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05Ahw1solution

# 05Ahw1solution - MATH[ILA rm 4 i gob(‘ﬁbn WW”— R4 4...

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Unformatted text preview: MATH [ILA rm 4% i gob/(‘ﬁbn WW”— R4 4. If z=1ii, than z2~2z+2n(1ii)2-—2(1ii)+2::i2im2\$2i+2=0ﬁ 1+2i 2—1” m (1+2i)(3+4i) +(2-i)(-5i)H——5+10i+ pm. (a) .+--—-—:-—— . . . . “5‘1“:2 ‘ 3m4z 5t (3m4z)(3+4z) (SIX-50 25 2.5 5 ! Si Si 5i 1 (1-—i)(2 w m3 —;) = (1—— 3i)(3 — i) m “10;: 2 (b) P . 2 . PU*1- (a) z1m21, 2223—: (C) Z; E(_3a1)a 2:: m: (114) EH 4. (a) Rewrite 32 ~— 1 + il=1 as 12 — (1 It is shown below~ —i)1:=1" This is the circle centered at I—i with radius 1” 5. (a) Write lz -4ii+lz+ 411:: 10 as izw 4il+iz ~— ("41312 10 to see that this is the locus of all points z such that the sum of the distances from z to 41' and ~4i is a constant. Such a curve is an ellipse with foci \$412 (1)) Write iz —ilmiz+il as 12: ~ll=lz -(wi)! to see that this is the locus of all points z such that the distance from 2 to 1 is always the same as the distance to win The curve is, then, the perpendicular bisector of the line segment from 1 to mi. RI?) . __ (I?) 121“ 'E'Jwiz; (c) (2+i)2=(§Tf)2=(2-i)2m4—4i+i2m4~4i-1=3-4i (d) |(22+5)(\/§wi)l=l2§f+5Hx5wil=lZz+5N2+1mx/ngz-t-SL 2. (a) Rewrite Rating) ”2 as Re[x+' .... “my #1)] 1*” 2, 01“ x = 2_ Th' - . , through the point z = 2, Shown below. IS is the vertical line PM I3 9“ (b) Rewﬁte lsziE: 4 as 2 i z w a i = 2“ This is the circle centeted at ~3- with 1' ———~ 34,0 2 21 r radius 2, shown below“ 8. In this problem, we shall use the inequalities (see Sec. 4) 1RezESIzI and Izl+zz+z3lslzli+lzzl+lsz Specifically, when 121\$ 1, iRe(2+z+z3)|s 52+§3+23l s 2+izi +5231: 2+Ezi+izE3S 2+1+ 1 u 4.. 15. Since 2:: 5-35— and y =%, the hyperbola x2 —y2 =1 can be written in the foliowing ways: + :1, 4 4 2.21”? _1 4 ‘I i at mar i—ar‘ "2—215 , ngm 25) g g( ) one value of arg( l J is 33-0375), or E Consequently, the principal value is e2 ._ 2: 2 4 4 é—TF—HZE, or wig. 4 4 (19) Since age/5 “if 2 We us), one value of arg(~\/§ - if is (SQ-g), or man So the principal value is “~71: + 27:, or 7L .2. (a) 1879'" 3., [0959+ {sweat “a, W6 1: l “’3‘” W _. - m.““ _-.~ 0 v9+ESTmCMQ (b) 6‘5 1 (cogEHLémB) " 9°59 wane—w eff; ) ) 5. We know from de Moivre's formula that (cos 9 + isin (9)3 = 00336 + isin39, or 0033 9 + 30031 9(isiu 9) + Boos 6(isin 9)2 + (isin 9)3 a 00339 + isinBB. That is, (cos3 9 ”3003 (:‘Isin2 6) + £60032 981m 9 —~ sin3 9) 2 00536 +isix139. By equaling real parts and then lmaginary parts here, we arrive at the desired uigonometxic identiﬁes: (a) 00336 = cos3 9 - 3eos liv‘sin2 B; ([7) sin 39 = 30052 95in 6 —~ sin3 6. 922 10. If Sml+z+zz+~-+zn, then 5-53 m<1+Z+Zz+‘”+Zn)—“(Z+Zz +zs+~---+z"+l) =2 1wz"+1" n+i lwz I , provided 2 #1. Thatis, w z Hence S =2 1 ~ Zn+1 1+z+zz+~~+zn 3 I—z Putting z = em (0 < 9 < 27:) in this identity, we have . . a . G 1_ei(n+i)9 beams—(2‘2 +«~r+e‘” 2W. lwela Now the real part of the 1eft~hand side here is evidentiy l+cos€+cos29+m+cosn9; and, to find the real pan of the right-hand side, we write that side in the form ex (—42) ex (wiEJMex [im(2n+l)6] l-exp[i(n+l)9] _ p 2 p 2 p 2 1 ._. 671909) exp(_i .2.) cxp£~i 2-] .., 6Xp(i g) which becomes cos—u—isin—mcos—wm 1 ———«--—-———— W1 . . 9 -’ ——213m-~— I 2 Of (2: #1). P2110 (COMM) The real pan of this is clearly 1 sin (21: +1)8 M... + MWJET__, 2 zanm“ 2 and we arrive at Lagrange 'S trigonometric identity: 8inan-+309 1+cosB+c0329+n~~+cosn6=%+W~—g§m—- (O< 9<27r). 28in“ 2 ...
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05Ahw1solution - MATH[ILA rm 4 i gob(‘ﬁbn WW”— R4 4...

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