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 ?
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James Sousa: The Product and Quotient of Complex Numbers in Trigonometric Form
Guidance
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.
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
Example A
Multiply where and .
Solution:
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
Rewriting 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 B
Using polar multiplication, find the product .
Solution:
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 C
Using polar division, find the quotient of given that and .
Solution:
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.
Concept question wrapup: 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 ? 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 

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Guided Practice
1) Find the product:
2) Find the quotient:
Answers
1) This one is easier than it looks: Recall
 a) ....... By substitution and multplication

b)
..... Substitute
 ..... Find common denominators
 ..... Simplify
is the product
2) First finding 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
Explore More
 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.