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Figure 6-3 Some points in the Cartesian complex-number
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We have learned that ( 1)1/2 = j You might now ask, What about the square root of a negative real number other than 1, such as 4 or 100 The positive square root of any negative real number is equal to j times the positive square root of the absolute value of that real number For example, ( 4)1/2 = j 41/2 = j2 and ( 100)1/2 = j 1001/2 = j10 We can also have negative square roots of negative reals That s because j is not the same quantity as j (You ll get a chance to prove this fact in Problem 1 at the end of this chapter) Negating the above examples, we get ( 4)1/2 = j 41/2 = j2 and ( 100)1/2 = j 1001/2 = j10
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Here s a challenge!
Demonstrate what happens when j is raised to successively higher positive-integer powers
Solution
Keep in mind that j is the negative square root of 1, which is ( 1)1/2 By definition, we know that j 2 = 1, so we can calculate the square of j as ( j )2 = ( 1 j )2 = ( 1)2 j 2 = 1 j2 = 1 ( 1) = 1 Now for the cube: ( j )3 = ( j )2 ( j ) = 1 ( j ) =j
How Complex Numbers Behave
The fourth power: ( j )4 = ( j )3 ( j ) = j ( j ) = j 2 = ( 1) =1 The fifth power: ( j )5 = ( j )4 ( j ) = 1 ( j ) = j The sixth power: ( j )6 = ( j )5 ( j ) = j ( j ) = ( j )2 = 1 Can you see what will happen if we keep going like this, increasing the integer power by 1 over and over We ll cycle endlessly through j, 1, j, and 1 If you grind things out, you ll see that j 7 = j, j 8 = 1, j 9 = j, j 10 = 1, and so on In general, if n is a positive integer, then ( j )n = ( j )n + 4
How Complex Numbers Behave
Complex numbers have properties that resemble those of the real numbers to some extent But there are some major differences as well Let s review the basic operations involving complex numbers, in case you ve forgotten them As we go along, we ll imagine two arbitrary complex numbers a + jb and c + jd where a, b, c, and d are real numbers, and j = ( 1)1/2
Complex Numbers and Vectors
Complex number sum When we want to find the sum of two complex numbers, we add the real and imaginary parts independently to get the real and imaginary components of the result The general formula is
(a + jb) + (c + jd ) = (a + c) + j(b + d )
Complex number difference We can find the difference between two complex numbers if we multiply the second complex number by 1, and then add it to the first complex number The general formula is
(a + jb) (c + jd ) = (a + jb) + [ 1(c + jd )] = (a c) + j(b d )
Complex number product When we want to multiply two complex numbers by each other, we can treat them individually as binomials We multiply the binomials and then simplify their product, remembering that j 2 = 1 The general formula works out as
(a + jb)(c + jd ) = ac + jad + jbc + j 2bd = (ac bd ) + j(ad + bc)
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