Complex numbers are found in real world calculations involving: quantum mechanics, signal analysis, fluid dynamics, control theory, and many other fields.

In electrical engineering, complex numbers are used for calculations involving *impedance* (the resistance to electric flow in a circuit).

Electrical engineers are familiar with the formula:

by comparing it to the similar expression below, which is explored in this lesson, can you identify the variable *j*?

### Product and Quotient Theorems

#### The Product Theorem

Since complex numbers can be transformed to polar form, the multiplication of complex numbers can also be done in polar form. Suppose we know *z*_{1} = *r*_{1} (cos *θ*_{1} + *i* sin *θ*_{1}) *z*_{2} = *r*_{2} (cos *θ*_{2} + *i* sin *θ*_{2})

To multiply the two complex numbers in polar form:

(Use *i*^{2} = -1, gather like terms, factor out , substitute the angle sum formulas for both sine and cosine)

This last equation states that the product of two complex numbers in polar form can be obtained by multiplying the polar r values of each of the complex numbers and then multiplying that value by cis of the sum of each of the two angles of the individual complex numbers. This is more concise than the rectangular form for multiplication of complex numbers.

#### The Quotient Theorem

Dividing complex numbers in polar form can be shown using a similar proof that was used to show multiplication of complex numbers. Here we omit the proof and give the result. For *z*_{1} = *r*_{1}(cos *θ*_{1} + *i* sin *θ*_{1}) and *z*_{2} = *r*_{2}(cos *θ*_{2} + *i* sin *θ*_{2}), then

### Examples

#### Example 1

Earlier, you were asked to identify the variable *j* in the following formula:

In electrical calculations, the letter *I* is commonly used to denote *current*, therefore imaginary numbers are identified with a *j*.

Note the similar usage of *i* in .

#### Example 2

Multiply where and .

For *z*_{1},

and

Note that is in the first quadrant since *a*, and *b* > 0.

For ,

and

Now we can use the formula

Substituting gives:

So we have

Re-writing in approximate decimal form:

5.656 (0.966 – 0.259i)

5.46 - 1.46i

If the problem was done using only rectangular units then

Gathering like terms and using i^{2} = -1

or

#### Example 3

Using polar multiplication, find the product .

Let and

and

and

For *θ*_{1}, first find

Since *x* > 0 and *y* < 0 we know that *θ*_{1} is in the in the 4^{th} quadrant:

For *θ*_{2},

Since *θ*_{2} is in the first quadrant,

Using polar multiplication,

subtracting 2π from the augment:

or in expanded form:

In decimal form this becomes: 55.426(0.866 + 0.500i) or 48 + 27.713i

Check:

#### Example 4

Using polar division, find the quotient of given that and .

For or and , so (4th quadrant)

For or and , so (3^{rd} quadrant)

Using the formula, or

Check by using the complex conjugate to do the division in rectangular form:

The two radically different approaches yield the same answer. The small difference between the two answers is a result of decimal rounding.

#### Example 5

Find the product: .

This one is easier than it looks: Recall .

....... By substitution and multplication

..... Substitute

..... Find common denominators

..... Simplify

is the product

#### Example 6

Find the quotient: .

First, find the quotient by polar multiplication:

since the angle is in the 1^{st} quadrant

θ_{1} = 1.107 radians

for ,

since *θ*_{2} is in the 4^{th} quadrant, between 4.712 and 6.282 radians (or 2*π*)

θ_{2} = 5.820 *radians*

Finally, using the division formula,

If we assume that , then

### Review

- Find the product using polar form:
- Multiply:
- Divide: 3cis(130
^{o}) ÷ 4cis(270^{o})

If and find:

If and find:

Find the products.

- Find the product using polar form:

Find the quotients.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 4.9.