Complex number ratio Suppose that we want to find the ratio (quotient) of two complex numbers

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(a + jb) / (c + jd ) Multiplying both the numerator and the denominator by (c jd ), we obtain [(a + jb)(c jd )] / (c + jd )(c jd ) which multiplies out to (ac jad + jbc j 2bd ) / (c 2 jcd + jcd j 2d 2) This expression can be simplified to [(ac + bd ) + j (bc ad )] / (c 2 + d 2) When we separate out the real and imaginary parts, we get [(ac + bd ) / (c 2 + d 2)] + j [(bc ad ) / (c2 + d 2)]

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How Complex Numbers Behave

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The square brackets, while technically superfluous, are included to visually set apart the real and imaginary parts of the result We have just derived a general complex-number ratio formula that we can always use: (a + jb)/(c + jd ) = [(ac + bd )/(c + d 2)] + j [(bc ad )/(c 2 + d 2)]

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For this formula to work, the denominator must not be equal to 0 + j0 That means we cannot have both c = 0 and d = 0 If both of these coefficients are 0, then we end up dividing by 0 That operation, unlike the square root of a negative real, remains undefined, at least as far as this book is concerned!

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Complex number raised to positive-integer power If a + jb is a complex number and n is a positive integer, then (a + jb)n is the result of multiplying (a + jb) by itself n times Complex conjugates Suppose we encounter two complex numbers that have the same coefficients, but opposite signs between the real and imaginary parts, as in

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a + jb and a jb We call any two such quantities complex conjugates They have some interesting properties When we add a complex number to its conjugate, we get twice the real coefficient In general, we have (a + jb) + (a jb) = 2a When we multiply a complex number by its conjugate, we get the sum of the squares of the coefficients In general, we have (a + jb)(a jb) = a2 + b2 Complex conjugates are often encountered in engineering They re especially useful in alternatingcurrent (AC) circuit, radio-frequency (RF) antenna, and transmission-line theories

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Sum example Let s find the sum of the two complex numbers 5 + j4 and 2 j3 When we add the real parts, we get

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5+2=7

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Complex Numbers and Vectors

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When we add the imaginary parts, we get j4 + ( j3) = j1 = j The sum can be expressed directly as (5 + j4) + (2 j3) = 7 + j The parentheses are not technically necessary, but they help to set the individual complexnumber addends apart on the left-hand side of the equation

Difference example To find the difference between 5 + j4 and 2 j3, we first multiply the second complex quantity by 1 That gives us

1 (2 j3) = 2 + j3 Now we can simply add 5 + j4 and 2 + j3 Adding the real parts, we obtain 5 + ( 2) = 3 Adding the imaginary parts gives us j4 + j3 = j 7 The difference can be expressed directly as (5 + j4) (2 j3) = 3 + j 7

Product example Let s multiply the complex numbers 5 + j4 and 2 j3 by each other When we treat them as binomials, the problem works out in a straightforward fashion, but we have to be careful with the signs We get

(5 + j4)(2 j3) = 5 2 + 5 ( j3) + j4 2 + j4 ( j3) = 10 + ( j15) + j8 + j ( j ) 4 3 = 10 + ( j 7) + 12 = 22 j 7 The product can be expressed directly as (5 + j4)(2 j3) = 22 j 7