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Ratio example When we find the ratio of a complex number to another complex number, we should expect some messy arithmetic Let s calculate
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(5 + j4) / (2 j3) Keeping track of the coefficients can be confusing when we use the formula for a ratio Here s the general formula again: (a + jb) / (c + jd ) = [(ac + bd ) / (c2 + d 2)] + j [(bc ad ) / (c2 + d 2)] The denominator in both addends is c2 + d 2 Here, c = 2 and d = 3, so we have c 2 + d 2 = 22 + ( 3)2 = 4 + 9 = 13 We can substitute 13 for the quantity c 2 + d 2 in our formula, giving us the expression [(ac + bd ) / 13] + j [(bc ad ) / 13] Knowing that a = 5, b = 4, c = 2, and d = 3, the above equation becomes [5 2 + 4 ( 3)] / 13 + j [4 2 5 ( 3)] / 13 which works out to 2/13 + j(23/13) Our ratio can be expressed directly as (5 + j4) / (2 j3) = 2/13 + j(23/13)
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Power example Let s find the cube of the complex number 2 j3 We square it first, multiplying by itself to get
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(2 j3)(2 j3) = 2 2 + 2 ( j3) + ( j3) 2 + ( j3) ( j3) = 4 + ( j6) + ( j6) + ( j ) ( j ) 3 3 = 4 + ( j12) + ( 9) = 5 j12 We multiply this result by the original quantity 2 j3, obtaining ( 5 j12)(2 j3) = 5 2 + ( 5) ( j3) + ( j12) 2 + ( j12) ( j3) = 10 + j15 + ( j24) + ( j ) ( j ) 12 3 = 10 + ( j 9) + ( 36) = 46 j 9
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The cube can be expressed directly as (2 j3)3 = 46 j 9
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When working with complex numbers, you should pay close attention to whether or not a numeral after the j operator is a superscript The two notations are perilously similar! For example, if you see 5 + j2 it means 5 plus twice j, which is a complex number that s neither pure real nor pure imaginary But if you see 5 + j2 it means 5 plus j squared, which can be simplified to 5 + ( 1) or 4, which is pure real
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Prove that the square of a complex number is equal to the square of the negative of that complex number That is, show that (a + jb)2 = ( a jb)2 for all real-number coefficients a and b
Solution
First, let s work out the square of a + jb We get (a + jb)2 = (a + jb)(a + jb) = a2 + jab + jba + j 2b2 = a2 + j 2ab b2 = a2 b2 + j 2ab Note that in the final term j2ab, the numeral 2 is a multiplier, not an exponent! Now let s find the square of a jb This is a nightmare of negatives, so we must be careful with the signs We have ( a jb)2 = ( a jb)( a jb) = ( a)2 + ( a)( jb) + ( jb)( a) + ( jb)2
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= a2 + jab + jba + ( j )2b2 = a2 + j2ab b2 = (a2 b2) + j2ab That s exactly what we got when we squared a + jb Therefore, we ve shown that (a + jb)2 = ( a jb)2