~I would like to give you a brief description of imaginary (immy for short) numbers. They are used extensively in higher math & in many engineering related problems.
~To understand them, you must understand the basic unit of the system,
namely the square root of -1. (which is given its own symbol)
~The square root of 1 is 1, since (1) (1) = 1, the square root of 4 is 2, since (2)(2)=4, the square root of 9 is 3, since (3)(3)=9, & so on. These are some of the "nice ones". (come out very nice).
~Notice that to be a square root of a number, the number times itself must equal the number. The square root of 2 doesn't come out nice, but is real. You can find it on your calculator. It is irrational (unending, non-repeating decimal). It can be approximated by 1.414. Likewise for the square roots of 3,5,6,7, & so on. All of these numbers are real, since they can be found on the real number line.
~However, the square root of -1 is not. You will not be able to find a real number such that when multiplied by itself gives -1. So, this number does not exist as a point on the real number line. So, we call it imaginary & give it the symbol i.
~This is the basic imaginary unit. All other imaginary numbers can be expressed in terms of i.
~The Complex Plane~
~A 2-D system (similar to the x & y axis system) is invented to represent these types of numbers. The horizontal axis consists of the real number line & the vertical axis is the imaginary axis, I.
~Since all numbers are complex numbers, i.e., form a + bi, where a is real part & bi is the imaginary part, they can be represented using this axis system.
~If b=0, then the complex number is real (no i's involved) & each number would be on the horizontal axis (real axis).
~If b is not 0, then the complex number is not real & is off the real axis.
~For example, to locate 3+2i on the axis system, you would start at the origin & move 3 units to the right on the real axis then move 2 units up in the imaginary direction.
~Any number off the horizontal real axis is imaginary. To locate -7-4i, you would move 7 units to the left of the origin on the real axis then move straight down 4 units in the imaginary direction. Therefore, all numbers of the form a+bi can be represented by a point in this system (called the complex plane).
~If a=0, then the complex number is of the form bi, or pure imaginary.
~To locate 3i, just move straight up from the origin 3 units. It will be on the imaginary axis. Likewise, -8i would be down 8 units from the origin on the imaginary axis.
~So, all pure imaginary numbers are on the imaginary axis, while the numbers 2+3i, -7-5i, -4+7i, 8-3i, & so on, are still imaginary, but not on the imaginary (vertical) axis.
~Working with complex numbers~
The rules for working with complex numbers follow the same principles as we used for the real numbers, with few exceptions. Some new rules must be established for multiplication:
(i)(i) = i2= -1, i3 = (i)(i)(i) = (i)2(i) = (-1)(i) = -i, i4 = (i)2(i)2 = (-1)(-1) = 1
~Since i2= -1 & i4= 1, imaginary numbers can give a real number result when multiplied.
~Note that i3=(i)(i2) = (i)(-1) = - i. Odd powers of i give an imaginary result.
~We see that there are only 4 powers of i, namely, i,-1,-i,1. Every power of i higher than 4 can be reduced to one of these, just divide the power by 4 & take the remainder for the power.
~If the remainder is 1, then the answer is i,
~If the remainder is 2, then the answer is -1,
~If the remainder is 3, then the answer is -i
~If the remainder is 0, then the answer is 1
~These are know as the 4 powers of i (i,-1,-i,1)
~For example, take i237. Divide 237 by 4 to get a quotient of 59 with a remainder of 1, So, i237 = i1 = i
~Here are some examples of adding , multiplying, subtracting, & dividing complex numbers.
~Ex: (7-3i) + (8+23i) = 15 +20i (just add the real parts & the immy parts separately)
~Ex: (2-3i)(5+2i) = 10 +4i -15i -6i2= 10 -11i -6(-1) = 10 -11i +6 = 16 -11i
~Ex: (4-9i)-(-5-6i) = 4-9i+5+6i=9-3i
~Ex: (2-5i)/(1-3i) = (2-5i)(1+3i)/(1-3i)(1+3i) [multiplying top & bottom by the conjugate of the bottom, i.e., (1+3i)].
That gives (2+6i-5i-15i2)/(1+3i-3i-9i2) = (2+i-15(-1))/(1-9(-1)) = (17+i)/10
~The uses for these numbers is beyond the scope of this course, however, for those of you who take more advance math & science courses, you will certainly encounter them.