# 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 (3*e*^{4j})(2*e*^{1.7j})

Answer

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

(3*e*^{4j})(2*e*^{1.7j})

= (3)(2)*e*^{4j+1.7j}

= 6*e*^{5.7j}

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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:

`r_1/_theta_1xxr_2/_theta_2=r_1r_2/_[theta_1+theta_2]`

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

To multiply complex numbers in polar form,

Multiplythepartsr

Addtheangleparts

### Example 2

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

Answer

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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## Division

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])`

or

`(r_1/_theta_1)/(r_2/_theta_2)=r_1/r_2/_[theta_1-theta_2]`

In words, this simply means...

To divide complex numbers in polar form,

Dividethepartsr

Subtracttheangleparts

**Example 3**

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

Answer

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)`

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**Example 4**

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

Answer

`(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)`

### Example 5

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

Answer

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

`8j = 8\ ∠\ 90°`

and

`7 + 2j = 7.28\ ∠\ 15.95°`

So

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

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### Exercises:

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

Answer

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

So

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

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

Answer

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

So

`(12/_320^"o")/(5/_210^"o")`

`=12/5angle(320^text(o)-210^text(o))`

`=2.4angle110^text(o)`

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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.

`(-2+5j)/(-1-j)`

Answer

#### Solution, Part (i)

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

`=5.39/1.41angle(112^text(o)-225^text(o))`

`=3.82angle247^text(o)`

`=-1.49-3.52j`

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

#### Part (ii) CHECK:

`(-2+5j)/(-1-j)`

`=((-2+5j)(-1+j))/((-1-j)(-1+j))`

`=(-3-7j)/2`

`=-1.5-3.5j`

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 + 5

j)(−1 +j)

This gives us

−2(−1 + *j*) + 5*j*(−1 + *j*)

= 2 − 2

j− 5j+ 5j^{2}= 2 − 2

j− 5j− 5= −3 − 7

j

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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