COMPLEX NUMBERS. a bi c di a c b d i. a bi c di a c b d i For instance, 1 i 4 7i i 5 6i

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(1)COMPLEX NUMBERS Im 2+3i _4+2i i 0 _i _2-2i. Re. 1. 3-2i. FIGURE 1. Complex numbers as points in the Argand plane. A complex number can be represented by an expression of the form a bi, where a and b are real numbers and i is a symbol with the property that i 2 1. The complex number a bi can also be represented by the ordered pair a, b and plotted as a point in a plane (called the Argand plane) as in Figure 1. Thus, the complex number i 0 1 i is identified with the point 0, 1. The real part of the complex number a bi is the real number a and the imaginary part is the real number b. Thus, the real part of 4 3i is 4 and the imaginary part is 3. Two complex numbers a bi and c di are equal if a c and b d, that is, their real parts are equal and their imaginary parts are equal. In the Argand plane the horizontal axis is called the real axis and the vertical axis is called the imaginary axis. The sum and difference of two complex numbers are defined by adding or subtracting their real parts and their imaginary parts: a bi c di a c b di a bi c di a c b di For instance, 1 i 4 7i 1 4 1 7i 5 6i The product of complex numbers is defined so that the usual commutative and distributive laws hold: a bic di ac di bic di ac adi bci bdi 2 Since i 2 1, this becomes a bic di ac bd ad bci EXAMPLE 1. 1 3i2 5i 12 5i 3i2 5i 2 5i 6i 151 13 11i Division of complex numbers is much like rationalizing the denominator of a rational expression. For the complex number z a bi, we define its complex conjugate to be z a bi. To find the quotient of two complex numbers we multiply numerator and denominator by the complex conjugate of the denominator. EXAMPLE 2 Express the number. 1 3i in the form a bi. 2 5i. SOLUTION We multiply numerator and denominator by the complex conjugate of 2 5i,. namely 2 5i, and we take advantage of the result of Example 1:. Im. 1 3i 1 3i 2 5i 13 11i 13 11 2 i 2 2 5i 2 5i 2 5i 2 5 29 29 z=a+bi. i 0. Re. Thomson Brooks-Cole copyright 2007. _i. The geometric interpretation of the complex conjugate is shown in Figure 2: z is the reflection of z in the real axis. We list some of the properties of the complex conjugate in the following box. The proofs follow from the definition and are requested in Exercise 18. Properties of Conjugates. z=a-bi –. zwzw. zw z w. zn zn. FIGURE 2 1.

(2) 2 ■ COMPLEX NUMBERS. . The modulus, or absolute value, z of a complex number z a bi is its distance from the origin. From Figure 3 we see that if z a bi, then. Im. @ z=a+bi „b„ „ + „@ œ „a. bi. = |z|. z sa. b. 0. Re. a. 2. b2. Notice that zz a bia bi a 2 abi abi b 2i 2 a 2 b 2. FIGURE 3. . zz z. and so. 2. This explains why the division procedure in Example 2 works in general: z w. . zw ww. . zw. w. 2. Since i 2 1, we can think of i as a square root of 1. But notice that we also have i2 i 2 1 and so i is also a square root of 1. We say that i is the principal square root of 1 and write s1 i. In general, if c is any positive number, we write sc sc i With this convention, the usual derivation and formula for the roots of the quadratic equation ax 2 bx c 0 are valid even when b 2 4ac 0: x. b sb 2 4ac 2a. EXAMPLE 3 Find the roots of the equation x 2 x 1 0. SOLUTION Using the quadratic formula, we have. x. 1 s1 2 4 1 1 s3 1 s3 i 2 2 2. We observe that the solutions of the equation in Example 3 are complex conjugates of each other. In general, the solutions of any quadratic equation ax 2 bx c 0 with real coefficients a, b, and c are always complex conjugates. (If z is real, z z, so z is its own conjugate.) We have seen that if we allow complex numbers as solutions, then every quadratic equation has a solution. More generally, it is true that every polynomial equation a n x n a n1 x n1 a 1 x a 0 0 of degree at least one has a solution among the complex numbers. This fact is known as the Fundamental Theorem of Algebra and was proved by Gauss. POLAR FORM. We know that any complex number z a bi can be considered as a point a, b and that any such point can be represented by polar coordinates r, with r 0. In fact,. Im. a+bi r. a r cos . Thomson Brooks-Cole copyright 2007. b. b r sin . ¨ 0. FIGURE 4. a. Re. as in Figure 4. Therefore, we have z a bi r cos r sin i.

(3) COMPLEX NUMBERS ■ 3. Thus, we can write any complex number z in the form z rcos i sin . . r z sa 2 b 2. where. tan . and. b a. The angle is called the argument of z and we write argz. Note that argz is not unique; any two arguments of z differ by an integer multiple of 2. EXAMPLE 4 Write the following numbers in polar form.. (a) z 1 i SOLUTION. (b) w s3 i. . (a) We have r z s12 12 s2 and tan 1, so we can take 4. Therefore, the polar form is Im. 1+i. z s2. œ„ 2 π 4. 0. _. π 6. . cos. i sin 4 4. . . (b) Here we have r w s3 1 2 and tan 1 s3 . Since w lies in the fourth quadrant, we take 6 and. Re. . 2. w 2 cos . œ„ 3-i. 6. i sin . 6. The numbers z and w are shown in Figure 5.. FIGURE 5. The polar form of complex numbers gives insight into multiplication and division. Let z1 r1cos 1 i sin 1 . z2 r2cos 2 i sin 2 . be two complex numbers written in polar form. Then Im. z™. z1 z2 r1r2cos 1 i sin 1 cos 2 i sin 2 . z¡. r1r2 cos 1 cos 2 sin 1 sin 2 isin 1 cos 2 cos 1 sin 2 . ¨™. Therefore, using the addition formulas for cosine and sine, we have. ¨¡ Re. ¨¡+¨™. z1z2 r1r2 cos1 2 i sin1 2 . 1. z¡z™. This formula says that to multiply two complex numbers we multiply the moduli and add the arguments. (See Figure 6.) A similar argument using the subtraction formulas for sine and cosine shows that to divide two complex numbers we divide the moduli and subtract the arguments.. FIGURE 6. Im. z. z1 r1 cos1 2 i sin1 2 z2 r2. r ¨ Thomson Brooks-Cole copyright 2007. 0. _¨ 1 r. Re. In particular, taking z1 1 and z2 z, (and therefore 1 0 and 2 ), we have the following, which is illustrated in Figure 7.. 1 z. If FIGURE 7. z2 0. z rcos i sin ,. then. 1 1 cos i sin . z r.

(4) 4 ■ COMPLEX NUMBERS. EXAMPLE 5 Find the product of the complex numbers 1 i and s3 i in polar form. SOLUTION From Example 4 we have. . i sin 4 4. 1 i s2 cos s3 i 2 cos . and. Im. z=1+i œ„ 2. zw. 2œ„2. So, by Equation 1,. π 12. 0. . 1 i(s3 i) 2s2 cos. Re. 2. 2s2 cos. w=œ„ 3-i FIGURE 8. 6. . . i sin . 4 6. 6. i sin. 4 6. . i sin 12 12. This is illustrated in Figure 8. Repeated use of Formula 1 shows how to compute powers of a complex number. If z r cos i sin then. z 2 r 2cos 2 i sin 2 . and. z 3 zz 2 r 3cos 3 i sin 3 . In general, we obtain the following result, which is named after the French mathematician Abraham De Moivre (1667–1754). 2. De Moivre’s Theorem If z r cos i sin and n is a positive integer, then. z n r cos i sin n r ncos n i sin n . This says that to take the nth power of a complex number we take the nth power of the modulus and multiply the argument by n. EXAMPLE 6 Find SOLUTION Since. 1 2. ( 12 12 i)10. 12 i 12 1 i, it follows from Example 4(a) that 12 12 i has the polar. form 1 1 s2 i 2 2 2. . cos. i sin 4 4. . So by De Moivre’s Theorem,. . . 1 1 i 2 2. 10. . Thomson Brooks-Cole copyright 2007. . 25 2 10. 10. s2 2. cos. cos. 10 10 i sin 4 4. 5 5 i sin 2 2. . . . 1 i 32. De Moivre’s Theorem can also be used to find the n th roots of complex numbers. An n th root of the complex number z is a complex number w such that wn z.

(5) COMPLEX NUMBERS ■ 5. Writing these two numbers in trigonometric form as w scos i sin . and. z r cos i sin . and using De Moivre’s Theorem, we get s ncos n i sin n r cos i sin The equality of these two complex numbers shows that sn r and. cos n cos . s r 1 n. or. sin n sin . and. From the fact that sine and cosine have period 2 it follows that n 2k. Thus. . w r 1 n cos. . 2k n. . 2k n. or. 2k n. i sin. . Since this expression gives a different value of w for k 0, 1, 2, . . . , n 1, we have the following. 3 Roots of a Complex Number Let z r cos i sin and let n be a positive integer. Then z has the n distinct n th roots. . wk r 1 n cos. 2k n. i sin. . 2k n. where k 0, 1, 2, . . . , n 1.. . Notice that each of the nth roots of z has modulus wk r 1 n. Thus, all the nth roots of z lie on the circle of radius r 1 n in the complex plane. Also, since the argument of each successive nth root exceeds the argument of the previous root by 2 n , we see that the n th roots of z are equally spaced on this circle. EXAMPLE 7 Find the six sixth roots of z 8 and graph these roots in the complex. plane. SOLUTION In trigonometric form, z 8cos i sin . Applying Equation 3 with n 6,. we get. . wk 8 1 6 cos. 2k 2k i sin 6 6. . Thomson Brooks-Cole copyright 2007. We get the six sixth roots of 8 by taking k 0, 1, 2, 3, 4, 5 in this formula:. . . w0 8 1 6 cos. i sin 6 6. w1 8 1 6 cos. i sin 2 2. w2 8 1 6 cos. 5 5 i sin 6 6. s2. . 1 s3 i 2 2. s2 i. s2 . . 1 s3 i 2 2.

(6) 6 ■ COMPLEX NUMBERS. . Im œ„2 i w¡ w™. 2 _œ„. w¸ 0. w∞ _œ„2 i. 7 7 i sin 6 6. w4 8 1 6 cos. 3 3 i sin 2 2. w5 8 1 6 cos. 11 11 i sin 6 6. œ„ 2 Re. w£ w¢. . w3 8 1 6 cos. s2 . . 1 s3 i 2 2. s2 i s2. . 1 s3 i 2 2. All these points lie on the circle of radius s2 as shown in Figure 9.. FIGURE 9. The six sixth roots of z=_8. COMPLEX EXPONENTIALS. We also need to give a meaning to the expression e z when z x iy is a complex number. The theory of infinite series as developed in Chapter 8 can be extended to the case 2 as our guide, where the terms are complex numbers. Using the Taylor series for e x (8.7.12) we define ez . 4. n0. zn z2 z3 1z n! 2! 3!. and it turns out that this complex exponential function has the same properties as the real exponential function. In particular, it is true that e z z e z e z. 5. 1. 2. 1. 2. If we put z iy, where y is a real number, in Equation 4, and use the facts that i 2 1,. i 3 i 2i i,. i 4 1, i 5 i, . . .. we get e iy 1 iy 1 iy . . 1. iy2 iy3 iy4 iy5 2! 3! 4! 5! y2 y3 y4 y5 i i 2! 3! 4! 5!. . . y2 y4 y6 y3 y5 i y 2! 4! 6! 3! 5!. cos y i sin y Here we have used the Taylor series for cos y and sin y (Equations 8.7.17 and 8.7.16). The result is a famous formula called Euler’s formula:. 6. e iy cos y i sin y. Thomson Brooks-Cole copyright 2007. Combining Euler’s formula with Equation 5, we get 7. e xiy e xe iy e x cos y i sin y.

(7) COMPLEX NUMBERS ■ 7. (a) e i. EXAMPLE 8 Evaluate: ■ ■. (b) e1i 2. SOLUTION. We could write the result of Example 8(a). as. (a) From Euler’s equation (6) we have. e i 1 0. e i cos i sin 1 i0 1. This equation relates the five most famous numbers in all of mathematics: 0, 1, e, i, and .. (b) Using Equation 7 we get. . i sin 2 2. e1i 2 e1 cos. . . 1 i 0 i1 e e. Finally, we note that Euler’s equation provides us with an easier method of proving De Moivre’s Theorem: r cos i sin n re i n r ne in r ncos n i sin n . EXERCISES A Click here for answers.. 25–28. Click here for solutions.. S. 5. 3. 2 5i 4 i. 4. 1 2i 8 3i . 5. 12 7i. 6. 2i ( 12 i ). 1 4i 7. 3 2i. 3 2i 8. 1 4i. 9.. 27. 3 4i. 2. (4 2 i) (9 2 i) 1. 1 1i. ■. ■. 11. i 3. 12. i 100. 13. s25. 14. s3 s12. ■. ■. 15–17. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. 32. z 4(s3 i ),. w 3 3i. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. 21. x 2x 5 0. 22. 2x 2x 1 0. 23. z z 2 0. 24. z z 0. 2. ■. 1 2. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. 5. ■. ■. ■. ■. ■. Find the indicated roots. Sketch the roots in the complex 38. The fifth roots of 32. ■. ■. ■. ■. 40. The cube roots of 1 i ■. ■. ■. 43. e i 3. 44. e i. 45. e 2i. 46. e i. ■. ■. ■. ■. ■. ■. ■. ■. ■. ■. Write the number in the form a bi. 42. e 2 i. ■. ■. ■. ■. ■. ■. ■. ■. 47. Use De Moivre’s Theorem with n 3 to express cos 3 and. 1 4. ■. ■. 36. 1 i 8. 5. i 2. 41. e. 20. x 1. ■. ■. 41–46. 19. 4x 9 0. 2. ■. 34. (1 s3 i ). 33. 1 i 20. ■. 4. 2. ■. Find the indicated power using De Moivre’s Theorem.. 39. The cube roots of i. Find all solutions of the equation.. 2. Thomson Brooks-Cole copyright 2007. ■. ■. (c) z z , where n is a positive integer [Hint: Write z a bi, w c di.] 2. ■. plane.. n. n. ■. ■. 37. The eighth roots of 1. ■. ■. w 1 s3 i. w 1 i. 37–40. 16. 1 2 s2 i. 18. Prove the following properties of complex numbers. (a) z w z w (b) zw z w. ■. ■. 31. z 2 s3 2i,. ■. 17. 4i. 19–24. ■. w 8i. 35. (2 s3 2i ). number.. ■. ■. 30. z 4 s3 4i,. 33–36. Find the complex conjugate and the modulus of the. 15. 12 5i. ■. 29. z s3 i,. ■. ■. 28. 8i ■. 29–32 Find polar forms for z w, z w, and 1 z by first putting z and w into polar form.. 3 4 3i. 10.. 26. 1 s3 i. 25. 3 3i. 1–14 Evaluate the expression and write your answer in the form a bi. 1. 5 6i 3 2i . Write the number in polar form with argument between 0. and 2.. ■. ■. sin 3 in terms of cos and sin ..

(8) 8 ■ COMPLEX NUMBERS. 48. Use Euler’s formula to prove the following formulas for cos x. and sin x : cos x . e ix eix 2. sin x . e ix eix 2i. 49. If ux f x itx is a complex-valued function of a real. Thomson Brooks-Cole copyright 2007. variable x and the real and imaginary parts f x and tx are differentiable functions of x, then the derivative of u is defined to be u x f x it x. Use this together with Equation 7 to prove that if Fx e rx, then F x re rx when r a bi is a complex number.. 50. (a) If u is a complex-valued function of a real variable, its. indefinite integral x ux dx is an antiderivative of u. Evaluate. ye. 1i x. dx. (b) By considering the real and imaginary parts of the integral in part (a), evaluate the real integrals. ye. x. cos x dx. and. ye. x. sin x dx. (c) Compare with the method used in Example 4 in Section 6.1..

(9) COMPLEX NUMBERS ■ 9. ANSWERS S. 1. 8 4i 1 2. (2 s2 )cos13 12 i sin13 12, 14 cos 6 i sin 6. 7.. 12 i. 11 13. . 10 13. 11. i. 37. 1, i, (1 s2 )1 i . i 13. 5i. 15. 12 5i; 13. 3. 23. 2 (s7 2)i. 21. 1 2i. 39. (s3 2) 2 i, i 1. Im. Im i. 19. 2 i. 17. 4i, 4. 35. 512 s3 512i. 33. 1024. 3. 13 18i. 5. 12 7i 9.. 31. 4 s2 cos7 12 i sin7 12,. Click here for solutions.. 1. 0. 1. 0. Re. 25. 3 s2 cos3 4 i sin3 4. [. 27. 5{cos tan. 1 4 3. 29. 4cos 2 i sin 2, cos 6 i sin 6,. Thomson Brooks-Cole copyright 2007. 1 2. _i. ( )] i sin[tan ( )]}. 1 4 3. cos 6 i sin 6. Re. 41. i. 43.. 1 2. (s3 2) i. 45. e 2. 47. cos 3 cos3 3 cos sin2, sin 3 3 cos2 sin sin3.

(10) 10 ■ COMPLEX NUMBERS. SOLUTIONS. 1. (5 − 6i) + (3 + 2i) = (5 + 3) + (−6 + 2)i = 8 + (−4)i = 8 − 4i. 3. (2 + 5i)(4 − i) = 2(4) + 2(−i) + (5i)(4) + (5i)(−i) = 8 − 2i + 20i − 5i2 = 8 + 18i − 5(−1) = 8 + 18i + 5 = 13 + 18i. 5. 12 + 7i = 12 − 7i. 1 + 4i 3 − 2i 3 − 2i + 12i − 8(−1) 11 10 11 + 10i 1 + 4i = · = = + i = 3 + 2i 3 + 2i 3 − 2i 32 + 22 13 13 13 1 1−i 1−i 1−i 1 1 1 = · = = = − i 9. 1+i 1+i 1−i 1 − (−1) 2 2 2 7.. 11. i3 = i2 · i = (−1)i = −i √ √ 13. −25 = 25 i = 5i. s √ √ 122 + (−5)2 = 144 + 25 = 169 = 13 t √ 17. −4i = 0 − 4i = 0 + 4i = 4i and |−4i| = 02 + (−4)2 = 16 = 4 t t 19. 4x2 + 9 = 0 ⇔ 4x2 = −9 ⇔ x2 = − 94 ⇔ x = ± − 94 = ± 94 i = ± 32 i.. Thomson Brooks-Cole copyright 2007. 15. 12 − 5i = 12 + 15i and |12 − 15i| =. 21. By the quadratic formula, x2 + 2x + 5 = 0 ⇔ s √ −2 ± 22 − 4(1)(5) −2 ± −16 −2 ± 4i x= = = = −1 ± 2i. 2(1) 2 2 s √ √ −1 ± 12 − 4(1)(2) −1 ± −7 1 7 2 23. By the quadratic formula, z + z + 2 = 0 ⇔ z = = =− ± i. 2(1) 2 2 2 s √ 3 25. For z = −3 + 3i, r = (−3)2 + 32 = 3 2 and tan θ = −3 = −1 ⇒ θ = 3π 4 (since z lies in the second √ 3π 3π quadrant). Therefore, −3 + 3i = 3 2 cos 4 + i sin 4 . √ 27. For z = 3 + 4i, r = 32 + 42 = 5 and tan θ = 43 ⇒ θ = tan−1 43 (since z lies in the first quadrant). Therefore, 3 + 4i = 5 cos tan−1 43 + i sin tan−1 43 . t√ √ 2 29. For z = 3 + i, r = 3 + 12 = 2 and tan θ = √13 ⇒ θ = π6 ⇒ z = 2 cos π6 + i sin π6 . √ √ For w = 1 + 3 i, r = 2 and tan θ = 3 ⇒ θ = π3 ⇒ w = 2 cos π3 + i sin π3 . Therefore, zw = 2 · 2 cos π6 + π3 + i sin π6 + π3 = 4 cos π2 + i sin π2 , z/w = 22 cos π6 − π3 + i sin π6 − π3 = cos − π6 + i sin − π6 , and 1 = 1 + 0i = 1(cos 0 + i sin 0) ⇒ 1/z = 12 cos 0 − π6 + i sin 0 − π6 = 12 cos − π6 + i sin − π6 . For 1/z, we could also use the formula that precedes Example 5 to obtain 1/z = 12 cos π6 − i sin π6 . t √ √ 2 √ = − √1 2 3 + (−2)2 = 4 and tan θ = 2−2 31. For z = 2 3 − 2i, r = 3 3 √ ⇒ θ = − π6 ⇒ z = 4 cos − π6 + i sin − π6 . For w = −1 + i, r = 2, √ 1 3π tan θ = −1 . Therefore, = −1 ⇒ θ = 3π ⇒ w = 2 cos 3π 4 4 + i sin 4 √ π √ π 3π 3π 7π zw = 4 2 cos − 6 + 4 + i sin − 6 + 4 = 4 2 cos 12 + i sin 7π , 12 π π 11π 11π 4 3π 3π 4 z/w = √2 cos − 6 − 4 + i sin − 6 − 4 = √2 cos − 12 + i sin − 12 √ 13π 13π = 2 2 cos 12 + i sin 12 , and 1/z = 14 cos − π6 − i sin − π6 = 14 cos π6 + i sin π6 . √ √ 33. For z = 1 + i, r = 2 and tan θ = 11 = 1 ⇒ θ = π4 ⇒ z = 2 cos π4 + i sin π4 . So by De Moivre’s Theorem, k√ l20 1/2 20 cos 204· π + i sin 204· π 2 cos π4 + i sin π4 = 2 (1 + i)20 = = 210 (cos 5π + i sin 5π) = 210 [−1 + i(0)] = −210 = −1024.

(11) COMPLEX NUMBERS ■ 11. t √ √ √ 2 2 2 3 + 22 = 16 = 4 and tan θ = 2√ 35. For z = 2 3 + 2i, r = = √13 ⇒ θ = π6 ⇒ 3 z = 4 cos π6 + i sin π6 . So by De Moivre’s Theorem, √ k √ 5 l √ 5 2 3 + 2i = 4 cos π6 + i sin π6 = 1024 − 23 + 12 i = −512 3 + 512i = 45 cos 5π + i sin 5π 6 6. 37. 1 = 1 + 0i = 1 (cos 0 + i sin 0). Using Equation 3 with r = 1, n = 8, and θ = 0, we have kπ kπ 0 + 2kπ 0 + 2kπ 1/8 cos wk = 1 + i sin = cos + i sin , where k = 0, 1, 2, . . . , 7. 8 8 4 4 w0 = 1(cos 0 + i sin 0) = 1, w1 = 1 cos π4 + i sin π4 = √12 + √12 i, = − √12 + √12 i, w2 = 1 cos π2 + i sin π2 = i, w3 = 1 cos 3π + i sin 3π 4 4 5π w4 = 1(cos π + i sin π) = −1, w5 = 1 cos 5π = − √12 − √12 i, 4 + i sin 4 = −i, w7 = 1 cos 7π = √12 − √12 i w6 = 1 cos 3π + i sin 3π + i sin 7π 2 2 4 4 39. i = 0 + i = 1 cos π2 + i sin π2 . Using Equation 3 with r = 1, n = 3, and θ = π π + 2kπ + 2kπ wk = 11/3 cos 2 + i sin 2 , where k = 0, 1, 2. 3 3 √ w0 = cos π6 + i sin π6 = 23 + 12 i √ 5π = − 23 + 12 i w1 = cos 5π 6 + i sin 6 9π = −i w2 = cos 9π 6 + i sin 6 41. Using Euler’s formula (6) with y =. π , 2. we have. π , 2. we have eiπ/2 = cos π2 + i sin π2 = 0 + 1i = i. √ π π 1 π 3 43. Using Euler’s formula (6) with y = , we have eiπ/3 = cos + i sin = + i. 3 3 3 2 2 45. Using Equation 7 with x = 2 and y = π, we have e2+iπ = e2 eiπ = e2 (cos π + i sin π) = e2 (−1 + 0) = −e2 .. 47. Take r = 1 and n = 3 in De Moivre’s Theorem to get. [1(cos θ + i sin θ)]3 = 13 (cos 3θ + i sin 3θ) (cos θ + i sin θ)3 = cos 3θ + i sin 3θ cos3 θ + 3 cos2 θ (i sin θ) + 3(cos θ)(i sin θ)2 + (i sin θ)3 = cos 3θ + i sin 3θ cos3 θ + 3 cos2 θ sin θ i − 3 cos θ sin2 θ − sin3 θ i = cos 3θ + i sin 3θ 3 cos θ − 3 sin2 θ cos θ + 3 sin θ cos2 θ − sin3 θ i = cos 3θ + i sin 3θ. Equating real and imaginary parts gives cos 3θ = cos3 θ − 3 sin2 θ cos θ. and sin 3θ = 3 sin θ cos2 θ − sin3 θ. 49. F (x) = erx = e(a+bi)x = eax+bxi = eax (cos bx + i sin bx) = eax cos bx + i(eax sin bx) ⇒ F 0 (x) = (eax cos bx)0 + i(eax sin bx)0. = (aeax cos bx − beax sin bx) + i(aeax sin bx + beax cos bx) = a [eax (cos bx + i sin bx)] + b [eax (− sin bx + i cos bx)] = aerx + b eax i2 sin bx + i cos bx. Thomson Brooks-Cole copyright 2007. = aerx + bi [eax (cos bx + i sin bx)] = aerx + bierx = (a + bi)erx = rerx.

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