qr code generator vb.net Simplify each of the following. a j 20 b 10j 47 c 1 j in Visual Studio .NET

Generator Code 128 Code Set B in Visual Studio .NET Simplify each of the following. a j 20 b 10j 47 c 1 j

Simplify each of the following. a j 20 b 10j 47 c 1 j
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answer.
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CHAPTER 6 Algebra of Complex Numbers
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Solutions (a) The four possible values of the powers of j repeat themselves over and over, endlessly, in the order j; 1; j; 1; since 4 goes into 20 exactly ve times, it follows that j 20 j 4 1; answer:
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(b) Since 4 goes into 47 11 times with 3 left over, we have that 10 j 47 10 j 3 10 j 10j; answer:
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(c) Multiply the numerator and denominator by j; thus j j j j; jj j 2 1 Thus we have the very useful fact that 1 j j answer:
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The fact that one over j is equal to minus j is used so often that it should be committed to memory. Problem 75 Express each of the following in terms of the imaginary unit j. a p 144 b q 100x4 y 10 c x2 4y2 z4 !1=2
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Problem 76 Simplify each of the following as much as possible (the dot means times ). a j j2 d e j 342 1 j3 f g 1 j 34 j 17
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b j 5 j 8 c j 31
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Problem 77 Solve the following for the unknown values of x. a x2 16 0 c 6x2 94:8 0
b x2 900 0
CHAPTER 6 Algebra of Complex Numbers
Complex Numbers. Addition and Multiplication
a jb
We begin with the de nition that a complex number is the algebraic SUM of a REAL number and an IMAGINARY number. Thus complex numbers have the general form where a is the real part and jb is the imaginary part of the complex number, with a and b both being real numbers. In working with complex numbers it s customary to write the real part rst; thus we generally write a jb, instead of jb a. The rst rule, in the algebra of complex numbers, concerns the SUM of two or more such numbers, and can be stated in the following way. REAL PART OF SUM SUM OF THE REAL PARTS of the numbers IMAGINARY PART OF SUM SUM OF THE IMAGINARY PARTS of the numbers Example
Find the sum of the complex numbers (6 j3), (10 j7), and ( 4 j9).
Solution 6 10 4 j 3 7 9 12 j5; answer:
Next, the PRODUCT of two complex numbers is found in exactly the same way as in the ordinary algebra of real numbers, except we must remember that j 2 1. Example 1
Find the product of the two complex numbers (5 j3) and (2 j5).
Solution Using the ordinary four-step rule for multiplying two binomials, we have
v j5 5 j3 2 v 10 j25 j6 15
10 15 j 25 6 25 j19; answer:
Find the value of j 6 j 3 j4 .
Example 2
Solution Here we have to multiply three quantities together, the three quantities being 1. the unit imaginary number j 2. the complex number (6 j) 3. the complex number (3 j4)
CHAPTER 6 Algebra of Complex Numbers
The procedure in such a case is to FIRST nd the product of ANY TWO of the three factors, then multiply that result by the remaining factor. In the above case let s rst multiply factors (1) and (2) together, giving j 6 j 1 j6 , which we must now multiply by factor (3), thus getting 1 j6 3 j4 3 j4 j18 24 21 j22; Problem 78 6 j5 8 j4 4 j3 Problem 79 j5 7 j 3 1 j10 j 2 4 3 10j 100 Problem 80 2 j3 6 j Problem 81 1 j 1 j2 3 j5 Problem 82 Given that a; b; c, and d are real numbers, a jb c jd Problem 83 6 j12 2 Problem 84 1 j 5 answer:
Conjugates and Division of Complex Numbers
Two complex numbers that di er ONLY IN THE SIGNS OF THEIR IMAGINARY PARTS are called conjugate complex numbers. Thus (a jb) and (a jb) are conjugate complex numbers, each being the conjugate of the other. The important fact that we are concerned with here is that the PRODUCT of two conjugate complex numbers is always a POSITIVE REAL NUMBER; thus a jb a jb a2 b2 140
which you should verify by direct multiplication. The relationship of eq. (140) is especially useful in nding the quotient of two complex numbers; this can be illustrated with the aid of eq. (141) below: c jd A jB a jb in which all the letters (except j) represent positive or negative real numbers. 141
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