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1 COMPLEX NUMBERS Thomson Brooks-Cole copyright 7 _+i _-i FIGURE Complex numbers as points in the Argand plane i _i FIGURE i _i +i z=a+bi z=a-bi -i 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. 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. 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 and the imaginary part is the real number b. Thus, the real part of i is and the imaginary part is. 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: For instance, The product of complex numbers is defined so that the usual commutative and distributive laws hold: a bic di ac di bic di Since i, this becomes EXAMPLE i EXAMPLE Express the number in the form a bi. 5i SOLUTION We multiply numerator and denominator by the complex conjugate of 5i, namely 5i, and we take advantage of the result of Example : 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 and are requested in Exercise 8. Properties of Conjugates a bi c di a c b di a bi c di a c b di i 7i 7i 5 i a bic di ac bd ad bci i 5i 5i i 5i 5i i 5 i i 5i z w z w ac adi bci bdi 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. 5i i 5i i zw z w z n z n

2 COMPLEX NUMBERS bi œ z = a@+b@ 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 and so 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 and so i is also a square root of. We say that i is the principal square root of and write s 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 bx c are valid even when b ac : EXAMPLE Find the roots of the equation x x. SOLUTION zz z z zw w ww x b sb ac a Using the quadratic formula, we have x s 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, 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 nx n a nx 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 and was proved by Gauss. zw w POLAR FORM Thomson Brooks-Cole copyright 7 FIGURE r a a+bi b 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. In fact, a r cos b r sin as in Figure. 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 and tan b a œ FIGURE 5 z z FIGURE z + _ r +i œ -i z The angle is called the argument of z and we write. Note that argz is not unique; any two arguments of z differ by an integer multiple of. EXAMPLE Write the following numbers in polar form. (a) z i (b) w s i SOLUTION (a) We have r z s s and tan, so we can take. Therefore, the polar form is (b) Here we have s and tan s. Since w lies in the fourth quadrant, we take and w cos i sin The numbers z and w are shown in Figure 5. The polar form of complex numbers gives insight into multiplication and division. Let be two complex numbers written in polar form. Then Therefore, using the addition formulas for cosine and sine, we have r w z r cos i sin z z r r cos i sin cos i sin r r cos cos sin sin isin cos cos sin This formula says that to multiply two complex numbers we multiply the moduli and add the arguments. (See Figure.) 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. z z z z r r cos i sin r cos i sin r z s cos i sin argz z r cos i sin z z Thomson Brooks-Cole copyright 7 FIGURE 7 _ r z In particular, taking z and z z, (and therefore and ), we have the following, which is illustrated in Figure 7. If z rcos i sin, then z r cos i sin.

4 COMPLEX NUMBERS œ FIGURE 8 z=+i œ zw w=œ -i EXAMPLE 5 Find the product of the complex numbers i and s i in polar form. SOLUTION and From Example we have So, by Equation, This is illustrated in Figure 8. then and peated use of Formula shows how to compute powers of a complex number. If In general, we obtain the following result, which is named after the French mathematician Abraham De Moivre (7 75). i(s i) s cos De Moivre s Theorem If z rcos i sin and n is a positive integer, then 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 Find. SOLUTION Since, it follows from Example (a) that i i i has the polar form So by De Moivre s Theorem, i s cos i sin s i cos i sin z n rcos i sin n r n cos n i sin n ( i) i s cos i sin z rcos i sin z r cos i sin z zz r cos i sin i sin s i cos i sin s cos i sin 5 cos 5 5 i sin i Thomson Brooks-Cole 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 and using De Moivre s Theorem, we get and z rcos i sin s n cos n i sin n rcos i sin The equality of these two complex numbers shows that Thomson Brooks-Cole copyright 7 and From the fact that sine and cosine have period it follows that ncos k k Thus w r i sin n n Since this expression gives a different value of w for k,,,..., n,we have the following. Roots of a Complex Number Let z rcos i sin and let n be a positive integer. Then z has the n distinct nth roots where k,,,..., n. s n r cos n cos n k ncos k k w k r i sin n 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. 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 with n, we get w k 8 cos i sin We get the six sixth roots of 8 by taking k,,,,, 5 in this formula: w 8 cos i sin i w 8 cos i sin 5 w 8cos or and k s s s i or s r n sin n sin k wk r n k i sin 5 s s i n

6 COMPLEX NUMBERS œ i w 7 w 8 cos i sin 7 s s i w w FIGURE 9 The six sixth roots of z=_8 w _œ œ w _œ i w All these points lie on the circle of radius s as shown in Figure 9. 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.) as our guide, we define 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 If we put z iy, where y is a real number, in Equation, and use the facts that we get w 8 cos w 5 8 cos e z i, e iy iy iy iy iy iy5!!! 5! iy y y! y! y cos y i sin y n z n i sin s i i sin s s i z z n!! e zz i i i i, e z e z i,! i y! y! i y 5 z! 5! i 5 i, Here we have used the Taylor series for cos y and sin y (Equations and 8.7.). The result is a famous formula called Euler s formula:...! iy y! y 5 5! e iy cos y i sin y Combining Euler s formula with Equation 5, we get Thomson Brooks-Cole copyright 7 7 e xiy e x e iy e x cos y i sin y

7 COMPLEX NUMBERS 7 e i EXAMPLE 8 Evaluate: (a) (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, and. SOLUTION (a) From Euler s equation () we have (b) Using Equation 7 we get e i cos i sin i Thomson Brooks-Cole copyright 7 A EXERCISES Evaluate the expression and write your answer in the form a bi.. 5 i i.. 5i i. i8 i 5. 7i. i( i) i i i i 9.. i i... s5. ss 5 7 Find the complex conjugate and the modulus of the number. 5. 5i. s i 7. i Click here for answers. 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 Find all solutions of the equation. 9. x 9. x. x x 5. x x. z z. z z S 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: Click here for solutions. ( i) (9 5 i) i rcos i sin n re i n r n e in r n cos n i sin n 5 8 Write the number in polar form with argument between and. 5. i. si 7. i 8. 8i 9 Find polar forms for zw, zw, and z by first putting z and w into polar form. 9. z s i,. z s i,. z s i, z (s i)., Find the indicated power using De Moivre s Theorem.. i. ( si) 5 5. (s i) 5. i 8 7 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. The cube roots of i Write the number in the form a bi. e i.. e i.. e i w si w 8i w i w i 5.. e i e i i e 7. Use De Moivre s Theorem with n to express cos and sin in terms of cos and sin.

8 8 COMPLEX NUMBERS 8. Use Euler s formula to prove the following formulas for cos x and sin x: sin x eix e ix cos x eix e ix i 5. (a) If u is a complex-valued function of a real variable, its indefinite integral x ux dx is an antiderivative of u. Evaluate y e i x dx 9. If ux f x itx is a complex-valued function of a real 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 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. (b) By considering the real and imaginary parts of the integral in part (a), evaluate the real integrals y e x cos x dx and y e x sin x dx (c) Compare with the method used in Example in Section.. Thomson Brooks-Cole copyright 7

9 COMPLEX NUMBERS 9 S ANSWERS Click here for solutions.. 8 i. 8i 5. 7i 7. i 9. i. i. 5i 5. 5i; 7. i, 9. i. i. (s7)i s cos i sin 5{cos[tan ( )] i sin[tan ( )]} 9. cos i sin, cos i sin, cos i sin. s cos7 i sin7, (s)cos i sin, cos i sin. 5. 5s 5i 7., i, (s) i 9. (s) i, i i. i. 5. e (s)i 7. cos cos cos sin, sin cos sin sin _i Thomson Brooks-Cole copyright 7

10 COMPLEX NUMBERS SOLUTIONS. (5 i)+(+i) =(5+)+( +)i =8+( )i =8 i. i 9+ 5 i =( 9) + 5 i = 5+( )i = 5 i. ( + 5i)( i) =()+( i)+(5i)() + (5i)( i) =8 i +i 5i =8+8i 5( ) = 8 + 8i +5=+8i. ( i)(8 i) =8 i i +( ) = 9i i = 7i. i i = i ( ) = + i i i = +i = i i +i = +i +i i i +i 8( ) + i = = = i + + i +i i = +i i +i +i +i +8( ) = = 5+i = 5 +i i +i = +i i i = i ( ) = i = i i = i +i + 9i = +i 9( ) = i. i = i i =( )i = i. i = i 5 =( ) 5 =. 5 = 5 i =5i. = i i = i = ( ) = 5. 5i =+5i and 5i = +( 5) = + 5 = 9 =. + i = i and + i = ( ) + = +8= 9= 7. i = i =+i =iand i = +( ) = = 8. Let z = a + bi, w = c + di. (a) z + w = (a + bi)+(c + di) =(a + c)+(b + d)i =(a + c) (b + d)i =(a bi)+(c di) =z + w (b) zw = (a + bi)(c + di) =(ac bd)+(ad + bc)i =(ac bd) (ad + bc)i. On the other hand, z w =(a bi)(c di) =(ac bd) (ad + bc)i = zw. (c) Use mathematical induction and part (b): Let S n be the statement that z n = z n. S is true because z = z = z. Assume S k is true, that is z k = z k.then z k+ = z +k = zz k = zz k [part (b) with w = z k ] = z z k = z +k = z k+, which shows that S k+ is true. Thomson Brooks-Cole copyright 7 Therefore, by mathematical induction, z n = z n for every positive integer n. Another proof: Use part (b) with w = z, and mathematical induction. 9. x +9= x = 9 x = 9 x = ± 9 = ± 9 i = ± i.

11 COMPLEX NUMBERS. x = x = x x + = x =or x += x = ± or x = ±i.. By the quadratic formula, x +x +5= x = ± ()(5) () = ± = ± i. x x += x = ( ) ± ( ) ()() () = ± i. = ± = ± i = ± i. By the quadratic formula, z + z += z = ± ()() = ± 7 (). z + z + = z +z += z = ± ()() = ± = ± i = () 8 8 ± i = ± 7 i. 5. For z = +i, r = ( ) + = and tan θ = = θ = (since z lies in the second quadrant). Therefore, +i = cos.. For z = i, r = + =and tan θ = = θ = 5 (since z lies in the fourth quadrant). Therefore, i = cos For z =+i, r = + =5and tan θ = θ =tan (since z lies in the first quadrant). Therefore, +i =5 cos tan tan. 8. For z =8i, r = +8 =8and tan θ = 8 is undefined, so θ = (since z lies on the positive imaginary axis). Therefore, 8i =8 cos. 9. For z = +i, r = + =and tan θ = θ = z = cos. For w =+ i, r =and tan θ = θ = w = cos. Therefore, zw = cos + + =cos, z/w = cos =cos,and=+i =(cos+isin ) /z = cos = cos.for/z, we could also use the formula that precedes Example 5 to obtain /z = cos i sin.. For z = i, r = +( ) = = 8 and tan θ = = θ = Thomson Brooks-Cole copyright 7 z =8 cos w =8 cos cos.forw =8i, r = +8 =8and tan θ = 8 is undefined, so θ =. Therefore, zw =8 8 cos + =cos + = cos, z/w = 8,and 8 =+i =(cos+isin ) /z = 8 cos = 8 cos. For /z, we could also use the formula that precedes Example 5 to obtain /z = 8 cos i sin.

12 COMPLEX NUMBERS. For z = i, r = +( ) =and tan θ = = θ = z = cos.forw = +i, r =, tan θ = = θ = w = cos. Therefore, zw = cos + z/w = cos /z = = cos cos + = cos 7 = i sin,and = cos. cos 7,. For z = +i = +i, r = + = = 8 and tan θ = = θ = z =8 cos.forw = i, r = ( ) +( ) = 8 = and tan θ = = θ = 5 w = cos 5 5. Therefore, zw =8 cos = cos 7 7, z/w = 8 cos 5 5 = cos,and /z = 8 cos i sin.. For z =+i, r = and tan θ = = θ = z = cos.soby De Moivre s Theorem, ( + i) = cos = / cos = (cos 5 5) = [ +i()] = =. For z = i, r = + =and tan θ = = θ = 5 z = cos 5 5. So by De Moivre s Theorem, 5. For z = +i, r = 5 i = cos = 5 cos = 5 cos = + i =+ i + = = and tan θ = = θ = z = cos. So by De Moivre s Theorem, 5 +i = cos 5 = 5 cos 5 5 = + i = 5 + 5i. For z = i, r = and tan θ = = θ = 7 z = cos 7 7 ( i) 8 = cos = cos Thomson Brooks-Cole copyright 7 =(cos ) =(+i) =

13 COMPLEX NUMBERS 7. =+i =(cos+isin ). Using Equation with r =, n =8,andθ =,wehave +k +k w k = cos /8 =cos k 8 8 k,wherek =,,,...,7. w =(cos+isin ) =, w = cos = + i, w = cos = i, w = cos = + i, w =(cos ) =, w 5 = cos 5 5 = i, w = cos = i, w7 = cos 7 7 = i 8. = + i =(cos+isin ). Using Equation with r =, n =5,andθ =,wehave +k +k w k = cos /5 = cos k k,wherek =,,,,. 5 w =(cos+isin ) = w = cos 5 5 w = cos 5 5 w = cos 5 5 w = cos i =+i = cos. Using Equation with r =, n =,andθ =,wehave w k = cos / +k +k,wherek =,,. w = cos = + i w = cos 5 5 = + i w = cos 9 9 = i. +i = cos. Using Equation with r =, n =,andθ =,wehave w k = / cos +k +k,wherek =,,. w = / cos w = / cos = / + i = / + / i w = / cos 7 7. Using Euler s formula () with y =,wehaveei/ =cos =+i = i.. Using Euler s formula () with y =,wehavee i =cos =.. Using Euler s formula () with y =,wehaveei/ =cos = + i.. Using Euler s formula () with y =,wehavee i =cos( )+i sin( ) =. Thomson Brooks-Cole copyright 7 5. UsingEquation7withx =and y =,wehavee +i = e e i = e (cos ) =e ( +)= e.. UsingEquation7withx = and y =,wehavee +i = e e i = e (cos ) = e cos + (e sin )i.

14 COMPLEX NUMBERS 7. Take r =and n =in De Moivre s Theorem to get [(cos θ θ)] = (cos θ θ) (cos θ θ) =cosθ θ cos θ + cos θ (i sin θ)+(cosθ)(i sin θ) +(isin θ) =cosθ θ cos θ + cos θ sin θ i cosθ sin θ sin θ i =cosθ θ cos θ sin θ cos θ + sinθ cos θ sin θ i =cosθ θ Equating real and imaginary parts gives cos θ =cos θ sin θ cos θ and sin θ =sinθ cos θ sin θ 8. Using Formula, Thus, cos x = eix + e ix. Similarly, e ix + e ix =(cosx x)+[cos( x)+i sin( x)] =cosx x +cosx i sin x =cosx e ix e ix =(cosx x) [cos( x)+i sin( x)] =cosx x cos x ( i sin x) =i sin x Therefore, sin x = eix e ix. i 9. F (x) =e rx = e (a+bi)x = e ax+bxi = e ax (cos bx 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 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 bx)] = ae rx + bie rx =(a + bi)e rx = re rx 5. (a) From Exercise 9, F (x) =e (+i)x F (x) =(+i)e (+i)x.so e (+i)x dx = F (x) dx = +i +i F (x)+c = i F (x)+c = i e (+i)x + C (b) e (+i)x dx = e x e ix dx = e x (cos x x) dx = e x cos xdx+ i e x sin xdx (). Also, i e(+i)x = e(+i)x ie(+i)x = ex+ix iex+ix = ex (cos x x) iex (cos x x) = ex cos x + ex sin x + iex sin x iex cos x = ex (cos x +sinx)+i ex (sin x cos x) () Equating the real and imaginary parts in ()and(), we see that e x cos xdx= ex (cos x +sinx)+c and e x sin xdx= ex (sin x cos x)+c. Thomson Brooks-Cole copyright 7