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# SSM-Ch13 - Chap 13 Sec 13.1 Complex Numbers and Functions...

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Chap . 13 Complex Numbers and Functions Sec . 13 . 1 Complex Numbers . Complex Plane Make sure that you really understand what is done in Eq. (7). Make also sure that you understand (8), the formulas for going from complex back to real. Problem Set . 13 . 1 . Page 606 1 . Powers of i . i 2 u003d u2212 1 and 1 i u003d u2212 i will be used quite frequently. A formal derivation of i 2 u003d u2212 1 from the multiplication formula is shown in the text. 1/ i u003d u2212 i / i ue0a2 u2212 i ue0a3 u003d u2212 i /1 u003d u2212 i follows from (7). 7 . Complex arithmetic . Here you first multiply complex numbers by real numbers, then add complex numbers, and finally square the result; note that the result of such operations on complex numbers can very well be a real number. 5 z 1 is obtained by multiplying the real part 2 by 5 and the imaginary part 3 by 5, 5 z 1 u003d 5 ue0a2 2 u002b 3 i ue0a3 u003d 5 u2219 2 u002b 5 u2219 3 i u003d 10 u002b 15 i . Similarly, 3 z 2 u003d 3 ue0a2 4 u2212 5 i ue0a3 u003d 12 u2212 15 i . Now you sum the results (the first two pairs of parentheses will not be needed; why?) 5 z 1 u002b 3 z 2 u003d ue0a2 10 u002b 15 i ue0a3 u002b ue0a2 12 u2212 15 i ue0a3 u003d 10 u002b 12 u002b ue0a2 15 u2212 15 ue0a3 i u003d 22 and square, 22 2 u003d 484. 9 . Division , real part . Do the division first and then take the real part. In the division you have a choice. You can divide first and then take the square of the result, or you square first and then divide (take the reciprocal). Try both ways, and you will see that the second way is simpler because 1/ z 1 gives fractions, which you then would have to square. So do z 1 2 u003d ue0a2 2 u002b 3 i ue0a3 2 u003d 4 u002b 2 u2219 2 u2219 3 i u2212 9 u003d u2212 5 u002b 12 i . Then do the division, using the recipe (7), 1 z 1 2 u003d u2212 5 u2212 12 i ue0a2 u2212 5 u002b 12 i ue0a3ue0a2 u2212 5 u2212 12 i ue0a3 u003d u2212 5 u2212 12 i 25 u002b 144 u003d u2212 5 u2212 12 i 169 . Now take the real part u2212 5/169.

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Chap. 13 Complex Numbers and Functions 129 19 . Division , conjugate . You can do the division for the given expression. Or you can calculate 1/ z u0304 and square the result. In both cases, taking the real part is the last step. In the first way, calculate 1 z u0304 2 u003d z 2 z 2 z u0304 2 u003d x 2 u002b 2 ixy u2212 y 2 ue0a2 zz u0304 ue0a3 2 u003d x 2 u2212 y 2 u002b 2 ixy ue0a2 x 2 u002b y 2 ue0a3 2 and then take the real part to get x 2 u2212 y 2 in the numerator and the denominator as before. In the second way, you calculate 1 z u0304 u003d z zz u0304 u003d x u002b iy x 2 u002b y 2 and then square the expression on the right, obtaining x 2 u002b 2 ixy u2212 y 2 ue0a2 x 2 u002b y 2 ue0a3 2 , of which you now take the real part as before. Sec . 13 . 2 Polar Form of Complex Numbers . Powers and Roots The role of polar coordinates is more important in complex analysis than in calculus. Notably, they help achieve a deeper understanding of multiplication and division. They also help in handling absolute values. Now the polar angle of a complex number is determined only up to integer multiples of 2 u03c0 . Often this is not essential, but where it matters, it can be handled by using the concept of the principal value (5).
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SSM-Ch13 - Chap 13 Sec 13.1 Complex Numbers and Functions...

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