Complex Numbers

Im A complex number can be represented by an expression of the form a bi , where a and 2+3i b are real numbers and i is a symbol with the property that i 2 1 . The complex num- _4+2i ber a bi can also be represented by the ordered pair a, b and plotted as a point in a i plane (called the Argand plane) as in Figure 1. Thus, the complex number i 0 1 i is identiﬁed with the point 0, 1 . 0 1 Re _i The real part of the complex number a bi is the real number a and the imaginary _2-2i 3-2i 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 FIGURE 1 is called the real axis and the vertical axis is called the imaginary axis. Complex numbers as points in The sum and difference of two complex numbers are deﬁned by adding or subtracting the Argand plane 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.

1 3i EXAMPLE 2 Express the number 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:

1 3i 1 3i 2 5i 13 11i 13 11 i 2 5i 2 5i 2 5i 22 52 29 29 Im z=a+bi The geometric interpretation of the complex conjugate is shown in Figure 2:z is the i 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. 0 Re _i Properties of Conjugates –z=a-bi z w z w zw z w z n z n FIGURE 2

Im The modulus, or absolute value, z of a complex number z a bi is its distance z=a+bi from the origin. From Figure 3 we see that if z a bi , then bi b@„ @+ œ„„„„ = a 2 2 |z| b z sa b

0 a Re Notice that FIGURE 3 zz a bia bi a 2 abi abi b 2i 2 a 2 b 2

and so zz z 2

This explains why the division procedure in Example 2 works in general:

z zw zw w ww 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 ax2 bx c 0 are valid even when b 2 4ac 0 :

b sb 2 4ac x 2a

EXAMPLE 3 Find the roots of the equation x 2 x 1 0 .

SOLUTION Using the quadratic formula, we have

1 s1 2 4 1 1 s3 1 s3 i x 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 ax2 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

n n1 an x an1 x a1 x a0 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

Im We know that any complex number z a bi can be considered as a point a, b and that a+bi any such point can be represented by polar coordinates r, with r 0 . In fact, r b a r cos b r sin ¨ 0 a Re as in Figure 4. Therefore, we have

Thus, we can write any complex number z in the form

z rcos i sin

b where r z sa 2 b 2 and tan 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 (b) w s3 i

SOLUTION (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 cos i sin œ„2 4 4 π 4 (b) Here we have r w s3 1 2 and tan 1s3 . Since w lies in the π 0 _ Re 6 fourth quadrant, we take 6 and 2 œ„3-i w 2cos i sin 6 6

FIGURE 5 The numbers z and w are shown in 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 r r i i z™ z¡ z 1z2 1 2 cos 1 sin 1 cos 2 sin 2 ¨™ r1r2 cos 1 cos 2 sin 1 sin 2 i sin 1 cos 2 cos 1 sin 2

¨¡ Therefore, using the addition formulas for cosine and sine, we have Re ¨¡+¨™

1 z1z2 r1r2cos1 2 i sin1 2

z¡z™ This formula says that to multiply two complex numbers we multiply the moduli and add FIGURE 6 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. Im z z1 r1 cos1 2 i sin1 2 z2 0 r z2 r2

¨ In particular, taking z1 1 and z2 z , (and therefore 1 0 and 2 ), we have the fol- 0 _¨ Re lowing, which is illustrated in Figure 7. 1 1 r z 1 1 If z rcos i sin , then cos i sin . z r FIGURE 7 Thomson Brooks-Cole copyright 2007 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

1 i s2 cos i sin 4 4

and s3 i 2cos i sin Im 6 6 z=1+i zw So, by Equation 1, œ„2 2œ„2 π 12 1 i(s3 i) 2s2 cos i sin 0 Re 4 6 4 6

2 2s2 cos i sin w=œ„3-i 12 12

FIGURE 8 This is illustrated in Figure 8.

Repeated use of Formula 1 shows how to compute powers of a complex number. If

z rcos 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 rcos i sin and n is a positive integer, then

z n rcos 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.

1 1 10 EXAMPLE 6 Find ( 2 2 i) . 1 1 1 1 1 SOLUTION Since 2 2 i 2 1 i , it follows from Example 4(a) that 2 2 i has the polar form

1 1 s2 i cos i sin 2 2 2 4 4 So by De Moivre’s Theorem,

1 1 10 s2 10 10 10 i cos i sin 2 2 2 4 4 25 5 5 1 cos i sin i 210 2 2 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

Writing these two numbers in trigonometric form as

w scos i sin and z rcos i sin

and using De Moivre’s Theorem, we get

s ncos n i sin n rcos i sin

The equality of these two complex numbers shows that

s n r or s r 1n

and cos n cos and sin n sin

From the fact that sine and cosine have period 2 it follows that

2k n 2k or n

2k 2k Thus w r 1ncos i sin n n

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 rcos i sin and let n be a positive integer. Then z has the n distinct n th roots

2k 2k w r 1ncos i sin k n n

where k 0,1,2,...,n 1 .

1n Notice that each of the n th roots of z has modulus wk r . Thus, all the n th roots of z lie on the circle of radius r 1n in the complex plane. Also, since the argument of each successive n th root exceeds the argument of the previous root by 2n , 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 2k 2k w 816cos i sin k 6 6 We get the six sixth roots of 8 by taking k 0, 1, 2, 3, 4, 5 in this formula:

s3 1 w 816cos i sin s2 i 0 6 6 2 2 w 816cos i sin s2 i 1 2 2

Im 7 7 s3 1 w 16 s 3 8 cos i sin 2 i œ„2i w¡ 6 6 2 2

w™ w¸ 3 3 w 816cos i sin s2 i 4 2 2 _œ„2œ0 „2 Re 11 11 s3 1 w 16 s £ ∞ 8 cos i sin 2 i w w 5 6 6 2 2

w¢ _œ„2i 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 where the terms are complex numbers. Using the Taylor series for e x (8.7.12) as our guide, we deﬁne

z n z2 z3 4 e z 1 z n0 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

5 e z1 z2 e z1e z2

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 iy2 iy3 iy4 iy5 e iy 1 iy 2! 3! 4! 5!

y 2 y 3 y 4 y 5 1 iy i i 2! 3! 4! 5!

y 2 y 4 y 6 y 3 y 5 1 iy 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

Combining Euler’s formula with Equation 5, we get

EXAMPLE 8 Evaluate: (a)e i (b) e 1 i 2

■■ We could write the result of Example 8(a) SOLUTION as (a) From Euler’s equation (6) we have e i 1 0 This equation relates the ﬁve most famous e i cos i sin 1 i0 1 numbers in all of mathematics: 0, 1, e, i, and . (b) Using Equation 7 we get

1 i e1i2 e1cos i sin 0 i1 2 2 e 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 ne in r ncos n i sin n

EXERCISES

25–28 Write the number in polar form with argument between 0 A Click here for answers. S Click here for solutions. and 2. s 1–14 Evaluate the expression and write your answer in the 25.3 3i 26. 1 3i form a bi. 27.3 4i 28. 8i

1 5 ■■■■■■■■■■■■■ 1.5 6i 3 2i 2. (4 2 i) (9 2 i) 29–32 Find polar forms for zw ,zw , and 1z by ﬁrst putting z 3.2 5i4 i 4. 1 2i8 3i and w into polar form. 5.12 7i 6. 2i( 1 i) 2 29. z s3 i, w 1 s3i 1 4i 3 2i 7. 8. 30. z 4s3 4i, w 8i 3 2i 1 4i 31. z 2s3 2i, w 1 i 1 3 9. 10. s 1 i 4 3i 32. z 4( 3 i), w 3 3i

■■■■■■■■■■■■■ 11.i 3 12. i 100 33–36 Find the indicated power using De Moivre’s Theorem. 13.s25 14. s3s12 33.1 i20 34. (1 s3i)5 ■■■■■■■■■■■■■ 35.(2s3 2i)5 36. 1 i8 15–17 Find the complex conjugate and the modulus of the ■■■■■■■■■■■■■ number. 37–40 Find the indicated roots. Sketch the roots in the complex 15.12 5i 16. 1 2s2 i plane. 17. 4i 37. The eighth roots of 1 38. The ﬁfth roots of 32

■■■■■■■■■■■■■ 39. The cube roots of i 40. The cube roots of 1 i 18. 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 41–46 Write the number in the form a bi . w i2 2i [Hint: Write z a bi,c di .] 41.e 42. e 19–24 Find all solutions of the equation. 43.e i3 44. e i 2 4 19.4x 9 0 20. x 1 45.e 2 i 46. e i 2 2 21.x 2x 5 0 22. 2x 2x 1 0 ■■■■■■■■■■■■■ 2 2 1 1 23.z z 2 0 24. z 2 z 4 0 47. Use De Moivre’s Theorem with n 3 to express cos 3 and

48. Use Euler’s formula to prove the following formulas for cos x 50. (a) If u is a complex-valued function of a real variable, its and sin x: indeﬁnite integral x ux dx is an antiderivative of u . Evaluate eix eix eix eix cos x sin x 1i x 2 2i y e dx

49. If ux f x itx is a complex-valued function of a real (b) By considering the real and imaginary parts of the integral variable x and the real and imaginary parts f x and tx are in part (a), evaluate the real integrals differentiable functions of x , then the derivative of u is deﬁned to be ux f x itx . Use this together with Equation 7 y e x cos x dx and y e x sin x dx to prove that if Fx e rx , then Fx re rx when r a bi is a complex number. (c) Compare with the method used in Example 4 in Section 6.1. Thomson Brooks-Cole copyright 2007 COMPLEX NUMBERS ■ 9

ANSWERS

31. 4s2 cos712 i sin712, S s 1 Click here for solutions. (2 2) cos 13 12 i sin 13 12 , 4 cos 6 i sin 6

1.8 4i 3. 13 18i 33.1024 35. 512s3 512i 11 10 s s 1 5.12 7i 7. 13 13 i 37.1, i, (1 2) 1 i 39. ( 3 2) 2 i, i 1 1 Im Im 9.2 2 i 11.i 13.5i 15. 12 5i; 13 i 3 17.4i, 4 19. 2 i

1 s 0 21.1 2i 23. 2 ( 7 2)i 1 Re 0 Re 25. 3s2 cos34 i sin34 _i 1 4 1 4 27. 5{cos[tan (3 )] i sin[tan (3)]} 41.i 43.1 (s32)i 45. e 2 29. 4cos2 i sin2, cos6 i sin6, 2 1 3 2 2 3 2 cos 6 i sin 6 47. cos 3 cos 3 cos sin , sin 3 3 cos sin sin Thomson Brooks-Cole copyright 2007 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 1+4i 3 2i 3 2i +12i 8( 1) 11 + 10i 11 10 7. = − = − − − = = + i 3+2i 3+2i · 3 2i 32 +22 13 13 13 − 1 1 1 i 1 i 1 i 1 1 9. = − = − = − = i 1+i 1+i · 1 i 1 ( 1) 2 2 − 2 − − − 11. i3 = i2 i =( 1)i = i · − − 13. √ 25 = √25 i =5i − 15. 12 5i =12+15i and 12 15i = 122 +( 5)2 = √144 + 25 = √169 = 13 − | − | − 17. 4i = 0 4i =0+4i =4i and 4is= 02 +( 4)2 = √16 = 4 − − |− | − t 19. 4x2 +9=0 4x2 = 9 x2 = 9 x = 9 = 9 i = 3 i. ⇔ − ⇔ − 4 ⇔ ± − 4 ± 4 ± 2 t t 21. By the quadratic formula, x2 +2x +5=0 ⇔ 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 23. By the quadratic formula, z2 + z +2=0 z = − ± − = − ± − = i. 2(1) 2 2 2 ⇔ s − ± 2 2 3 3π 25. For z = 3+3i, r = ( 3) +3 =3√2 and tan θ = 3 = 1 θ = 4 (since z lies in the second − − − − ⇒ quadrant). Therefore, s3+3i =3√2 cos 3π + i sin 3π . − 4 4 2 2 4 1 4 27. For z =3+4i, r = √3 +4 =5and tan θ = θ =tan− (since z lies in the ﬁrst quadrant). 3 ⇒ 3 1 4 1 4 Therefore, 3+4i =5 cos tan− 3 + i sin tan− 3 .

2 29. For z = √3+i, r = √3 +12 =2and tan θ =1 θ = π z =2 cos π + i sin π . √3 ⇒ 6 ⇒ 6 6 For w =1+√3 i, r =2t andtan θ = √3 θ = π w =2 cos π + i sin π . Therefore, ⇒ 3 ⇒ 3 3 zw =2 2 cos π + π + i sin π + π =4cos π + i sin π , · 6 3 6 3 2 2 z/w = 2 cos π π + i sin π π =cos π + i sin π ,and1=1+0i =1(cos0+i sin 0) 2 6 − 3 6 − 3 − 6 −6 ⇒ 1/z = 1 cos 0 π + i sin 0 π = 1 cos π + i sin π .For1/z,wecouldalsousetheformulathat 2 − 6 − 6 2 − 6 − 6 precedes Example 5 to obtain 1/z = 1 cos π i sin π . 2 6 − 6 2 2 2 1 31. For z =2√3 2i, r = 2 √3 +( 2) =4and tan θ = − = − − 2√3 − √3 θ = π z =4tcos π + i sin π .Forw = 1+i, r = √2, ⇒ − 6 ⇒ − 6 − 6 − 1 3π 3π 3π tan θ = 1 = 1 θ= 4 w = √2cos 4 + i sin 4 . Therefore, − − ⇒ ⇒ zw =4√2 cos π + 3π + i sin π + 3π =4 √2 cos 7π + isin 7π , − 6 4 − 6 4 12 12 z/w = 4 cos π 3π + i sin π 3π = 4 cos 11π + i sin 11π √2 − 6 − 4 − 6 − 4 √2 − 12 − 12 13π 13π =2√2 cos 12 + i sin 12 ,and 1/z = 1 cos π i sin π = 1 cos π + i sin π . 4 − 6 − − 6 4 6 6 33. For z =1+ i, r = √2 andtanθ = 1 =1 θ = π z = √2 cos π + i sin π .Soby 1 ⇒ 4 ⇒ 4 4 De Moivre’s Theorem, 20 20 20 √ π π 1/2 20 π 20 π (1 + i) = 2 cos 4 + i sin 4 = 2 cos 4· + i sin 4· k l =210(cos 5π + i sin 5π)=210[ 1+i(0)] = 210 = 1024 − − − Thomson Brooks-Cole copyright 2007 COMPLEX NUMBERS ■ 11

2 35. For z =2√3+2i, r = 2 √3 +22 = √16 = 4 and tan θ = 2 = 1 θ = π 2√3 √3 ⇒ 6 ⇒ π π t z =4 cos 6 + i sin 6 . So by De Moivre’s Theorem,

5 5 2 √3+2i = 4 cos π + i sin π =45 cos 5π + i sin 5π =1024 √3 + 1 i = 512 √3+512i 6 6 6 6 − 2 2 − k l 37. 1=1+0i =1(cos0+ i sin 0). Using Equation 3 with r =1, n=8,andθ =0,wehave 0+2kπ 0+2kπ kπ kπ w =11/8 cos + i sin =cos + i sin ,wherek =0, 1, 2,...,7. k 8 8 4 4 w =1(cos0+i sin 0) = 1, w =1 cos π + i sin π = 1 + 1 i, 0 1 4 4 √2 √2 π π 3π 3π 1 1 w2 =1 cos + i sin = i, w3 =1 cos + i sin = + i, 2 2 4 4 − √2 √2 5π 5π 1 1 w4 =1(cos π + i sin π)= 1, w5 =1 cos + i sin = i, − 4 4 − √2 − √2 3π 3π 7π 7π 1 1 w6 =1 cos + i sin = i, w7 =1 cos + i sin = i 2 2 − 4 4 √2 − √2

π π π 39. i =0+i =1 cos 2 + i sin 2 . Using Equation 3 with r =1, n =3,andθ = 2 ,wehave π +2kπ π +2kπ w =11/3 cos 2 +i sin 2 ,wherek =0, 1, 2. k 3 3 π π √3 1 w0 = cos 6 + i sin 6 = 2 + 2 i 5π 5π √3 1 w1 = cos + i sin = + i 6 6 − 2 2 9π 9π w2 = cos + i sin = i 6 6 −

π iπ/2 π π 41. Using Euler’s formula (6) with y = 2 ,wehavee =cos2 + i sin 2 =0+1i = i. π π π 1 √3 43. Using Euler’s formula (6) with y = ,wehaveeiπ/3 =cos + i sin = + i. 3 3 3 2 2 45. UsingEquation7withx =2and y = π,wehavee2+iπ = e2eiπ = e2(cos π + i sin π)=e2( 1+0)= e2. − − 47. Take r =1and n =3in 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 =cos3θ + i sin 3θ cos3 θ +3 cos2 θ (i sin θ)+3(cosθ)(i sin θ)2 +(i sin θ)3 =cos3θ + i sin 3θ

cos3 θ + 3cos 2 θ sin θ i 3cosθ sin2 θ sin3 θ i =cos3θ + i sin 3θ − − cos3 θ 3sin2 θ cos θ + 3sinθ cos2 θ sin3 θi =cos3θ + i sin 3θ − − Equating real and imaginary parts gives cos 3θ =cos3 θ 3sin2 θ cosθ and sin 3θ =3sinθ 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) ⇒ ax ax F 0(x)=(e cos bx)0 + i(e 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