Chapter 2 Section 4: Complex Numbers

Today we learned about Complex Numbers in Section Four.

Before we did that, though, we reviewed the realm of Real numbers. Real numbers include both Rational numbers (e.g. 4, -7, -7/3, 0), and Irrational numbers (e.g. ,  , 5+)

Next, Mr.Wilhelm started the section on complex numbers.  He described them as having a real part and an imaginary part.

Standard Form of a complex number:  a+bi (where a and b are both real numbers and i represents the imaginary).

We were reminded of the value of an imaginary unit:

                           i=

Addition and Subtraction of Complex Numbers:

When adding complex numbers, one treats the imaginary unit like any other variable.

                        (2+4i)+(6-5i)= 8-i

the same is true when subtracting..

                        (2+4i)-(6-5i)= -4+9i

Multiplication of Complex Numbers:

When multiplying complex numbers, the important thing to remember is that i^2= -1

                        (2+4i)(6-5i)=

                      12 +14i-20i^2=         you multiply -20 by -1 and get +20

                        12+14i+20=

                            32+4i

(always simplify and make sure your answer is in standard form)

Dividing Complex Numbers:

When dividing Complex numbers, the key thing to remember is to multiply the numerator and denominator of your problem by the conjugate of the denominator.

                         now multiply by the conjugate

next, distribute and get... 

next, simplify the i^2s to -1s...

next, combine like terms and get...

finally, convert into standard form...

After learning about operations of Complex Numbers, Mr.Wilhelm talked a little bit about how even if a graph (e.g. a parabola) does not touch the x-axis, it still has imaginary zeros.

Then, we talked about taking i to higher powers.

 i^1= i

i^2=-1

i^3= -i  (because root -1 times root -1 times root -1 is equal to negative root -1 which is equal to -i)

i^4= 1 (because root -1 times root -1 times root -1 is equal to negative root -1, and when you multiply that by another root -1 you get -i^2 which is equivalent to 1)

we also talked about how this applies to other higher powers in a cycle of fours but I don't really understand that yet so I probably shouldn't confuse all of you..

We finished off the day talking about imaginary numbers as zeros, converting the zeros into polynomials and vice versa.

*If something like 5+3i is a zero, then its conjugate (5-3i) is also a zero

Example (creating a polynomial out of zeros and imaginary zeros):

Zeros: 2, 5+3i, and therefore 5-3i

the long way is...

f(x)= (x-2)(x-(5+3i))(x-(5-3i))

=(x-2)(x-5-3i)(x-5+3i)

=(x-2)[x^2-5x+3ix-5x+25-15i-3ix+15i-9i^2]

=(x-2)[x^2-10x+25-9i^2]

=(x-2)[x^2-10x+25+9]

=x^3-10x^2+34x-2x^2+20x-68

=x^3-12x^2+54x-68

the shorter way is...

(starting from the second step of the last example)

f(x)=(x-2)(x-5-3i)(x-5+3i)

(x-5-3i) represents (a-b) and (x-5+3i) represents (a+b), and when you take (a-b)(a+b) you get a^2-b^2

so in actuality you have f(x)=(x-2)(x-5)^2 -(3i)^2 (because x-5 represents a and 3i represents b)

this simplifies to

f(x)=(x-2)(x^2-10x+34)

=x^3-10x^2+34x-2x^2+20x-68

=x^3-12x+54x-68