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 _4+i _i FIGURE Complex numbers as points in the Arg plane i _i +i i A complex number can be represented by an expression of the form a bi, where a b are real numbers i is a symbol with the property that i. The complex number a bi can also be represented by the ordered pair a, b plotted as a point in a plane (called the Arg plane) as in Figure. Thus, the complex number i i is identified with the point,. The real part of the complex number a bi is the real number a the imaginary part is the real number b. Thus, the real part of 4 i is 4 the imaginary part is. Two complex numbers a bi c di are equal if a c b d, that is, their real parts are equal their imaginary parts are equal. In the Arg plane the horizontal axis is called the real axis the vertical axis is called the imaginary axis. The sum difference of two complex numbers are defined by adding or subtracting their real parts their imaginary parts: a bi c di a c b di For instance, a bi c di a c b di i 4 7i 4 7i 5 6i The product of complex numbers is defined so that the usual commutative distributive laws hold: a bic di ac di bic di ac adi bci bdi Since i, this becomes a bic di ac bd ad bci EXAMPLE i 5i 5i i 5i 5i 6i 5 i 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 denominator by the complex conjugate of the denominator. i EXAMPLE Express the number in the form a bi. 5i SOLUTION We multiply numerator denominator by the complex conjugate of 5i, namely 5i, we take advantage of the result of Example : Thomson BrooksCole copyright 7 i _i FIGURE z=a+bi z=abi i 5i The geometric interpretation of the complex conjugate is shown in Figure : 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 are requested in Exercise 8. Properties of Conjugates i 5i z w z w 5i i 5i i zw z w z n z n
2 COMPLEX NUMBERS bi œ z = z=a+bi b z The modulus, or absolute value, of a complex number z a bi is its distance from the origin. From Figure we see that if z a bi, then z sa b a Notice that FIGURE zz a bia bi a abi abi b i a b so zz z This explains why the division procedure in Example works in general: Since i, we can think of i as a square root of. But notice that we also have i i so i is also a square root of. We say that i is the principal square root of write s i. In general, if c is any positive number, we write sc sc i With this convention, the usual derivation formula for the roots of the quadratic equation ax bx c are valid even when b 4ac : EXAMPLE Find the roots of the equation x x. SOLUTION z w zw ww x b sb 4ac a Using the quadratic formula, we have zw w x s 4 s s i We observe that the solutions of the equation in Example are complex conjugates of each other. In general, the solutions of any quadratic equation ax bx c with real coefficients a, b, 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 n x n a x a of degree at least one has a solution among the complex numbers. This fact is known as the Fundamental Theorem of Algebra was proved by Gauss. POLAR FORM Thomson BrooksCole copyright 7 FIGURE 4 r a a+bi b We know that any complex number z a bi can be considered as a point a, b that any such point can be represented by polar coordinates r, with r. In fact, a r cos b r sin as in Figure 4. Therefore, we have z a bi r cos r sin i
3 COMPLEX NUMBERS Thus, we can write any complex number z in the form z rcos i sin where r z sa b tan b a œ FIGURE 5 4 _ 6 +i œ i The angle is called the argument of z we write. Note that argz is not unique; any two arguments of z differ by an integer multiple of. EXAMPLE 4 Write the following numbers in polar form. (a) z i (b) w s i SOLUTION (a) We have r z s s tan, so we can take. Therefore, the polar form is z s cos (b) Here we have r w s tan s. Since w lies in the fourth quadrant, we take w cos i sin i sin argz The numbers z w are shown in Figure 5. The polar form of complex numbers gives insight into multiplication division. Let 4 4 z r cos i sin z r cos i sin be two complex numbers written in polar form. Then z z z z r r cos i sin cos i sin r r cos cos sin sin isin cos cos sin + Therefore, using the addition formulas for cosine sine, we have z z r r cos i sin z z FIGURE 6 r z This formula says that to multiply two complex numbers we multiply the moduli add the arguments. (See Figure 6.) A similar argument using the subtraction formulas for sine cosine shows that to divide two complex numbers we divide the moduli subtract the arguments. z r cos i sin z r z Thomson BrooksCole copyright 7 FIGURE 7 _ r z In particular, taking z z z, ( therefore ), we have the following, which is illustrated in Figure 7. If z rcos i sin, then z r cos i sin.
4 4 COMPLEX NUMBERS EXAMPLE 5 Find the product of the complex numbers i s i in polar form. œ z=+i œ zw w=œ i SOLUTION From Example 4 we have So, by Equation, i s cos s i cos i sin 6 i(s i) s cos 4 s cos i sin 4 i sin i sin 4 6 FIGURE 8 This is illustrated in Figure 8. peated use of Formula shows how to compute powers of a complex number. If z rcos i sin then z r cos i sin z zz r cos i sin In general, we obtain the following result, which is named after the French mathematician Abraham De Moivre ( ). De Moivre s Theorem If z rcos i sin n is a positive integer, then z n rcos i sin n r n cos n i sin n This says that to take the nth power of a complex number we take the nth power of the modulus multiply the argument by n. EXAMPLE 6 Find. SOLUTION Since, it follows from Example 4(a) that i i i has the polar form So by De Moivre s Theorem, ( i) i s cos i 4 i sin 4 s cos i sin cos 5 5 i sin i Thomson BrooksCole copyright 7 De Moivre s Theorem can also be used to find the nth roots of complex numbers. An n th root of the complex number z is a complex number w such that w n z
5 COMPLEX NUMBERS 5 Writing these two numbers in trigonometric form as w scos i sin using De Moivre s Theorem, we get z rcos i sin s n cos n i sin n rcos i sin The equality of these two complex numbers shows that s n r or s r n cos n cos sin n sin From the fact that sine cosine have period Thus n k it follows that ncos k k w r i sin n n or Since this expression gives a different value of w for k,,,..., n,we have the following. k n Roots of a Complex Number Let z rcos i sin let n be a positive integer. Then z has the n distinct nth roots where k,,,..., n. ncos k k w k r i sin n n w k r n Notice that each of the nth roots of z has modulus. Thus, all the nth roots of z lie on the circle of radius r n in the complex plane. Also, since the argument of each successive nth root exceeds the argument of the previous root by n, we see that the n th roots of z are equally spaced on this circle. Thomson BrooksCole copyright 7 EXAMPLE 7 Find the six sixth roots of z 8 graph these roots in the complex plane. SOLUTION In trigonometric form, z 8cos i sin. Applying Equation with n 6, we get w k 8 6cos i sin 6 6 We get the six sixth roots of 8 by taking k,,,, 4, 5 in this formula: w 8 6cos 6 i sin s 6 s i w 8 6cos i sin s i 5 5 w 86cos i sin s s 6 6 i k k
6 6 COMPLEX NUMBERS w _œ œ w œ i _œ i w w FIGURE 9 The six sixth roots of z=_8 w w All these points lie on the circle of radius s as shown in Figure 9. COMPLEX EXPONENTIALS 7 7 w 86cos i sin s s 6 6 i w 4 86cos i sin s i w 5 86cos i sin s s 6 6 i 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 where the terms are complex numbers. Using the Taylor series for e x (8.7.) as our guide, we define 4 e z n z n n! z z! z! it turns out that this complex exponential function has the same properties as the real exponential function. In particular, it is true that 5 e zz e z e z If we put z iy, where y is a real number, in Equation 4, use the facts that i, i i i i, i 4, i 5 i,... we get e iy iy iy! iy! iy4 4! iy5 5! iy y! i y! y 4 4! i y 5 5! y! y 4 4! y 6 6! iy y! y 5 5! cos y i sin y Here we have used the Taylor series for cos y sin y (Equations ). The result is a famous formula called Euler s formula: 6 e iy cos y i sin y Combining Euler s formula with Equation 5, we get Thomson BrooksCole copyright 7 7 e xiy e x e iy e x cos y i sin y
7 COMPLEX NUMBERS 7 EXAMPLE 8 Evaluate: (a) e i (b) e i We could write the result of Example 8(a) as e i This equation relates the five most famous numbers in all of mathematics:,, e, i,. SOLUTION (a) From Euler s equation (6) we have (b) Using Equation 7 we get e i cos i sin i e i e cos i sin e i i e Finally, we note that Euler s equation provides us with an easier method of proving De Moivre s Theorem: rcos i sin n re i n r n e in r n cos n i sin n Thomson BrooksCole copyright 7 A EXERCISES Click here for answers. i8 i 4 Evaluate the expression write your answer in the form a bi.. 5 6i i. (4 i) (9 5 i). 5i4 i i( i) 7. 4i i 8. i 4i 9.. i 4 i. i. i. s5 4. ss 5 7 Find the complex conjugate the modulus of the number. 5. 5i 6. s i 7. 4i 8. Prove the following properties of complex numbers. (a) z w z w (b) zw z w (c) z n z n, where n is a positive integer [Hint: Write z a bi, w c di.] 9 4 Find all solutions of the equation. 9. 4x 9. x 4. x x 5. x x. z z 4. z z 4 S Click here for solutions. 5 8 Write the number in polar form with argument between. 5. i 6. si 7. 4i 8. 8i 9 Find polar forms for zw, zw, z by first putting z w into polar form. 9. z s i, w si. z 4s 4i, w 8i. z s i, w i. z 4(s i), w i 6 Find the indicated power using De Moivre s Theorem.. i 4. ( si) 5 5. (s i) 5 6. i Find the indicated roots. Sketch the roots in the complex plane. 7. The eighth roots of 8. The fifth roots of 9. The cube roots of i 4. The cube roots of i 4 46 Write the number in the form a bi. 4. e i 4. e i 4. e i 44. e i e i i e 47. Use De Moivre s Theorem with n to express cos sin in terms of cos sin.
8 8 COMPLEX NUMBERS 48. Use Euler s formula to prove the following formulas for cos x sin x: sin x eix e ix cos x eix e ix i 49. If ux f x itx is a complexvalued function of a real variable x the real imaginary parts f x tx are differentiable functions of x, then the derivative of u is defined to be ux f x itx. Use this together with Equation 7 to prove that if Fx e rx, then Fx re rx when r a bi is a complex number. 5. (a) If u is a complexvalued function of a real variable, its indefinite integral x ux dx is an antiderivative of u. Evaluate y e i x dx (b) By considering the real imaginary parts of the integral in part (a), evaluate the real integrals y e x cos x dx y e x sin x dx (c) Compare with the method used in Example 4 in Section 6.. Thomson BrooksCole copyright 7
9 COMPLEX NUMBERS 9 ANSWERS S Click here for solutions.. 8 4i. 8i 5. 7i 7. i 9. i. i. 5i 5. 5i; 7. 4i, 4 9. i. i. (s7)i s cos4 i sin4 5{cos[tan ( 4 )] i sin[tan ( 4 )]} 4cos i sin, cos6 i sin6, cos6 i sin6. 4s cos7 i sin7, (s)cos i sin, 4 cos6 i sin s 5i 7., i, (s) i 9. (s) i, i i 4. i 4. (s)i 45. e 47. cos cos cos sin, sin cos sin sin _i Thomson BrooksCole copyright 7
10 COMPLEX NUMBERS SOLUTIONS. (5 6i)+(+i) =(5+)+( 6+)i =8+( 4)i =8 4i. ( + 5i)(4 i) =(4)+( i)+(5i)(4) + (5i)( i) =8 i +i 5i i = 7i =8+8i 5( ) = 8 + 8i +5=+8i +4i +i = +4i +i i i +i 8( ) + i = = = i + + i +i = +i i i = i ( ) = i = i. i = i i =( )i = i. 5 = 5 i =5i 5. 5i =+5i 5i = +( 5) = = 69 = 7. 4i = 4i =+4i =4i 4i = +( 4) = 6 = x +9= 4x = 9 x = 9 x = ± 94 = ± 9 i = ± i By the quadratic formula, x +x +5= x = ± 4()(5) () = ± 6 = ± 4i = ± i.. By the quadratic formula, z + z += z = ± 4()() () = ± 7 = ± 7 i. 5. For z = +i, r = ( ) + = tan θ = = θ = 4 (since z lies in the second quadrant). Therefore, +i = cos + i sin For z =+4i, r = +4 =5 tan θ = 4 θ =tan 4 (since z lies in the first quadrant). Therefore, +4i =5 cos tan 4 + i sin tan For z = +i, r = + = tan θ = θ = 6 z = cos 6 + i sin 6. For w =+ i, r = tan θ = θ = w = cos + i sin. Therefore, zw = cos i sin + 6 =4 cos + i sin, z/w = cos 6 + i sin 6 =cos 6 + i sin 6,=+i =(cos+i sin ) /z = cos 6 + i sin 6 = cos 6 + i sin 6.For/z,wecouldalsousetheformulathat precedes Example 5 to obtain /z = cos i sin 6 6. Thomson BrooksCole copyright 7. For z = i, r = +( ) =4 tan θ = = θ = 6 z =4 cos 6 + i sin 6.Forw = +i, r =, tan θ = = θ = 4 w = cos + i sin 4 4. Therefore, zw =4 cos i sin =4 cos i sin, z/w = 4 cos i sin 6 4 = 4 cos + i sin = cos + i sin, cos 6 i sin 6 = 4 /z = 4 cos 6 + i sin 6.. For z =+i, r = tan θ = = θ = 4 z = cos 4 + i sin 4.Soby De Moivre s Theorem, ( + i) = cos + i sin 4 4 = / cos + i sin 4 4 = (cos 5 + i sin 5) = [ +i()] = = 4
11 COMPLEX NUMBERS 5. For z = +i, r = + = 6 = 4 tan θ = = θ = 6 z =4 cos + i sin 6 6. So by De Moivre s Theorem, 5 +i = 4 cos + i sin 5 =4 5 cos 5 + i sin =4 + i = 5 +5i 7. =+i =(cos+i sin ). Using Equation with r =, n =8,θ =,wehave +k +k w k = cos /8 + i sin =cos k i sin k,wherek =,,,...,7. 4 w =(cos+i sin ) =, w = cos + i sin 4 4 = + i, w = cos + i sin = i, w = cos + i sin 4 4 = + i, w 4 =(cos + i sin ) =, w 5 = cos 5 + i sin = i, w 6 = cos + i sin = i, w7 = cos 7 + i sin = i 9. i =+i = cos + i sin. Using Equation with r =, n =,θ =,wehave w k = cos / +k + i sin +k,wherek =,,. w = cos + i sin 6 6 = + i w = cos 5 + i sin = + i w = cos 9 + i sin = i 4. Using Euler s formula (6) with y =,wehaveei/ =cos + i sin =+i = i. 4. Using Euler s formula (6) with y =,wehaveei/ =cos + i sin = + i. 45. UsingEquation7withx = y =,wehavee +i = e e i = e (cos + i sin ) =e ( +)= e. 47. Take r = n =in De Moivre s Theorem to get [(cos θ + i sin θ)] = (cos θ + i sin θ) (cos θ + i sin θ) =cosθ + i sin θ cos θ + cos θ (i sin θ)+(cosθ)(i sin θ) +(i sin θ) =cosθ + i sin θ cos θ + cos θ sin θ i cosθ sin θ sin θ i =cosθ + i sin θ cos θ sin θ cos θ + sinθ cos θ sin θ i =cosθ + i sin θ Equating real imaginary parts gives cos θ =cos θ sin θ cos θ sin θ =sinθ cos θ sin θ 49. F (x) =e rx = e (a+bi)x = e ax+bxi = e ax (cos bx + i sin bx) =e ax cos bx + i(e ax sin bx) F (x) =(e ax cos bx) + i(e ax sin bx) =(ae ax cos bx be ax sin bx)+i(ae ax sin bx + be ax cos bx) = a [e ax (cos bx + i sin bx)] + b [e ax ( sin bx + i cos bx)] = ae rx + b e ax i sin bx + i cos bx = ae rx + bi [e ax (cos bx + i sin bx)] = ae rx + bie rx =(a + bi)e rx = re rx Thomson BrooksCole copyright 7
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