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6. Products and Quotients of Complex Numbers

by M. Bourne

When performing addition and subtraction of complex numbers, use rectangular form. (This is because we just add real parts then add imaginary parts; or subtract real parts, subtract imaginary parts.)

When performing multiplication or finding powers and roots of complex numbers, use polar and exponential forms. (This is because it is a lot easier than using rectangular form.)

We start with an example using exponential form, and then generalise it for polar and rectangular forms.

Example 1

Find (3e4j)(2e1.7j), where `j=sqrt(-1).`


Here we are multiplying two complex numbers in exponential form. It is no different to multiplying whenever indices are involved.


= (3)(2)e4j+1.7j

= 6e5.7j

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Multiplying Complex Numbers in Polar Form

We can generalise the example we just did, as follows:

`(r_1\ e^(\ theta_1j))(r_2\ e^(\ theta_2j))=r_1r_2\ e^((theta_1+\ theta_2)j`

From this, we can develop a formula for multiplying using polar form:

`r_1(cos\ theta_1+j\ sin\ theta_1)` `xxr_2(cos\ theta_2+j\ sin\ theta_2)`

`=r_1r_2(cos[theta_1+theta_2]` `{:+j\ sin[theta_1+theta_2])`

or with equivalent meaning:


In words, all this confusing-looking algebra simply means...

To multiply complex numbers in polar form,

Multiply the r parts

Add the angle parts

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Example 2

Find 3(cos 120° + j sin 120°) × 5(cos 45° + j sin 45°)


3(cos 120° + j sin 120°) × 5(cos 45° + j sin 45°)

= (3)(5)(cos(120° + 45°) +j sin(120° + 45°)

= 15 [cos(165°) +j sin(165°)]

In this example, the r parts are 3 and 5, so we multiplied them. The angle parts are 120° and 45°, and we added them.

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As we did before, we do an example in exponential form first, then generalise it for polar form.

Example in Exponential Form:

`8\ e^(\ 3.6j)-:2\ e^(\ 1.2j)` `=4\ e^(\ 3.6j-1.2j)` `=4\ e^(\ 2.4j)`

[We divided the number parts, and subtracted the indices, just using normal algebra.]

From the above example, we can conclude the following:

`(r_1(costheta_1+j\ sintheta_1))/(r_2(costheta_2+j\ sintheta_2))` `=r_1/r_2(cos[theta_1-theta_2]+j\ sin[theta_1-theta_2])`



In words, this simply means...

To divide complex numbers in polar form,

Divide the r parts

Subtract the angle parts

Example 3

Find `(2/_90^"o")/(4/_75^"o")`


Here, the r parts are 2 and 4, so we divide them, and the angle parts are `90°` and `75°`, so we subtract them.

`(2 angle 90^text(o))/(4 angle 75^text(o))=2/4angle(90^text(o)-75^text(o))` `=1/2angle15^text(o)`

Example 4

Find `(3/_20^"o")/(9/_60^"o")`


`(3 angle 20^text(o))/(9 angle 60^text(o))` `=3/9angle(20^text(o)-60^text(o))=1/3angle-40^text(o)`

In this section, we normally leave our answer as a positive angle, so since -40° = 360° − 40° = 320°, we write :

`(3 angle 20^text(o))/(9 angle 60^text(o))=1/3angle320^text(o)`

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Example 5

Find `(8j)/(7+2j)` using polar form.


First, we express `8j` and `7 + 2j` in polar form.

`8j = 8\ ∠\ 90°`


`7 + 2j = 7.28\ ∠\ 15.95°`


`(8j)/(7+2j)` `=(8angle90^text(o))/(7.28 angle 15.95^text(o))` `=1.10angle74.05^text(o)`

Easy to understand math videos:


1. Evaluate: `(0.5 ∠ 140^"o")(6 ∠ 110^"o")`


`0.5xx6=3` and `140^"o"+110^"o"=250^"o"`


`(0.5 ∠ 140^"o")(6 ∠ 110^"o")=3 ∠ 250^"o"`

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2. Evaluate: `(12/_320^"o")/(5/_210^"o")`


`12/5=2.4`; and ` 320^"o"-120^"o"=110^"o"`





3. (i) Evaluate the following by first converting numerator and denominator into polar form.

(ii) Then check your answer by multiplying numerator and denominator by the conjugate of the denominator.



Solution, Part (i)

`(-2+5j)/(-1-j) = (5.39 angle 112^text(o))/(1.41 angle 225^text(o))`




Note: `112^"o" - 225^"o" = -113^"o"` is equivalent to positive `247^"o"`.

Part (ii) CHECK:





Here's an explanation of what happened in the expansion of terms in the above answer.

The top of the fraction (the numerator) is:

(−2 + 5j)(−1 + j)

This gives us

−2(−1 + j) + 5j(−1 + j)

= 2 − 2j − 5j + 5j2

= 2 − 2j − 5j − 5

= −3 − 7j

The bottom part of the fraction is

(−1 − j)(−1 + j)

= (−1)2 − (j)2

= 1 − (−1)

= 1 + 1

= 2

[There is a rounding error in Part (i) since we used decimal approximations throughout. The answer in Part (ii) is exact.]


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