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- Question : 1S - (a) (,fi -i)-i(l-..fii) = ..fi -i-i -..fi =-2i; (b) (2,-3)(-2,1) = (-4 + 3,6 + 2) = (-1,8); (e) (3,1)(3,-1)(} 1~) = (10,0>(1", 1 ~) = (2,1).
- Question : 2S - a) Re(iz)=Re[i(x+iy)]=Re(-y+ix)=-y=-Imz; (b) Im(iz) = Im[i(x + iy)] = Im(-y + ix) = x = Rez.
- Question : 3S - (1 + z)2 = (1 + z)(l + z) = (1 + z)
- Question : 4S - If z =
- Question : 5S - To prove tllat multiplication is commutati.ve, write Z1"2 = (x1,Yi)(XvY2) = (x,xz -Y
- Question : 6S - (a) To verify the associative law for addition, write (z1 +z2)+Z3 =[(Xi,Y1)+(x2,Y2)]+(x3,Y3)= (x1 +x2,Y1 + Y2)+(x3,y3) = ((X1 + Xz) + X3, (Y1 + Yz)+ Y3) =(X
- Question : 7S - Toe problem here is to solve the equation z2 + z + 1 = O for z = (x,y) by writing (x,y)(x,y) + (x,y) + (1,0) = (0,0). Since (x2 - y2 +X+ 1, 2.xy +y)= (0,0), it follows that x 2 -y2 + x + 1 = O and 2xy + y = O. By writing the second of these equations as (2x + l)y = O, we see that either 2x + 1 = O or y= O. If y= O, the first equation becomes x2 + x + 1 = O, which has no real roots (according to the quadratic formula). Hence 2x + 1 = O, or x = -1/2. In that case, the first equation reveals that y2 = 3/4, or y=
- Question : 1S - (a) 1 + 2i + 2 - i = (1 + 2i)(3 + 4i) + (2 - i)(-5i) = -5 + IOi + -5 - IOi = _ 2. 3- 4i Si (3- 4i)(3 + 4i) (5i)(-5i) 25 25 5' Si Si Si 1 (b) (1-i)(2-i)(3-i) = (1-3i)(3-i) = --1-0i = - 2; (e) (I-i)4 =[(1-i)(l-i)]2 =(-2i)2 =-4.
- Question : 2S - (a) (-I)z=-z since z+(-l)z=z[l+(-l)]=z
- Question : 3S - (Z.ZjXZjZ*) = ZJZjCZjZ,)] = z1[(z2z3)z
- Question : 4S - Z1Z =(~J(~)=(~Jz(.!.)=(~)(zz-1)=(~)
- Question : 1S - (a) z1 = 2i, (b) z,=(-V3,l), Z2=(V3,0) (e) Z1 = (-3,1), z2 = (1,4) Z
- Question : 2S - Inequalities (3), Sec. 4, are Rez S IRezl S lzl and Imz S llmzl S lzl. These are obvious if we write them as x
- Question : 3S - In order to verify the inequality "V21zl ~ IRezl + llmzl, we rewrite it in the following ways: "'12~x2 + y2 ~ lxl + lyl, 2(x2 + y2) ~ lxl2 + 21xllyl + lyl2, lxl2 -21xllyl + lyl2 ~ O, (lxl-lyl)2 ~ O. This last form of the inequality to be verified is obviously true since the left-hand side is a perfect square.
- Question : 4S - (a) Rewrite lz -1 + il= 1 as ~ -(1-i)I = l. This is the circle centered at 1-i with radius l. It is shown below.
- Question : 5S - (a) Write lz-4il+lz+ 4il= 10 as lz-4il+lz-(-4i)I= 10 to see that this is the locus of all points z such that the sum of the distances from z to 4i and -4i is a constant. Such a curve is an ellipse with foci
- Question : 1S - (a) z+3i=z+3i=z-3i; (b) iz = iz = -iz; (e) (2+i)2 =(2+i)2 =(2-i)2 =4-4i+i2 =4-4i-1=3-4i; (d) 1(2.z + 5)( "'2 - i)l=l2Z + 511"'2 - il=l2z + 5 h/2 + 1 = "1312z + 51.
- Question : 2S - (a) Rewrite Re{z - i) = 2 as Re[x + i(-y -1)] = 2, or x = 2. This is the vertical line through the point z = 2, shown below. (b) Rewrite 12z-il= 4 as 21z- ~ 1 = 4, or lz- ~ 1 = 2. This is the circle centered at i with radius 2, shown below.
- Question : 3S - w rite Z
- Question : 4S - (a) ZfaZs = (ZYZ2 )z3 = ZyZ2 Z3 = (*! Z2 )^3 = *1 *2 *3 J (b) Z4 = zV = z2z2 = zzzz = (z z)(z z) = zzzz = z4.
- Question : 5S
- Question : 6S - In this problem, we shall use the inequalities (see Sec. 4) Specifically, when lzl:S; l, 1Re(2 + z + z3)1 :S; 12 + z + z31 :S; 2+1.zl +lz31 = 2+1zl+lzl3 :S; 2 + 1 + 1 = 4.
- Question : 7S - First write z4 - 4z2 + 3 = (z2 - l)(z2 - 3). Then observe that when lzl= 2, and Thus, when lzl = 2, lz4 -4z2 + 31=1z2 -ll
- Question : 8S - (a) Prove that z is real ~ z = z. ( ) Suppose that z is real, so that z = x + iO. Toen z = x -iO = x + iO = z. (b) Prove that z is either real or pure imaginary ~ z2 = z2. () Suppose that z is either real or pure imaginary. If z is real, so that z = x, then z2 = x2 = z2. If Z is pure imaginary, so that Z = iy, then z2 = ( -iy )2 = (iy )2 = z2
- Question : 9S - (a) We shall use mathematical induction to show that (n = 2,3, ... ). This is known when n = 2 (Sec. 5). Assuming now that it is true when n = m, we may write Z1 + Z2 +
- Question : 10S - Toe identities (Sec. 5) zz =lzl2 and Rez = z + z enable us to write lz-z01= Ras 2 (z-z0 )(z-z0 ) = R2 , ZZ - ( ZZo + ZZo) + ZoZo = R2 , 1 zl2 - 2 Re( zz0 ) + 1 z/ = R2
- Question : 11S - Since x = z; z and y = z ;i z, the hyperbola x2 -y2 = 1 can be written in the following ways:
- Question : 1S - (a) Since 2z 2 + 2z 2 = 1, 4 z2 +z2 = 2. arg( i .)=argi-arg(-2-2i), -2-2z one value of arg( i .) is 1C -(- 3,c), or Stc. Consequently, the principal value is -2-2z 2 4 4 51C 31C --2,c or -- 4 ' 4
- Question : 2S - Toe solution 9 = ,r of the equation l/9 -11= 2 in the interval O :s; 9 < 2,r is geometrically evident if we recall that e;9 lies on the circle I zl = 1 and that 1 / 9 - 11 is the distance between the points e
- Question : 3S - We know from de Moivre's formula that (cos 9+isin 9)3 = cos39 +isin39, or cos3 9 + 3cos2 9(isin 9) + 3cos 9(isin 9)2 + (isin 9)3 = cos39 + isin39. Thatis, (cos3 9-3cos 9sin2 9)+i(3cos2 9sin 9-sin3 9) = cos39+ isin39. By equating real parts and then imaginary parts here, we arrive at the desired trigonometric identities: (a) cos39=cos3 9-3cos9sin2 9; (b) sin39=3cos2 9sin9-sin3 9.
- Question : 4S - Here z = re;9 is any nonzero complex number and na negative integer (n = -1,-2, ... ). Also, m = -n = l, 2,... . By writing and (z -l)m --[- 1e i (-9)]m -- (- 1) m ei (-m9) -_ -1e i (-m9) , r r ,m we see that (zmr1 = (z-1r. Thus the definition zn = (z-1r can also be written as zn = (zmrl.
- Question : 5S - First of ali, given two nonzero complex numbers z1 and z2 , suppose that there are complex numbers c1 and c2 such that z1 = c1c2 and Zi = c1c2
- Question : 1S - (a) Since 2i = 2exp[{; + 2kn)] (k = 0,
- Question : 2S - (a) Since -16=16exp[i(n+2kn)] (k=0,
- Question : 3S - (a) By writing -1 = lexp[i(n+ 2kn)] (k = 0,
- Question : 4S - (a) Let a denote any fixed real number. In order to find the two square roots of a+ i in exponential fonn, we write A=la+il ="1a2 +1 and a=Arg(a+i). Since a+ i = Aexp[i(a + 2kn)] (k = 0,
- Question : 5S - The four roots of the equation z4 + 4 = O are the four fourth roots of the number -4. To find those roots, we write -4 = 4exp[i(n+ 2kn)] (k = 0,
- Question : 6S - Let e be any nth root of unity other than unity itself. With the aid of the identity (see Exercise 10, Sec. 7), we find that
- Question : 7S - Observe first that and ( -1)1/m * i(-(} + 2kn) 1 i(-{}) i(2kn) Z = -exp = mCexp--exp , r m '\/T m m where k = O,l,2, ... ,m-1. Since the set is the same as the set i(-2kn) exp--..;.. m i(2k1C) exp----'- m but in reverse order, we find that (z11mr 1 = (z-1 tm.
- Question : 1S - (a) Write lz-2 + ilS 1 as lz -(2 -i)IS 1 to see that this is the set of points inside and on the circle centered at the point 2 - i with radius l. It is nota domain. y o X (b) W rite l 2z + 31 > 4 as I z -( - ~ )1 > 2 to see that the set in question consists of all points exterior to the circle with center at - 3/2 and radius 2. It is a domain. Write Imz > l as y> l to see that this is the half plane consisting of all points lying above the horizontal line y = l. It is a domain. o X (d) Toe set Imz = l is simply the horizontal line y= l. lt is nota domain. y y=l o X (e) Toe set O~ argz ~ : (z :1: O) is indicated below. It is nota domain. y 0 X (fJ Toe set I z - 41~ 1 zl can be written in the form (x - 4 )2 + y2 ~ x2 + y2, which reduces to x ~ 2. This set, which is indicated below, is not a domain. The set is also geometrically evident since it consists of all points z such that the distance between z and 4 is greater than or equal to the distance between z and the origin. X
- Question : 2S - (a) The closure of the set -n < argz < n (z :t: O) is the entire plane. (b) We first write the set IRezl O, or lyl> O. Hence the closure of the set IRezl O can be written as y2 < x2 , or lyl
- Question : 3S - Toe set S consists of all points z such that lzl< 1 or lz-21< 1, as shown below. y X Since every polygonal line joining z1 and z2 must contain at least one point that is not in S, it is clear that S is not connected.
- Question : 4S - We are given that a set S contains each of its accumulation points. The problem here is to show that S must be closed. We do this by contradiction. We let Zo be a boundary point of S and suppose that it is nota point in S. The fact that z0 is a boundary point means that every neighborhood of z0 contains at least one point in S; and, since z0 is not in S, we see that every deleted neighborhood of S must contain at least one point in S. Thus Zo is an accumulation point of S, and it follows that z0 is a point in S. But this contradicts the fact that Zo is not in S. We may conclude, then, that each boundary point Zo must be in S. That is, S is closed.

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