7+j)/10`

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b. Simplify:

`(1-sqrt(-4))/(2+9j)`

Answer

We multiply the top and bottom of the fraction by the conjugate of the bottom (denominator).

`(1-sqrt(-4))/(2+9j)`

` =(1-2j)/(2+9j) xx (2-9j)/(2-9j)`

`=(2-9j-4j+18j^2)/(4-81j^2)`

`=(-16-13j)/(4+81)`

`=(-16-13j)/85`

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Exercises

1. Express in the form a + bj:

`(4+sqrt(-16))+(3-sqrt(-81))`

Answer

`(4+sqrt(-16))+(3-sqrt(-81))`

`=(4+4j)+(3-9j)`

`=7-5j`

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2. Express in the form a + bj.

`sqrt(-4)/(2+sqrt(-9))`

Answer

We multiply the top and bottom of the fraction by the conjugate of the bottom (denominator).

`(2j)/(2+3j) xx (2-3j)/(2-3j)`

` = (4j-6j^2)/(4+9)`

`=(6+4j)/13`

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