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Math Notes
Introduction Before we introduce the notion of a complex number we must first introduce one of the elements of a complex number, an imaginary number. Imaginary numbers are numbers which are given by the
In order to help us deal with these imaginary numbers, we are going to first get rid of the square roots. Whenever we find a negative under the square root, we can always separate out the factor of as we did in the example above. Then we are going to replace that with the letter i. This letter i stands for the imaginary number, the . So that now can be written as 2i. (It should be noted here that in some disciplines, other characters are used. In engineering for instance, it's common to represent the imaginary number with a j, because i is used for current. However, we will use i throughout our discussion here.) Because i is defined to be , i also has another important property: i2 = -1. We can use this property to determine other powers of i, as well as the value one obtains when multiplying numbers containing i. Worked Examples A. Rewrite the following number as an imaginary number containing i. Reduce the square roots as much as possible.
Remember, first factor out the ,
giving us
B. Add the following imaginary numbers: 2i and 3i. Well, one adds 2i + 3i just like one adds 2x + 3x. They are like terms, so add the coefficients. 2i + 3i = (2 + 3)i = 5i. C. What is i3? i4? i17? i95? i3 = i2*i = (-1)*i = -i. Use the definition of i2 = -1 to reduce this problem. i4 = i2*i2 = (-1)*(-1) = 1. Again, we used the definition of i2 = -1. We'll use this property in the next example, too. i17 = i16*i = (i4)4*i = (1)4* i = i. We have an odd number, so separate out one of the i's to make one of the factors even. See how the exponent divides by 4? We can use here the property from the previous example, that i4 = 1 to reduce the problem further. i95 = i94*i = (i2)47*i = (-1)47*i = (-1)*i = -i. As with the previous problem, we have an odd exponent, so we factor out one of the i's and attempt to reduce what remains. (Notice that we are only reducing using even powers.) The exponent 94 isn't divisible by 4, so we will use the original definition to reduce this term to a power of i2. (-1) to any odd power is still -1. Notice that powers of i go through a cycle, i, -1, -i, 1, i, -1, -i, 1, ... (each number in this list is i, i2, i3, i4, i5, ...) D. Find the value of (3i)*(4i). Just as though we were multiplying 3x*4x, we will multiply the coefficients and the 'variables'. So 3i*4i = (3*4)*(i*i) = 12i2 = 12*(-1) = -12. That last step used the definition i2 = -1. Problem Solving Tips
Additional Problems with Answers 1. Rewrite the following as imaginary numbers. a.
2. Add the following imaginary numbers.
a. i + 3i
b. 2i - 4i
c.
3. Multiply the following imaginary numbers.
a. i*4i
b. (-i)(-6i) c.
i2*i5
d.
4. Simplify the following imaginary numbers. a. (i + 2i)(3i) b. (-i - 7i)(i2) c. (4i)(i2 - 3i) 5. Reduce the following imaginary numbers as much as possible, a. i6 b. i13 c. i23 d. i56 e. i81 f. i107 g. i4025 Answers 1. a. 3i,
b. 8i, c.
Links to Outside Sources Math
Forum: Ask Dr. Math FAQ: Imaginary Numbers Links to Supporting Topics Complex Numbers in Cartesian Form (a
+ bi) History
Question Corner -- The Origin of Complex Numbers |
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