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p /2
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(2p /3, 4/5)
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Product = (5p /6, 2/5)
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(p /6, 1/2)
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Add the angles
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and multiply the magnitudes
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Figure 6-8 Product of the polar complex vectors (p /6,1/2) and
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(2p /3,4/5) Each radial division represents 01 unit
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Polar complex vector product example Figure 6-8 shows the polar complex vectors (p /6,1/2) and (2p /3,4/5), along with their product Each radial division is 01 unit When we add the angles, we get
p /6 + 2p /3 = 5p /6 When we multiply the magnitudes, we get 1/2 4/5 = 2/5 so the product vector is (5p /6,2/5)
Polar complex vector ratio example Figure 6-9 shows the ratio of the polar complex vectors (7p /4,8) and (p,2) Each radial division is 1 unit When we subtract the angles, we get
7p /4 p = 3p /4
Complex Vectors 107
p /2
Ratio = (3p /4, 4)
(p, 2)
(7p /4, 8)
Subtract the angles
3p /2
and divide the magnitudes
Figure 6-9 Ratio of the polar complex vector (7p /4,8) to the polar
complex vector (p,2) Each radial division represents 1 unit
When we divide the magnitudes, we get 8/2 = 4 so the ratio vector is (3p /4,4)
De Moivre s theorem The above schemes for finding products, ratios, and powers of polar complex numbers can be summarized in a famous theorem attributed to the French mathematician Abraham De Moivre, (pronounced De Mwahvr ), who lived during the late 1600s and early 1700s This theorem can be found in two different versions, depending on which text you consult The first, and more general, version of De Moivre s theorem involves products and ratios Suppose we have two polar complex numbers c1 and c2, where
c1 = r1 cos q1 + j(r1 sin q1) and c2 = r2 cos q2 + j(r2 sin q2)
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where r1 and r2 are real-number polar magnitudes, and q1 and q2 are real-number polar direction angles in radians Then the product of c1 and c2 is c1c2 = r1r2 cos (q1 + q2) + j [r1r2 sin (q1 + q2)] If r2 is nonzero, the ratio of c1 to c2 is c1/c2 = (r1/r2) cos (q1 q2) + j [(r1/r2) sin (q1 q2)] The second, and more commonly known, version of De Moivre s theorem can be derived from the first version Suppose that we have a complex number c such that c = r cos q + j(r sin q) where r is the real-number polar magnitude and q is the real-number polar direction angle Also suppose that n is an integer Then c to the nth power is cn = rn cos (nq) + j[r n sin (nq)] I recommend that you enter this version of De Moivre s theorem into your brain storage, and save it there forever!
Are you confused
Do you wonder why we haven t described how to find a root of a complex vector You might think, It ought to be simple, just like finding a power backward Can t we divide the polar angle by the index of the root, and then take the root of the magnitude That s a good question Doing that will indeed give us a root But there are often two or more complex roots for any given complex number We re about to see an example of this
Here s a challenge!
Cube the polar complex vectors (2p /3,1) and (4p /3,1) Here s a warning: The solution might come as a surprise! What do you suppose these results imply
Solution
To cube a polar complex vector, we multiply the direction angle by 3 (the value of the exponent) and cube the magnitude Let s do this with the vectors we ve been given here In the case of (2p /3,1)3, we get an angle of (2p /3) 3 = 2p That s outside the allowed range of angles, but if we subtract 2p, we get 0, which is okay We get a magnitude of 13 = 1 Now we know that (2p /3,1)3 = (0,1)
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