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Note that in the expression j 2ab, the numeral 2 is not an exponent! Now let s find the square of the complex number a jb Paying careful attention to the signs, we get (a jb)2 = (a jb)(a jb) = a2 + a( jb) + ( jb)a + ( jb)2 = a2 jab jba + ( j )2b2 = a2 j 2ab b2 = (a2 b2) j 2ab The two final products we ve derived are (a2 b2) + j 2ab and (a2 b2) j 2ab which, by definition, are complex conjugates 6 First, let s find the product of the polar complex vectors (p /4,21/2) and (3p /4,21/2) We must add the direction angles and multiply the magnitudes The sum of the angles is p /4 + 3p /4 = p The product of the magnitudes is 21/2 21/2 = 2 Therefore, the product vector is (p,2) The angle q is equal to p, and the magnitude r is equal to 2 To convert this polar vector (q,r) = (p,2) to the complex-number form a + jb where a and b are real-number coefficients, we use the formula for that purpose, getting a + jb = r cos q + j(r sin q) = 2 cos p + j(2 sin p) = 2 ( 1) + j 2 0 = 2 + j0 = 2 The product of the two original polar complex vectors (p /4,21/2) and (3p /4,21/2) is a vector representing the pure real number 2 7 Let s convert the polar vector (q,r) = (p /4,21/2) to a complex number in the traditional real-plus-imaginary form The conversion formula tells us that r cos q + j(r sin q) = 21/2 cos (p /4) + j[21/2 sin (p /4)] = 21/2 21/2/2 + j(21/2 21/2/2) = 1 + j Repeating the process with the polar vector (q,r) = (3p /4,21/2), we get r cos q + j(r sin q) = 21/2 cos (3p /4) + j[21/2 sin (3p /4)] = 21/2 ( 21/2/2) + j(21/2 21/2/2) = 1 + j
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When we multiply these two complex numbers as binomials, we get (1 + j )( 1 + j ) = 1 ( 1) + 1 j + j ( 1) + j j = 1 + j + ( j ) + ( 1) = 2 This agrees with the solution to Problem 6 It should! We ve been multiplying the same two vectors, representing the same two complex numbers, all along If we hadn t gotten identical results using the polar method and the Cartesian method, we d have made a mistake somewhere 8 Let s convert the polar vector (q,r) = (2p /3,1) to the real-plus-imaginary complexnumber form The conversion formula tells us that r cos q + j(r sin q) = cos (2p /3) + j sin (2p /3)] = 1/2 + j(31/2/2) If you don t remember why sin (2p /3) = 31/2/2, you might want to verify it for extra credit (Here s a hint: Use the Pythagorean theorem to solve for the height of a right triangle whose base is 1/2 unit wide and whose hypotenuse is 1 unit long) Now let s repeat the process with the polar vector (q,r) = (4p /3,1) The conversion formula gives us r cos q + j(r sin q) = cos (4p /3) + j sin (4p /3)] = 1/2 + j( 31/2/2) = 1/2 j(31/2/2) 9 Figure A-6 is a graph of the three cube roots of 1 as polar complex vectors Each radial division represents 1/5 unit
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