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

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Unformatted text preview: 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 u u 1 and 1 i u u i will be used quite frequently. A formal derivation of i 2 u u 1 from the multiplication formula is shown in the text. 1/ i u u i / i u u i U u u i /1 u u 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 u 5 u 2 U 3 i U u 5 U 2 U 5 U 3 i u 10 U 15 i . Similarly, 3 z 2 u 3 u 4 u 5 i U u 12 u 15 i . Now you sum the results (the first two pairs of parentheses will not be needed; why?) 5 z 1 U 3 z 2 u u 10 U 15 i U U u 12 u 15 i U u 10 U 12 U u 15 u 15 U i u 22 and square, 22 2 u 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 u u 2 U 3 i U 2 u 4 U 2 U 2 U 3 i u 9 u u 5 U 12 i . Then do the division, using the recipe (7), 1 z 1 2 u u 5 u 12 i u u 5 U 12 i Uu u 5 u 12 i U u u 5 u 12 i 25 U 144 u u 5 u 12 i 169 . Now take the real part u 5/169. 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 u and square the result. In both cases, taking the real part is the last step. In the first way, calculate 1 z u 2 U z 2 z 2 z u 2 U x 2 ¡ 2 ixy u y 2 u zz u U 2 U x 2 u y 2 ¡ 2 ixy u x 2 ¡ y 2 U 2 and then take the real part to get x 2 u y 2 in the numerator and the denominator as before. In the second way, you calculate 1 z u U z zz u U x ¡ iy x 2 ¡ y 2 and then square the expression on the right, obtaining x 2 ¡ 2 ixy u y 2 u x 2 ¡ y 2 U 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 u . 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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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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

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