Complex Numbers 30 Example 4 Express cos30 in terms of cos and sin. Solution.By Eq.(16)(with n=3)we have cos30 Re (cos30+isin 30)=Re(cos0+i sin@)3. (17) According to the binomial formula, (a+b)3=a3+3a2b+3ab2+b3. Thus,making the obvious identificationsa=cosb=isin in(17),we deducc cos30 Re cos0+3co20(i sin0)+3cos0 (-sin20)-i sin30 (18) =cos30-3cos6sin29.■ Example 5 Compute the integral by using the representation(1)together with the binomial formula(see Exercises 1.1 Prob.27). Solution.We can express the integrand as a--e. and expanding via the binomial formula gives cosg=(e0+4e3ne-19+6e2i9e-29+4e0e-30+e4io） =京(e0+4e20+6+4e-20+eo) =76+8c0s29+2c0s49). Thus g8e产 2z6+8cos20+2cos40d0 =z69+4sm29+n491- 10=22m=.■
1.4 The Complex Exponential EXERCISES 1.4 In Problems I and 2 write each of the given numbers in the form a+bi. 1.(a)e-tr4 (6)e43a e-1+1π/2 (c)e 2.a6-e-31 2 (b)2e3+Hm/6 (c)ez,wherez=4e!/3 In Problems 3 and 4 write each of the given numbers in the polar form rei. 3a倒 3 (b)-8x(1+√5) (c)(I+i)5 a(号+122 21 -3+i 3 5.Show that e=e and arge+is=y+2k(.). 6.Show that,for real e泪-e-泪 (a倒am9=e9+e-i可 2 （b）csc9=e0-a1a-ee+页 7.Show thatforall (The exponenial function is periodicwith period 2πi) 8.Show that,for allz. (a)eital =-e: (b)2= 9.Show that (e)=en for any integer n. 10.Show that le|≤1 if Re z≤0. 11.Determine which of the following properties of the real exponential function remain true for the complex exponential function(that is,for replaced by ) (a)is never zero. (b)e*is a one-to-one function. (c)e is defined for all. (d)e-x=I/e. 12.Use De Moivre's formula together with the binomial formula to derive the following identities (a)sin 30 =3cos20sin-sin30 (b)sin40 =4cos3 0 sin0-4cos0 sin30 13.Show how the following trigonometric identities follow from Eqs.(11)and (12). (a)sin2日+cos20=1 (b)cos(01+02)=cos0 cos02-sin01 sin02
Complex Numbers 32 14.Does De Moivre's foru hold for negative integers? 15.(a)Show that the multiplicative law(1)follows from Definition 5 (b)Show that the division rule ()follows from Definition5. 16.Let z=r.().Show that exp(Inr+i)=z. traversed in the counter describe each of the following curves az0=3e,0≤t≤2n b)z0=2e+i.0≤1≤2 z0=22.0si≤12间200=3e-+2-1.0s1≤2 18.Sketch the curves that are given for2 by (a)z()=e(l+ (b)z(1)=e(l-i)r (g20=e-l+加 (dz)=e-1-r 19.Let nbe a positive integer greater than 2.Show that the points1, n-1,form the vertices of a regular polygon. 20.Prove that ifz≠1，then 1*z+2+…+”=-1 2-1 Use this result and De Moivre's formula to establish the following identities. 2sin(/2) (）in6+sih20+…+sim49=in0m0/2sim0m+1)9/2 sin(8/2) where 0<0<2r. 21.Prove that ifn is a positive integer,then g2sn9≠0.±2m,±4。 [HINT:Argue first that if=ei,then the left-hand side equals ("/(1-)1] 22.Show that ifn is an integer.the 广erw=广w0+m6响-{石≠8 2丛yn0Ww的org咖 (a)cosd b)sin5(29)d0. thsacoavenieaeinpriaingwesoietneswriecpta)insteadofe
1.5 Powers and Roots 1.5 Powers and Roots In this section we shall derive formulas for the nth power and the mth roots of a com- Letz=re=r(cos+i sin)be the polar form of the complex number z.By taking =22=z in Eg.(13)of Sec.1.4,we obtain the formula z2=y2e20 Since=z2,we can apply the identity a second time to deduce that z3=r3e39 Continuing in this manner we arrive at the formula for the nth power of z: =reime=r(cosn+isin ne). (1) Clearly this is just an extension of De Moivre's formula,discussed in Example 3 of Sec.1.4. Equation(1)is an appealing formula for raising a complex number to a positive erpower.Il is casy to see that the identity is also valid for negative integers is an mth root of z satistying ses whether the formula will work for n=1/m,so that 3m=z. (2) Certainly if we define 5=relelm (3) (wheredenotes the customary,positive,o)we computea comple umber satisfying E()[as is easily seen by applyingE(But the ma er is more complicated than this;the number 1.for instance,has two square roots:1 and -1. And each of these has,in turn,two square roots-generating four fourth roots of 1, namely,1,1,and To see how the the scheme of thinget'swork outhe polar description of the equation=I for each of these numbers: 14=1e04=14e0=1, 4=(ae24=1e2m=l, (-1)4=(1e)4=14e4=1, (-i)1=(1e3m/2)4=14e6=1. It is instructive to trace the consecutive powers of these roots in the Argand diagram. Thus Fig1.1 shows before (1),()(-1)and ()go around twice:the powers of ()go around three times counterclockwise,and of course 1,12,13,and 14 never move
Complex Numbers 34 4e8 Powers o 2=-1 =2=3=4 Powers of=- Powers of= Figure 1.16 Sucessive powers of the fourth ootsof. Clearly,the multiplicity of roots is tied to the ambiguity in representing in polar form,setc Thus to compute all themth roots of aumberwe must apply formula(3)tovepolar representation ofFor the cube roots of ity,for example,we would compute as shown in the table opposite.Obviously the roots recur in sets of three,since3 whenevermm3. Geeralnwecn see that there are xactly mdistinct mthfuity.de- moted bym and they are given by 1/m=efkn/m cos 2k k=0,1,2,m-1). (4) The arguments of these roots are 2/m radians apart,and the roots themselves form the vertices of a regular polygon (Fig.1.17). Taking1inEq.(4)weobtain the roott “m=t2m=cos2 分+1sn2 and it is easy to see that the complete set of roots can be displayed as 1m,02,w- L2人&点6宝c1d华1o长一