Math 20B. Lecture Examples.

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1 Math 20B. Lecture Eamples. (7/8/08) Comple eponential functions A comple number is an epression of the form z = a + ib, where a an b are real numbers an i is the smbol that is introuce to serve as a square root of. The real part of z = a + ib is the real number a, an the imaginar part of z = a + ib is the real number b. Real numbers are consiere to be comple numbers with zero imaginar parts. a + ib is the rectangular representation of the comple number. Comple numbers are ae an multiplie with the proceures use for real numbers with the aitional rule that i 2 =. For eample, (2 + 3i) + (4 5i) = (2 + 4) + (3 5)i = 6 2i (2 + 3i)(4 5i) = 2(4) + [2( 5) + 3(4)]i + 3( 5)i 2 = 8 + 2i 5i 2 = 8 + 2i 5( ) = i The comple conjugate of a comple number z = a + ib is z = a ib. The magnitue or absolute value of the comple number z = a + ib is z = a 2 + b 2. Notice that zz = (a + ib)(a ib) = a 2 i 2 b 2 = a 2 + b 2 = z 2. A comple number z = a+ib can plotte in an -plane at the point with coorinates = a, = b (Figure ). Then the comple conjugate z = a ib is the mirror image of z = a + ib about the -ais, an the magnitue of z = a + ib is its istance z = a 2 + b 2 to the origin. b z z = a + ib a FIGURE b z = a ib To fin the rectangular representation of the quotient z /z 2 of two comple numbers where z 2 0, ou can multipl the numerator an enominator of the ratio b the conjugate z 2 of the enominator to make the enominator a real number. For eample, + 2i ( + 2i)(3 4i) (3) + [2(3) (4)]i 2(4)i2 = = 3 + 4i (3 + 4i)(3 4i) i 2 = 3 + 2i i = = i. Lecture notes to accompan a Math 20B Supplement

2 Math 20B. Lecture Eamples. Comple Eponential Functions (7/8/08) p. 2 Eample Calculate (a) z + z 2, (b) z z 2, an (c) z /z 2 for z = i an z 2 = 3 + i. Answer: (a) z + z 2 = 4 (b) z z 2 = 4 2i (c) z z 2 = i Comple eponentials As ou will see later in Math 20B, the functions = e, = cos, an = sin can be given for all real b the infinite series, e = cos = sin = If we set = ib in the first formula, we obtain e ib = n! n = ! 3 + 4! 4 + 5! 5 + (2n)! 2n = ! 5 (2n + )! 2n+ = 3! 3 + 5! 5. n! (ib)n = + ib + 2 (ib)2 + 3! (ib)3 + 4! (ib)4 + 5! (ib)5 + = + ib 2 b2 3! b3 i + 4! b4 + 5! b5 i + = ( 2 b2 + 4! b5 ) + i(b 3! b3 + 5! b5 ) which gives e ib = cos b + i sin b. () This leas us to the efinition of the comple eponential function, e a+ib = e a e ib = e a (cosb + i sin b). (2) Comple eponential functions satisf the rules e z e z 2 = e z +z 2 (3) ( e z ) z2 = e z z 2 (4) satisfie b real eponential functions.

3 p. 3 Math 20B. Lecture Eamples. Comple Eponential Functions (7/8/08) Polar forms of comple numbers If we set b = in (), we obtain e i = cos + i sin. (5) This shows that e i is the point with polar coorinates r = an = (Figure 2). sin e i = cos + i sin FIGURE 2 cos Moreover, if the comple number z = a + ib has polar coorinates [r, ], then its -coorinates are a = r cos, b = r sin an it can be written (Figure 3) a + ib = r cos + ir sin = r(cos + i sin ) = re i. This is the polar form of the comple number. Recall that r = z is the magnitue of the number. The angle is calle its argument. b = r sin r = z z = a + ib = re i FIGURE 3 a = r cos Eample 2 Plot the point z = 2 2i an fin a polar form for it. Answer: Figure A2 One answer: r = 2 2 = 5 4 π 2 r Figure A2 2 2i 2

4 Math 20B. Lecture Eamples. Comple Eponential Functions (7/8/08) p. 4 Eample 3 Give the rectangular representation of the comple number e 3+4i. Answer: e 3+4i = e 3 cos(4) + ie 3 sin(4) With the polar representations, we can write for z = r e i an z 2 = r 2 e i 2 ( )( ) z z 2 = r e i r 2 e i 2 = r r 2 e i( + 2 ) (6) ( ) z = r e i z 2 r 2 e i = r e i( 2 ). (7) 2 r 2 The first equation shows that to multipl two comple numbers, ou multipl their mouli an a their arguments. The secon shows that to ivie one comple number b another, ou ivie the magnitue of the first b the magnitue of the secon an subtract the argument of the secon from the argument of the first. Eample 4 (a) Calculate z 2 for z = i. (b) Fin the magnitues an arguments of z an z 2 an eplain the results. Answer: (a) z 2 = 36i (b) z = 6 z 2 = 36 [Argument of z] = 4 π [Argument of z2 ] = 2 π The magnitue of z 2 is the square of the magnitue of z. The argument of z 2 is 2 times the argument of z. Trigonometric ientities We can replace b in formula (5) to obtain e i = cos( ) + i sin( ) = cos i sin. Then we have the two equations, { e i = cos + i sin e i = cos i sin which can be solve for cos an sin. (Do it.) The result is formulas, cos = ei + e i 2 sin = ei e i 2i (6) (7) that epress cos an sin in terms of comple eponential function. These can be use to erive properties of the trigonometric functions from properties of the eponential functions. Eample 5 Use comple eponential functions to erive the trigonometric ientit, Answer: Omitte. The answer is the solution. sin 2 = 2 [ cos(2)]. Integrals The ifferentiation an integration formulas for comple eponential functions are the same as for real eponential functions: ( e (a+ib)) = (a + ib)e (a+ib) (8) an for nonzero a + ib, e (a+ib) = a + ib e(a+ib) + C (9)

5 p. 5 Math 20B. Lecture Eamples. Comple Eponential Functions (7/8/08) Eample 6 Use comple eponential functions to fin the erivative, (sin ). Eample 7 Answer: (sin ) = cos Use comple eponential functions to fin the antierivative, sin(3). Answer: sin(3) = 3 cos(3) + C Eample 8 Use comple eponential functions to fin the perform the integration, e cos. Answer: e cos = 2 e (cos + sin ) + C