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Study Notes 038:
Trigonometry 12:
By William E. Steinman:
Let’s look at complex numbers.
We can start with an example.
Using this standard form simplifies operations on imaginary numbers.
A complex number then is of the form x + yi where i is the imaginary unit and x and y are real numbers.
The first term, x is the real part of the complex number and the second term yi is called the imaginary part.
Any real number can be expressed as a complex number in which y = 0.
So 7 can be expressed as 7 + 0i.
This makes the set of real numbers a subset of the set of complex numbers.
If the real part x is equal to 0 and y is not equal to zero we call that a pure imaginary number.
We have a two general rule about the powers of i.
The odd powers of i will be equal to i or –i.
For example:
i3 = i2i = -1i = -i.
The even powers will equal either 1 or -1.
For example:
i4 = i2i2 = -1 -1 = 1
Using the standard form of x + yi allows us to make algebraic manipulations on complex numbers.
Like this:
Remember, we defined i2 as being equal to -1.
We can get into trouble with these manipulations if we start to believe that imaginary numbers follow the same rules we use for real numbers.
We could get an erroneous result for the above problem if we believed that and did the following pseudo solution.
Okay!
This means we can do ordinary algebraic manipulations with complex numbers while treating i separately.
For addition, we add the real parts and the imaginary parts.
For example:
(2 + 3i) + (4 – 5i) = (2 + 4) + (3 – 5)i = 6 – 2i
For subtraction, the rule is the same. We subtract the real parts, then the imaginary parts.
For example:
(2 + 3i) – (4 – 5i) = (2 – 4) + (3 –(-5))i = -2 + 8i
To multiply two complex umbers, carry out the multiplication as though the numbers were ordinary binomials and replace i2 with -1.
For example:
(2 + 3i)(4 – 5i) = 8 + 2i – 15i2 = 8 - 15(-1) + 2i = 23 + 2i
To divide multiply both the numerator and the denominator of the fraction by the conjugate of the denominator.
That means we are simply multiplying by 1.
Like this:
(2 + 3i)/(4 – 5i) = (2 + 3i)(4+ 5i)/(4 – 5i)(4 + 5i) = (8 – 15) + (10 + 12i)/(16 + 25)
All of that = -7/41 + (22/41)i
Notice that the denominator applies equally to the real and imaginary part of this final fraction.
That’s all of it for now.
Next time we can summarize the identities.