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7.2 Imaginary concepts -- complex numbers

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This module introduces the concept of complex numbers in Algebra.

A “complex number” is the sum of two parts: a real number by itself, and a real number multiplied by i . It can therefore be written as a + bi , where a and b are real numbers.

The first part, a , is referred to as the real part . The second part, bi , is referred to as the imaginary part .

Examples of complex numbers a + bi ( a is the “real part”; bi is the “imaginary part”)
3 + 2i a = 3 , b = 2
π a = π , b = 0 (no imaginary part: a “pure real number”)
-i a = 0 , b = -1 (no real part: a “pure imaginary number”)

Some numbers are not obviously in the form a + bi . However, any number can be put in this form.

Example 1: Putting a fraction into a + bi form ( i in the numerator)
3 − 4i 5 size 12{ { {3 - 4i} over {5} } } {} is a valid complex number. But it is not in the form a + bi , and we cannot immediately see what the real and imaginary parts are.
To see the parts, we rewrite it like this:
3 − 4i 5 size 12{ { {3 - 4i} over {5} } } {} = 3 5 size 12{ { {3} over {5} } } {} – 4 5 size 12{ { {4} over {5} } } {} i
Why does that work? It’s just the ordinary rules of fractions, applied backward. (Try multiplying and then subtracting on the right to confirm this.) But now we have a form we can use:
3 − 4i 5 size 12{ { {3 - 4i} over {5} } } {} a = 3 5 size 12{ { {3} over {5} } } {} , b = – 4 5 size 12{ { {4} over {5} } } {}
So we see that fractions are very easy to break up, if the i is in the numerator. An i in the denominator is a bit trickier to deal with.
Example 2: Putting a fraction into a + bi form ( i in the denominator)
1 i size 12{ { {1} over {i} } } {} = 1 ⋅ i i ⋅ i size 12{ { {1 cdot i} over {i cdot i} } } {} Multiplying the top and bottom of a fraction by the same number never changes the value of the fraction: it just rewrites it in a different form.
= i − 1 size 12{ { {i} over { - 1} } } {} Because i • i is i 2 , or –1.
= -i This is not a property of i , but of –1. Similarly, 5 − 1 size 12{ { {5} over { - 1} } } {} = –5 .
1 i size 12{ { {1} over {i} } } {} : a = 0 , b = -1 since we rewrote it as -i , or 0 - 1i

Finally, what if the denominator is a more complicated complex number? The trick in this case is similar to the trick we used for rationalizing the denominator: we multiply by a quantity known as the complex conjugate of the denominator .

Definition of complex conjugate

The complex conjugate of the number a + bi is a - bi . In words, you leave the real part alone, and change the sign of the imaginary part.

Here is how we can use the “complex conjugate” to simplify a fraction.

Example: Using the Complex Conjugate to put a fraction into a + bi form
5 3 − 4i size 12{ { {5} over {3 - 4i} } } {} The fraction: a complex number not currently in the form a + b i
= 5 ( 3 + 4i ) ( 3 − 4i ) ( 3 + 4i ) size 12{ { {5` \( 3+4i \) } over { \( 3 - 4i \) \( 3+4i \) } } } {} Multiply the top and bottom by the complex conjugate of the denominator
= 15 + 20 i 3 2 − ( 4i ) 2 size 12{ { {"15"+"20"i} over {3 rSup { size 8{2} } - \( 4i \) rSup { size 8{2} } } } } {} Remember, ( x + a ) ( x – a ) = x 2 –a 2
= 15 + 20 i 9 + 16 size 12{ { {"15"+"20"i} over {9+"16"} } } {} ( 4i ) 2 = 4 2 i 2 = 16 ( –1 ) = –16 , which we are subtracting from 9
= 15 + 20 i 25 size 12{ { {"15"+"20"i} over {"25"} } } {} Success! The top has i , but the bottom doesn’t. This is easy to deal with.
= 15 25 size 12{ { {"15"} over {"25"} } } {} + 20 i 25 size 12{ { {"20"i} over {"25"} } } {} Break the fraction up, just as we did in a previous example.
= 3 5 size 12{ { {3} over {5} } } {} + 4 5 size 12{ { {4} over {5} } } {} i So we’re there! a = 3 5 size 12{ { {3} over {5} } } {} and b = 4 5 size 12{ { {4} over {5} } } {}

Any number of any kind can be written as a + bi . The above examples show how to rewrite fractions in this form. In the text, you go through a worksheet designed to rewrite − 1 3 size 12{ nroot { size 8{3} } { - 1} } {} as three different complex numbers. Once you understand this exercise, you can rewrite other radicals, such as i size 12{ sqrt {i} } {} , in a + bi form.
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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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