Solution Set I for 18.075-Fall 2004

SOLUTION SET I FOR 18.075–FALL 2004 10. Functions of a Complex Variable 10.1. Introduction. The Complex Variable. . 3. Establish the following results: (a) Re(z1 + z2 ) = Re(z1 ) + Re(z2 ), but Re(z1 z2 ) ≤= Re(z1 )Re(z2 ) in general; (b)Im(z1 + z2 ) = Im(z1 ) + Im(z2 ), but Im(z1 z2 ) ≤= Im(z1 )Im(z2 ) in general; (c)|z1 z2 | = |z1 ||z2 |, but |z1 + z2 | = ≤ |z1 | + |z2 | in general; (d)z1 + z2 = z1 + z2 and z1 z2 = z1 z2 . Solution. (a) We want to show that Re(z 1 + z2 ) = Re(z1 ) + Re(z2 ). Let z1 = a1 + ib1 , z2 = a2 + ib2 , then z1 + z2 = (a1 + a2 ) + i(b1 + b2 ). Hence Re(z1 + z2 ) = a1 + a2 and clearly Re(z1 ) + Re(z2 ) = a1 + a2 . Let us show that in general Re(z1 z2 ) ≤= Re(z1 )Re(z2 ). We have therefore z1 z2 = (a1 + ib1 )(a2 + ib2 ) = (a1 a2 − b1 b2 ) + i(a1 b2 + a2 b1 ), On the other hand Re(z1 z2 ) = a1 a2 − b1 b2 . Re(z1 )Re(z2 ) = a1 a2 ≤= a1 a2 − b1 b2 in general. (b) We want to show that Im(z1 + z2 ) = Im(z1 ) + Im(z2 ). From part (a) we have that z1 + z2 = (a1 + a2 ) + i(b1 + b2 ). Hence Im(z1 + z2 ) = b1 + b2 and clearly Im(z1 ) + Im(z2 ) = b1 + b2 . Let us show that in general Im(z1 z2 ) ≤= Im(z1 )Im(z2 ).We have therefore z1 z2 = (a1 + ib1 )(a2 + ib2 ) = (a1 a2 − b1 b2 ) + i(a1 b2 + a2 b1 ), Im(z1 z2 ) = a1 b2 + a2 b1 . Date: September 13, 2004. 1 2 SOLUTION SET I FOR 18.075–FALL 2004 On the other hand Im(z1 )Im(z2 ) = b1 b2 ≤= a1 b2 + a2 b1 in general. (c) We want to show that |z1 z2 | = |z1 ||z2 |. From part (a) we have Hence z1 z2 = (a1 + ib1 )(a2 + ib2 ) = (a1 a2 − b1 b2 ) + i(a1 b2 + a2 b1 ). � (a1 a2 − b1 b2 )2 + (a1 b2 + a2 b1 )2 � = (a21 a22 − 2a1 a2 b1 b2 + b21 b22 ) + (a12 b22 + 2a1 b2 a2 b1 + a22 b21 ) � = a21 a22 + b21 b22 + a21 b22 + a22 b21 . � � On the other hand |z1 | = a21 + b21 and |z2 | = a22 + b22 . Therefore � � � � 2 2 2 2 2 2 2 2 |z1 ||z2 | = ( a1 + b1 )( a2 + b2 ) = (a1 + b1 )(a2 + b2 ) = a21 a22 + b21 b22 + a21 b22 + a22 b21 |z1 z2 | = which is equal to |z1 z2 |. Let us show that |z1 + z2 | ≤= |z1 | + |z2 | in general. From part (a) we have z1 + z2 = (a1 + a2 ) + i(b1 + b2 ). Hence On the other hand |z1 + z2 | = � |z1 | + |z2 | = and (a1 + a2 )2 + (b1 + b2 )2 . � (a1 + a2 )2 + (b1 + b2 )2 � � (a1 + a2 )2 + (b1 + b2 )2 ≤= (a1 + a2 )2 + (b1 + b2 )2 in general (choose for example a1 = 1, b1 = 0, a2 = 0, b2 = 1). (d) We want to show that z1 + z2 = z1 + z2 . From part (a) we have z1 + z2 = (a1 + a2 ) + i(b1 + b2 ) then On the other hand z1 + z2 = (a1 + a2 ) − i(b1 + b2 ). z1 + z2 = (a1 − ib1 ) + (a2 − ib2 ) = (a1 + a2 ) − i(b1 + b2 ) which is equal to z1 + z2 . Let us show that z1 z2 = z1 z2 . From part (a) we have that z1 z2 = (a1 a2 − b1 b2 ) + i(a1 b2 + a2 b1 ). Hence On the other hand, z1 z2 = (a1 a2 − b1 b2 ) − i(a1 b2 + a2 b1 ). z1 z2 = (a1 − ib1 )(a2 − ib2 ) = (a1 a2 − b1 b2 ) − i(a1 b2 + a2 b1 ) which is equal to z1 z2 . SOLUTION SET I FOR 18.075–FALL 2004 4. Establish the following results: (a)z + z = 2Rez, (b)z − z = 2iImz, (c)z1 z2 + z1 z2 = 2(z1 z2 ) = 2Re(z1 z2 ), (d)Rez � |z |, (e)Imz � |z |, (f)|z1 z2 + z1 z2 | � 2|z1 z2 |, (g)(|z1 | − |z2 |)2 � |z1 + z2 |2 � (|z1 | + |z2 |)2 . [Use part (f).] Solution.(a) Let z = a + ib, then z + z = (a + ib) + (a − ib) = 2a = Rez; (b) Let z = a + ib, then z − z = (a + ib) − (a − ib) = 2ib = 2Imz; (c) Let z1 = a1 + ib1 and z2 = a2 + ib2 , then z1 z2 = (a1 + ib1 )(a2 − ib2 ) = (a1 a2 + b1 b2 ) + i(b1 a2 − a1 b2 ), z1 z2 = (a1 − ib1 )(a2 + ib2 ) = (a1 a2 + b1 b2 ) − i(b1 a2 − a1 b2 ) and z1 z2 + z1 z2 = 2(a1 a2 + b1 b2 ). Hence 2Re(z1 z2 ) = 2Re(z1 z2 ) = z1 z2 + z1 z2 = 2(a1 a2 + b1 b2 ); (d) Let z = a + ib, then |z| = � � a2 + b2 � a2 = |a| � a = Rez; (e) Let z = a + ib, then |z| = � a2 + b2 � � b2 = |b| � b = Imz; (f)Let z1 = a1 + ib1 and z2 = a2 + ib2 , then using part (c) and part (d) we get, |z1 z2 + z1 z2 | = 2|Re(z1 z2 )| � 2|z1 z2 |; (g) Let z1 = a1 + ib1 and z2 = a2 + ib2 , then (|z1 | − |z2 |)2 = (|z1 |2 + |z2 |2 ) − 2|z1 ||z2 |; |z1 + z2 |2 = = = = (z1 + z2 )(z1 + z2 ) (z1 + z2 )(z1 + z2 ) z 1 z1 + z 1 z2 + z 2 z1 + z 2 z2 (|z1 |2 + |z2 |2 ) + (z1 z2 + z2 z1 ); (|z1 | + |z2 |)2 = (|z1 |2 + |z2 |2 ) + 2|z1 ||z2 |. 3 4 SOLUTION SET I FOR 18.075–FALL 2004 To simplify let A = (|z1 |2 +|z2 |2 ). We ant to show that(|z1 |−|z2 |)2 � |z1 +z2 |2 � (|z1 |+|z2 |)2 . The above identities imply that this is equivalent to showing A − 2|z1 z2 | � A + (z1 z2 + z2 z1 ) � A + 2|z1 z2 |. Hence we have to prove that −2|z1 z2 | � (z1 z2 + z2 z1 ) � 2|z1 z2 |. Part (f) implies that moreover it is always true that |z1 z2 + z1 z2 | � 2|z1 z2 |, −|z1 z2 + z1 z2 | � (z1 z2 + z1 z2 ) � |z1 z2 + z1 z2 |. Thus we conclude that −2|z1 z2 | � −|z1 z2 + z1 z2 | � (z1 z2 + z1 z2 ) � |z1 z2 + z1 z2 | � 2|z1 z2 | which is what we wanted to show. 5. Express the following quantities in the form a + ib, where a and b are real: 1+i , (c) e�i/2 , (a) (1 + i)3 , (b) 1−i � (d) e2+�i/4 , (e) sin( 4 + 2i), (f) cosh(2 + �i 4 ). Solution.(a) (1 + i)3 = 1 + 3i + 3i2 + i3 = 1 + 3i − 3 − i = −2 + 2i; 1+i 1+i 1+2i+i2 2i (b) 1−i = 1+i 1−i ( 1+i ) = 1−i2 = 2 = 0 + i; (c) e�i/2 = cos �2 + i sin �2 = 0 + i; � � � � �i �i (d)e2= 4 = e2 (e 4 ) = e2 (cos �4 + i sin �4 ) = e2 ( 2/2 + i 2/2) = e2 2/2 + ie2 2/2; � � (e)sin( �4 + 2i) = (sin �4 cosh 2) + i(cos �4 sinh 2) = cosh 2/ 2 + i sinh 2/ 2; � � � � (f)cosh(2 + �i 4 ) = (cosh 2 cos 4 ) + i(sinh 2 sin 4 ) = cosh 2/ 2 + i sinh 2/ 2. 9. Prove that ez possesses no zeros, that the zeros of sin z and cos z all lie on the real axis, and that those of sinh z and cosh z all lie on the imaginary axis. Solution. We have to show that ez ≤= 0 for all z. Let z = x + iy, then ez = e(x+iy) = ex eiy = ex (cos y + i sin y). Since ex ≤= 0 for all x real, ez = 0 if and only if cos y = sin y = 0 for a real number y, which is not possible. Hence ez ≤= 0 for all z. Now we want to prove that the zeros of sin z and cos z are real. Let z = x + iy, then (identity (32) pg 544) sin(x + iy) = sin x cosh y + i(cos x sinh y). Therefore sin(x + iy) = 0 if and only if (1) sin x cosh y = 0 SOLUTION SET I FOR 18.075–FALL 2004 5 and (2) cos x sinh y = 0. Observe that ey + e−y ≤ 0, = 2 for all y, therefore if (1) holds, we must have sin x = 0 which implies cos x = ≤ 0. Thus, if (2) holds, we must have ey − e−y sinh y = =0 2 which implies y = 0. Hence the zeros of sin z are real. Analogously, (identity (32) pg 544) cosh y = cos(x + iy) = cos x cosh y + i(sin x sinh y). Therefore if we assume cos(x + iy) = 0, we must have (1)� cos x cosh y = 0 and (2)� sin x sinh y = 0. ≤ 0, for all y if (1)’ holds, we must have cos x = 0 which implies sin x = ≤ 0. Since cosh y = Thus, if (2)’ holds, we must have sinh y = 0 which implies y = 0.Hence the zeros of cos z are real. Finally we want to show that the zeros of sinh z and cosh z are imaginary. The following identity (identity (32)pg 544) sinh(x + iy) = sinh x cos y + i(cosh x sin y). implies that sinh(x + iy) = 0 if and only if (3) sinh x cos y = 0 and (4) cosh x sin y = 0. ≤ 0, for all x, (4) implies sin y = 0 and so cos y ≤= 0. Hence, if (3) holds, Since cosh x = we must have sinh x = 0 which implies x = 0.Hence the zeros of sinh z are imaginary. Analogously, (identity (32) pg 544) cosh(x + iy) = cosh x cosh y + i(sinh x sin y). Therefore cosh(x + iy) = 0 if and only if (3)� cosh x cos y = 0 and (4)� sinh x sin y = 0. Since cosh x = ≤ 0, for all x, (3)’ implies that cos y = 0 and so sin y = ≤ 0. Thus, if (4)’ holds, we must have sinh x = 0 which implies x = 0.Hence the zeros of cosh z are imaginary. 6 SOLUTION SET I FOR 18.075–FALL 2004 10.3. Other Elementary Functions. . 12. Show that the nth roots of unity are of the form � nk (k = 0, 1, . . . , n − 1), where �n = cos(2�/n) + i sin(2�/n). Solution. We need to solve the equation z n = 1. The principal value of the argument � of unity is �P = 0. Hence, from the formula derived in class, �k z = |1|1/n ei(�P +2k�)/n = ei2�/n = (�n )k , k = 0, 1, 2, . . . , n − 1, where �n = ei2�/n = cos(2�/n) + i sin(2�/n). 13. Determine all possible values of the following quantities in the form a + ib, and in each case give also the principal value, assuming the definition (39): 1 3 (a) log (1 + i), (b) (i) 4 , (c) (1 + i) 2 . � Solution. (a) log (1 + i) = log 2 + i(2k� + �4 ), where k is an integer. The principal value � is log 2 + i �4 . � � 4 3 3� 3� k� k� 3� k� (b) (i) 4 = 4 (i)3 = ei( 2 +2k�) = ei( 8 + 2 ) = cos ( 3� 8 + 2 ) + i sin ( 8 + 2 ), where k = 0, 1, 2, 3. The principal value is cos 38� + i sin 3� 8�. � 1 1 1 log (1+i) (log 2+i(2k�+ �4 )) 2 2 2 (c) (1 + i) = e =e = 4 2(cos ( �8 + k�) + i sin ( �8 + k�)), where � k = 0, 1. The principal value is 4 2(cos �8 + i sin �8 ).