Chapter 3 Elementary Functions

Exercises Pages 89-901. Show that(a) exp (2� 3�i) = �e2;(b) exp

�2+�i4

�=p

e2 (1 + i) ;

(c) exp (z + �i) = � exp z:

Solution (1a): e2�3�i = e2e�3�i = e2 [cos (�3�) + i sin (�3�)]= e2 [cos (3�)� i sin (3�)] = e2 (1� i0) = �e2Solution (1b): e

2+�i4 = e

12 ei

i4 =

pe�cos �4 + i sin

�4

�=peh1p2+ i 1p

2

i=p

e2 (1 + i)

Solution (1c): ez+�i = ezei� = ez (cos� + i sin�) = ez (�1 + i0) = �ez:3. Use the Cauchy-Riemann equations and the theorem in Section 20 to

show that the function f(z) = exp z is not analytic anywhere.Solution: ez = exe�iy = ex cos y � iex sin y: C-R does not hold.6. Show that

��exp �z2��� � exp�jzj2� :Solution:

��exp �z2��� = ��exp �x2 � y2�+ i2xy��=��exp �x2 � y2��� jexp (i2xy)j

=��exp �x2 � y2���

= exp�x2 � y2

�:

Since exp is a monotonically increasing function, we see that this is �exp

�x2 + y2

�= exp

���z2��� ; since x2 � y2 � x2 + y2:8. Find all values of z such that(a) ez = �2;(b) ez = 1 +

p3i;

(c) exp(2z-1) = 1.Solution 8b: ez = exeiy = 2ei�=3 so x = ln 2 and y = �=3 + 2n� where n is

any integer.

Solution (8a): z = ln j�2j+ arg (�2) = ln (2) + i (� + 2n�) ; n 2 Z:Solution (8b): z = ln

��1 +p3i��+i arg �1 +p3i� = ln (2)+i (�=3 + 2n�) ; n 2Z:Solution (8c): 2z � 1 = ln j1j + arg (1) = arg (1) = 2n�i; n 2 Z: Thus

z = 1=2 + n�i; n 2 Z:

10. (a) Show that if ez is real, then Im z = n� (n = 0;�1;�2; : : :) :

(b) If ez is pure imaginary, what restriction is placed on z?

Solution (10a): z = x + iyez = exeiy = ex (cosx+ i sin y)

1

ez is real =) sin y = 0 =) y = n� (n = 0;�1;�2; : : :)Solution (10b): z = x + iyez = exeiy = ex (cosx+ i sin y)ez is pure imaginary=) cosx = 0 =) x = Re z = �

2 (2n+ 1) (n = 0;�1;�2; : : :) :11. Describe the behavior of ez = exeiy as (a) x tends to -1; (b) y tends

to 1:

Solution (11a): Since jezj = jexj ; so jezj ! 0 as x ! �1, which meansez ! 0 as x! �1:Solution (11b): If x is �xed and y !1, then jezj = ex is �xed, but arg (ez)

increases continuously. We see that ez surrounds the circle fjzj = exg in thecounterclockwise direction.

Exercises Pages 94-951. Show that (a) Log (�ei) = 1� �

2 i;(b) Log (1� i) =12 ln 2�

�4 i:

Solution (1a): Log (�ei) = ln (e) + iArg (�ei) = 1� �=2i

Solution (1a): z = �ei = e (�i) = ee�i�2r = e; � = ��

2Log (�ei) = ln e� i�2 = 1�

�2 i

Solution (1b): z = 1� i =p2e�i

�4

r =p2; � = ��

4

Log (1� i) = lnp2� �

4 i =12 ln 2�

�4 i:

2. Verify that when n = 0;�1;�2; � � � ;(a) log e = 1 + 2n�i;(b) log i =

�2n+ 1

2

��i;

(c) log��1 +

p3i�= ln 2 + 2

�n+ 1

3

��i:

Solution (2a): Since ln jej = 1 and Arg (e) = 0, so log e = 1+ i (0 + 2n�) =1 + 2n�i; n 2 Z:Solution (2b): Since ln jij = 0 and Arg (i) = �=2, solog i = 0 + i (�=2 + 2n�) = (2n+ 1=2)�i; n 2 Z:Solution (2c): Since ln

���1 +p3i�� = ln 2 and Arg��1 +

p3i�= 2�=3, so

log��1 +

p3i�= ln 2 + i (2�=3 + 2n�) = ln 2 + 2 (n+ 1=3)�i:

3. Show that(a) Log (1 + i)2 = 2Log (1 + i) ;(b) Log (�1 + i)2 6= 2Log (�1 + i) :

Solution (3a): Log (1 + i)2= Log (2i) + ln (2) + i�=2 while 2Log (1 + i) =

2�lnp2 + i�=4

�= ln (2) i�=2:

Solution (3b): Log (�1 + i)2 = Log (�2i) = ln (2)�i�=2 while 2Log (�1 + i) =2�lnp2 + i3�=4

�and these are not equal.

4. Show that

2

(a) log�i2�= 2 log i when log z = ln r + i�

�r > 0; �4 < � <

9�4

�;

(b) log�i2�6= 2 log i when log z = ln r + i�

�r > 0; 3�4 < � <

11�4

�:

Solution: In part (a), log z is de�ned with �=4 < � < 9�=4: Thereforelog (�1) = i�; log (i) = i�=2; and log

�i2�= 2 log i:

In part (b), log z is de�ned with 3�=4 < � < 11�=4: Therefore log (�1) =i�; log (i) = 10i�=4; and log

�i2�6= 2 log i:

9. Show that(a) the function Log (z � i) is analytic everywhere except on the half line y

= 1 (x � 0) ;(b) the function Log(z+4)

z2+i is analytic everywhere except at the points

� (1� i) =p2 and on the portion x � �4 of the real axis.

Solution (9a): If z is not on the half line y = 1, x � 0 then z-i is not onnonpositive real axis. This means that Log (z � i) is analytic on the complementof this half line.

Solution: (a): Log (z � i) is de�ned and analytic provided (z � i) is not 0or a negative real number. This condition holds everywhere except of the halfline y = 1 and x � 0:(b): The function Log(z+4)

(z2+i) is analytic provided the numerator and denom-inator are analytic and the denominator is nonzero. The numerator is analyticif z+4 is not on the negative real line and not zero. Therefore z is not allowed tobe on the portion x � �4 of the real line. The denominator is always analyticand is zero if z2 = i: Using the polar form of i, we can �nd the two square rootsof i to be � (1� i) =

p2: Therefore these two values of z are also disallowed.

10. Show in two ways that the function ln�x2 + y2

�is harmonic in every

domain that does not contain the origin.

Solution:For the �rst way, we directly check via the de�nition.

@x ln�x2 + y2

�= 2x

(x2+y2) ; @2x ln

�x2 + y2

�= 2

(y2�x2)(x2+y2)2

:

@y ln�x2 + y2

�= 2y

(x2+y2) ; @2y ln

�x2 + y2

�= 2

(x2�y2)(x2+y2)2

:

Thus, clearly, if H = ln�x2 + y2

�; we have that Hxx + Hyy = 0 provided

x2 + y2 6= 0: That is, that H is harmonic away from (0,0).For a constant c 6= 0; it is clear that H is harmonic if and only if cH

is harmonic. Thus, it will su¢ ce to show that H (x; y) = 12 ln

�x2 + y2

�=

lnpx2 + y2: It is clear, however, that H (x; y) = Re f (z) with f (z) = log (z) :

If we choose an � branch, we see that f is analytic away from the branch cut.Thus, H is harmonic there. We may, however, choose a di¤erent branch andsee that H is still the real part, and thus, is harmonic away from the new branchcut. The only point in common to every branch cut is the branch point (0,0).

3

Exercises Pages 96-97Solution: z = rei�

Log (z) = ln jzj+ iArg (z)z1 = jz1j eiArg(z1) = r1ei�1 ; z2 = jz2j eiArg(z2) = r2ei�2 (�� � �1 � �;�� � �2 � �)Log (z1) = ln r1 + i�1Log (z2) = ln r2 + i�2Log (z1) + Log (z2) = ln r1 + ln r2 + i (�1 + �2) = ln r1r2 + i (�1 + �2)z1z2 = r1r2e

i(�1+�2)

Log (z1z2) = ln jz1z2j+ iArg (z1z2)

Arg (z1z2) =

8<: �1 + �2;�� � �1 + �2 � ��1 + �2 � 2�; � < �1 + �2 � 2��1 + �2 + 2�;�2� � �1 + �2;��

9=; :That is Arg (z1z2) = �1 + �2 + 2�N (N has one of the values 0,� 1) :HenceLog (z1z2) = ln r1r2 + i (�1 + �2) + i (2�N) = Log (z1) +Log (z2) + i (2�N) :4. By choosing speci�c nonzero values of z1 and z2, show that expression

(4), Section 31, for log (z1=z2) is not always valid when log is replaced by Log.

Solution: Choose z1 = �i; z2 = i. Then Log (z1)�Log (z2) = ��=2��=2 =�� while Log (z1=z2) = Log (�1) = �:

Exercises Pages 99-1001. Show that when n = 0;�1;�2; � � � ;(a) (i+ 1)i = exp

���4 + 2n�

�exp

�i2 ln 2

�;

(b) (�1)1=n = e(2n+1)i:

Solution (1a): (1 + i)i = ei log(1+i) = ei(lnp2�arg(1+i)) = ei ln(2)=2e��=4+2n�:

Solution (1a): Note that log (1 + i) = ln j1 + ij + i arg (1 + i) = ln�p2�+

i (�=4 + 2n�) = ln(2)2 + i (�=4 + 2n�) ; for n 2 Z: Thus (1 + i)i =

exp (i log (1 + i)) = exp�� (�=4 + 2n�) + i ln(2)2

�=

exp (��=4� 2n�) exp�i ln(2)2

�:

Solution (2b): From log (�1) = ln j�1j+ i arg (�1) = i arg (�1) =i (� + 2n�), we have (�1)1=� = exp (log (�1) =�) = exp (i (� + 2n�) =�) =

exp (i (2n+ 1)) :

2. Find the principal value of (a) ii;(b)�e2

��1�

p3i��3�i

;(c) (1� i)4i :

Solution (2a): ii =�cos �2 + i sin

�2

�i=�ei

�2

�i= e

�2

Solution (2b):�e2

��1�

p3i��3�i

=he (�1)

�12 +

p32 i�i3�i

=�ee�i�ei

�3

�3�i=hee�i

2�3

i3�i= e2�

2

e3�i = �e2�2

4

Solution (2c): (1� i)4i =�p2e�i

�4

�4i=�eln

p2e�i

�4

�4i=�e12 ln 2e�i

�4

�4i= e�e

i2 ln 2

= e��cos

�2p2�+ i sin

�2p2��:

5. Show that the principal nth root of a nonzero complex number z0, de�nedin Section 8, is the same as the principal value of z1=n0 , de�ned in Section 32.

Solution: Write z0 = r0ei�0 ; where �0 = Argz0: Then, the principal nth

root is npr0e

i�0=n: On the other hand, using the de�nition in Section 32, wehaveP:V:z

1=n0 = P:V: exp [(1=n) log z0]

= exp [(1=n) (ln r0 + i�0)]= exp [(1=n) ln r0] exp (i�0=n) = n

pr0e

i�0=n:

6. Show that if z 6= 0 and a is a real number, then jzaj = exp (a ln jzj) =

jzja ;where the principal value of jzja is to be taken.

Solution: The logarithm of a complex number z is de�ned as every complexnumber w which satis�es the equation ew = z: The complex logarithm isde�ned for all z 6= 0: This is denoted by log (z) = wza = eln(z

a) = ea ln z = ea(lnjzj+iArg(z)) = ea lnjzjeiaArg(z)

jzaj = ea lnjzj = alnjzja = jzja :

Exercises Pages 103-1042. Show that Euler�s formula (Section 6) continues to hold when � is replaced

by z: eiz = cos z + i sin z:Suggestion: To verify this, start with the right-hand side.Solution: cos (z) + i sin (z) =

�eiz + e�iz

�=2 + i

�eiz � e�iz

�=2i

=�eiz � ieiz

�=2 +

�eiz + ie�iz

�=2 = eiz:

5. (a) Verify the identity sin2 z + cos2 z = 1 using the identity

cos (z1 + z2) = cos z1 cos z2� sin z1 sin z2 and the relations sin (�z) = � sin zand cos (�z) = cos z; (b) Verify the identity sin2 z + cos2 z = 1 using the factthat the entire function f (z) = sin2 z + cos2 z � 1 has zero values along the xaxis and the following lemma.Lemma: Suppose that(i) a function f is analytic throughout a domain D;(ii) f(z) = 0 at each point z of a domain or line segment contained in D.Then f (z) � 0 in D; that is , f(z) is is identically equal to zero throughout

D.Solution 5a: (sin z)2 + (cos z)2 = 1The number z = x + iy is a complex number. The expression

5

(sin z)2+ (cos z)

2= 1 is written as sin2 (x+ iy) + cos2 (x+ iy) = 1:

sin (x+ iy) = sinx (cosh y) + i cosx (sinh y)cos (x+ iy) = cosx (cosh y)� i sinx (sinh y)sin2 (x+ iy) = (sinx cosh y)

2+ (i cosx sinh y)

2+ 2i sinx cosh y cosx sinh y

cos2 (x+ iy) = (cosx cosh y)2+ (i sinx sinh y)

2 � 2i cosx cosh y sinx sinh ysin2 (x+ iy) + cos2 (x+ iy) = (sinx cosh y)

2 � (cosx sinh y)2

+(cosx cosh y)2 � (sinx sinh y)2

sin2 (x+ iy) + cos2 (x+ iy) = sin2 x�cosh2 y � sinh2 y

�+cos2 x

�cosh2 y � sinh2 y

�sin2 (x+ iy) + cos2 (x+ iy) = sin2 x+ cos2 xsin2 (x+ iy) + cos2 (x+ iy) = 1

This completes the proof that (sin z)2 + (cos z)2 = 1:Note: cosh2 y� sinh2 y = 1 and sin2 x+cos2 x = 1 because x and y are realnumbers.

Solution 5b: The identity sin2 z + cos2 z = 1 can be derived by using thefunction f (z) = sin2 z + cos2 z � 1: We replace sin2 2z + cos2 2z by 1, sof (z) = 1� 1 = 0: Since f(2) has a zero value, f(z) is analytic throughout a

domain D. The function f(z) will be zero for all values of z along the z-axis.

Solution:(a) Putting z1 = z and z2 = �z in the identitycos (z1 + z2) = cos z1 cos z2 � sin z1 sin z2; we havecos z cos (�z)� sin z sin (�z) = cos (z � z))cos z � cos z + sin z � sin z = cos 0; since sin (�z) = � sin z andcos (�z) = cos z ) sin2 z + cos2 z = 1:(b) Let f (z) = sin2 z + cos2 z � 1 for all z 2 C:We have f 0 (x) = sin2 z + cos2 z � 1 for all z 2 C: We havef 0 (x) = 2 sinx cosx+ 2 cosx � (� sinx)� 0 = 0 for all x 2 R: Therefore we

have f(x) = C, where C is a constant. We have C = f (0) = sin2 0+cos2 0�1 =0 + 1� 1 = 0: Thus we have f(x) = 0 for all x 2 R: Let D = C. Then R is aline contained in D. We have just seen that f(z) = 0 at each point z of the realline R contained in D. Thus f(z) satis�es the condition (ii) of the lemma givenin the question. Also we know that sin z and cos z are analytic functions ateach point of C. Since the square of the analytic function is analytic and thesum of analytic functions are analytic, we have that f(z) is also analytic at eachpoint of C. Therefore f(z) also satis�es the condition (i) of the lemma given inthe question. Hence we have that f(z) = 0 for all z 2 D = C; which impliesthat sin2 z + cos2 z = 1 for all z 2 C:

10. Point out how it follows from the expressions

jsin zj2 = sin2 x + sinh2 y; jcos zj2 = cos2 x + sinh2 y for jsin zj2 and jcos zj2that (a) jsin zj � jsinxj ; (b) jcos zj � jcosxj :

6

Solution:��sin2 z�� = sin2 x+ sinh2 y��cos2 z�� = cos2 x+ sinh2 yThe real sine has the property that jsinxj � 1 for all real x. The real cos

has the property that jcosxj � 1 for all real x. The sin x and cos x are boundedregions with a modulus between 0 and 1.(a) So jsin zj2 has the additonal term sinh2 y: If we let z = x + iy in the

identity��sin2 z�� = sin2 x+ sinh2 y; then

limy!1

jsin (x+ iy)j2 = sin2 x+ limy!1

sinh2 y =1:The function jsin zj is unbounded and greater than jsinxj ; where x is a real

number.(b) So jcos zj2 has the additonal term sinh2 y: If we let z = x +iy in the

identity��cos2 z�� = cos2 x+ lim

y!1sinh2 y =1:

The function jcos zj is unbounded and greater than jcosxj ; where x is a realnumber.11. With the aid of expressions (15) and (16) in Section 33 for jsin zj2 and

jcos zj2, show that(a) jsinh yj � jsin zj � cosh y;(b) jsinh yj � jcos zj � cosh y:

Solution: First note that since sinh 0 = 0 and cosh 0 - 1, these yield theusual bounds for sin and cos when z is replaced by a real number.(a) The lower bound follows easily, jsin zj2 = sin2 x + sinh2 y � sinh2 y:

To get the upper bound, we write jsin zj2 = sin2 x + sinh2 y =�sin2 x� 1

�+�

sinh2 y + 1�= � cos2 x+ cosh2 y � cosh2 y; which yields the result.

(b) The proof here is almost identical. We only need to interchange sinand cos.12. (a) Use the de�nitions sin z = (eiz�e�iz)

2i ; cos z =(eiz+e�iz)

2 toshow that 2 sin (z1 + z2) sin (z1 � z2) = cos 2z2 � cos 2z1:(b) With the aid of the identity obtained in part (a), show that if cos z1 =

cos z2, then at least one of the numbers z1+z2 and z1�z2 is an integral multipleof 2�:

Solution (12a): 2 sin (z1 + z2) sin (z1 � z2)� 2(2i)(2i)

�ei(z1+z2) � e�i(z1+z2)

� �ei(z1�z2) � e�i(z1�z2)

�= � 1

2

�eiz1 � eiz2 � e�i2z2 + e�i2z1

�= � 1

2 (2 cos (2z1)� 2 cos (2z2)) = cos (2z2)� cos (2z1) :Solution (12b): From (a), we have cos (z2)� cos (z1) =2 sin

�z1+z22

�sin�z1�z22

�: So if cos (z2) = cos (z1), then either sin

�z1+z22

�= 0

or sin�z1�z22

�= 0: In the �rst case, we have z1+z2

2 = k� for some k 2 Z, whichimplies z1+ z2 = 2k�: In the second case, we have z1+z2

2 = k� for some k 2 Z,which implies z1 � z2 = 2k�:

7

16. Show that(a) cos (iz) = cos (iz) for all z;(b) sin (iz) = sin (iz) if and only if z = n�i (n = 0;�1;�2; : : :) :

Solution:(a) Show that cos (iz) = cos (iz) for all z.cos (iz) = cos [i (x+ iy)] = cos (xi� y)= cos (ix) cos y + sin y sin (ix) = coshx cos y + i sin y sinhx= coshx cos y � i sin y sinhxcos (iz) = cos [i (x� iy)] = cos (ix+ y)cos (ix+ y) = cos (ix) cos y � sin (ix) sin y = coshx cos y � i sinhx sin yHence cos (iz) = cos (iz) ; so it is proved for all z.(b) Show that sin (iz) = cos (iz) for all z.sin (iz) = sin [i (x+ iy)] = sin (xi� y)= sin (ix) cos y � sin y cos (ix) = i sinhx cos y � sin y coshx= � sin y coshx� i sinhx cos y = � (sin y coshx+ i sinhx cos y)sin (iz) = sin [i (x+ iy)] = sin (ix+ y)sin (ix+ y) = sin (ix) cos y + cos (ix) sin y = coshx sin y + i sinhx cos yHence sin (iz) = sin (iz) if and only if z = n�i (n = 0;�1;�2; � � � ) :If z 6= n�i; then sin (iz) = � sin (iz) :

17. Find all roots of the equation sin z = cosh 4 by equating the real partsand the imaginary parts of sin z and cosh 4.Answer:

��2 + 2n�

�� 4i (n = 0;�1;�2; � � � ) :

Solution:We note that sin z = cosh 4 is equivalent to (

eiz�e�iz)2i =

(e4�e�4)2 : The left

side is (eixe�y�e�ixey)

2i = i cosx(ey�e�y)

2 + sinx(ey+e�y)

2 : Thus, we must have

cosx

�(ey�e�y)

2

�= 0 and sinx (

ey+e�y)2 =

(e4�e�4)2 :

For the �rst equation, the possible solutions are thatx = �

2 + �k or that y = 0. If y =0, however, then the second equation

becomes sinx = (e4�e�4)2 ; which has no solution as the right side is larger than

1.Thus, we must have that x = �

2+�k: If k is an odd number, then sinx = �1;and the second equation above again has no solution as (

ey+e�y)2 > 0: Thus,

we must have x = (�=2) + 2�k for k 2 Z: This gives sinx = 1; and it remainsto solve cosh y = cosh 4, which implies that y = �4: Thus, z =

��2 + 2k�

��4i:

18. Find all roots of the equation cos z = 2.

Solution:

cos z =(eiz+e�iz)

2 = 2

8

e2iz � 4eiz + 1 = 0 from factorization eiz = 2�p3

eiz = eixey=(cos x+i sin x)ey=2�

p3; so ey cosx = 2�

p3; ey sinx = 0

x = 2n� and �y = ln�2�

p3�

x = 2n� + i ln�2�

p3�for n = 0;�1;�2; � � �

Exercises Pages 107-1082. (a) Prove that sinh 2z = 2 sinh z cosh z by starting with the hyperbolic

sine and the hyperbolic cosine of a complex variable are de�ned as they are witha real variable; that is, sinh z = ez�e�z

2 ; cosh z = ez+e�z

2 :(b) Prove that sinh 2z = 2 sinh z cosh z by starting with the identitysin 2z = 2 sin z cos z and using the relations�i sinh (iz) = sin z; cosh (iz) = cos z;�i sin (iz) = sinh z; cos (iz) = cosh z:

Solution:

Prove that sinh z = 2 sinh z cosh z:Method 1sinh 2z = sinh (z + z)This is in the form sinh (a+ b) = sinh a cosh b+ cosh a sinh b; where a = b.So sinh (z + z) = sinh z cosh z + cosh z sinh z:sinh (2z) = 2 sinh z cosh z:Method 2

By de�nition sinh z = (e2z�e�2z)2 and cosh z = (ez+e�z)

2 : So

sinh 2z =(e2z�e�2z)

2 by substitution of 2z in place of z.

2 sinh z cosh z = 2

�(ez�e�z)

2

(ez+e�z)2

�=(ez�e�z)(ez+e�z)

2

=

h(ez)2�(e�z)

2i

2 =(e2z�e�2z)

2 :This completes the proof that sinh z = 2 sinh z cosh z:

Solution:(a) We have sinh 2z = e2z�e�2z

2

=(ez)2�(e�z)

2

2

= 2 � (ez�e�z)2 � (e

z+e�z)2

= 2 sinh z cosh z:(b) We have sinh 2z = �i sin (2iz) = �2i sin (iz) cos (iz)= 2 [�i sin (iz)] [cos (iz)]= 2 sinh z cosh z:Thus we have proved sinh 2z = 2 sinh z cosh z by starting with the identities

in part (a) as well as in part (b).

5. Verify expression (12), Section 34, for jcosh zj2 :

9

Solution (5): Since cosh (z) = cosh (x) cos (y) + i sinh (x) sin (y) : Sojcosh (z)j2 = cosh2 (x) cos2 (y) + sinh2 (x) sin2 (y) =�sinh2 (x) + 1

�cos2 (y) + sinh2 (x) sin2 (y) =

sinh2 (x) cos2 (y) + sinh2 (x) sin2 (y) + cos2 (y) =sinh2 (x) + cos2 (y) :

7. Show that

(a) sinh (z + �i) = � sinh z;Show that(b) cosh (z + �i) = � cosh z;Show that(c) tan z (z + �i) = tanh z:

Solution:(a) Show that sinh (z + �i) = � sinh z:sinh (z + �i) = sinh (x+ iy + �i) = sinh [x+ i (y + �)] :We use the formula sinh (a+ ib) = sinh a cos b+ cosh a sin b with a = x andb = y + �:sinh [x+ i (y + �)] = sinhx cos (y + �) + coshx sin (� + y)= � sinhx cos y � coshx sin y becausesin (� + a) = � sin a and cos (� + a) = � cos a:sinh [x+ i (y + �)] = � (sinhx cos y + coshx sin y) = � sinh (x+ iy)sinh (z + �i) = � sinh z:This completes the proof that sinh (z + �i) = � sinh z:(b) Show that cosh (z + �i) = � cosh z:cosh (z + �i) = cosh (x+ iy + �i) = cosh [x+ i (y + �)] :We use the formula cosh (a+ ib) = cosh a cos b+ sinh a sin b with a = x andb = y + �:cosh [x+ i (y + �)] = coshx cos (y + �) + sinhx sin (� + y)= � coshx cos y � sinhx sin y becausesin (� + a) = � sin a and cos (� + a) = � cos a:cosh [x+ i (y + �)] = � (coshx cos y + sinhx sin y) = � cosh (x+ iy)cosh (z + �i) = � cosh z:This completes the proof that cosh (z + �i) = � cosh z:(c) Show that tan (z + �i) = tanh z:tanh (x+ iy + �i) = tanh [x+ i (y + �)]

We use the formula tanh (a+ ib) = (tanh a+i tan b)(1+i tanh a tan b) with a = x and b = y+�:

tanh [x+ i (y + �)] = [tanh x+i tan(�+y)][1+i tanh x tan(�+y)] because tan (� + a) = tan a

tanh [x+ i (y + �)] = [tanh x+i tan(y)](1+i tanh x tan(y))

tanh [x+ i (y + �)] = tanh (x+ iy)This completes the proof that tan (z + �i) = tanh z:

8. Give details showing that the zeros of sinh z and cosh z are as in the

10

following statements:(a) sinh z = 0 if and only if z = n�i (n = 0;�1;�2; : : :) and

(b) cosh z = 0 if and only if z =��2 + n�

�i (n = 0;�1;�2; : : :) :

Solution:

(a) sinh z = 0

sinh z =(ez�e�z)

2 = 0ez � e�z = 0ez � 1

ez = 0e2z � 1 = 0e2z = 1We have sinh z = �i sin (iz) :Now if sinh z = 0) sinh z = �i sin (iz) = 0) sin (iz) = 0) z = n�i since

sin � = 0) � = n�:(b) Now we have cosh z = cos (iz) :Now if cosh z = 0) cos (iz) = 0) iz =

��2 + n�

�izi =

��2 + n�

�ii2

z = ���2 + n�

�i) z =

��2 + n�

�i:

We do not have to consider -sine since cos(-x) = cos x.

15. (a) By using one of the identities sinh z = sinhx cos y + i coshx sin y;

cosh z = coshx cos y + i sinhx sin y; �nd all roots of the equation sinh z = i;

(b) By using one of the identities sinh z = sinhx cos y + i coshx sin y;

cosh z = coshx cos y+ i sinhx sin y; �nd all roots of the equation cosh z = 12 :

Solution:

(a) sinh z = isinh z = sinhx cos y + i cosx sin yFrom sinh z = i and sinh z = sinhx cos y + i cosx sin y,i = sinhx cos y + i cosx sin y0 + i = sinhx cos y + i coshx sin yComparing both sides 0 = sinhx cos yi = i coshx sin ysinhx cos y = 0

sinhx = 0 or cos y = 0(ex�e�x)

2 = sinhx cos y = 0

11

(ex�e�x)2 = 0 cos y = (2n� 1) �2

x = 0 y = (2n� 1) �2ii =

ii coshx sin y

1 = coshx sin y

1 =�ex+e�x

2

�sin y

(b) cosh z = 1/2sinh z = sinh x cos y +i cosh x sin ycosh z = cosh x cos y + i sinh x sin yHere cosh z = 1

2 =12 + i � 0

) coshx cos y = 12 ; sinhx sin y = 0

From sinhx sin y = 0) sinhx = 0 or sin y = 0) ex�e�x

2 = 0

) ex�1=ex2 = 0

) e2x � 1 = 0) w2x = 1 = e0

) x = 0When x = 0, cosh x = 1) cos y = 1

2From coshx cos y = 1

2 ) y = �3

or generally y = 2n� ���3

�; n 2 Z

So z = 0 + i�2n� � �

3

�; n 2 Z

or z = i�2n� � �

6

�; n 2 Z

Again from sinhx sin y = 0; if we take sin y = 0, then y = 0.) cos y = cos 0 = 1) coshx cos y = 1=2) coshx = 1=2

) ex+e�x

2 = 12

) ex + e�x = 1) ex + 1

ex = 1) e2x + 1 = ex

) e2x + e�x + 1 = 0ex = 1�

p1�42

) ex is imaginary, which is not possible. So the second case does not yieldany solution.16. Find all roots of the equation cosh z = �2: (Compare this exercise

with Exercise 18, Section 33.)

Solution: First note that cosh is always positive for real numbers. Thus,we are giving an example of a property of hyperbolic trig functions that doesnot carry over to C.We �rst use the identity coshx = coshx cos y + i sinhx sin y:Thus, for this to be = -2, we must have coshx cos y = �2; sinhx sin y = 0:

12

The second equation tells us that either sinh x = 0 or sin y = 0. That is,either x = 0 or y=n�; n 2 Z: That former is not possible as cosh 0 = 1 andcos y can never be equal to -2 as would be required. Thus, we must have thaty = n� if there are any solutions.Plugging y = n� in, we have coshx cos (n�) = (�1)n coshx = �2: Since

cosh is always positive for real numbers, we must have that n is odd. Weshall write the odd n as 2m+1 where m ranges over all integers. It remains,

then, to �nd the x solving cosh x = 2. This is equivalent to (ex+e�x)

2 = 2 ore2x � 4ex + 1 = 0:Solving this quadratic equation, we have ex = (4�

p12)

2 = 2 �p3: Thus,

x = ln�2�

p3�; and z = ln

�2�

p3�+ (2m+ 1)�i: This may appear dif-

ferent form the books answer, but note that � ln�2 +

p3�= ln 1

(2+p3)

=

ln

�(2�

p3)

(2+p3)(2�

p3)

�= ln 2�

p3:

Exercises Pages 1101. (a) Find all the values of tan�1 (2i) ;

(b) Find all the values of tan�1 (1 + i) ;(c) Find all the values of cosh�1 (�1) ;(d) Find all the values of tanh�1 0:

Solution a:tan�1 (2i) = i

2 logh(i�2i)(i+2i)

i) tan�1 (z) = i

2 logh(i�z)(i+z)

i) i

2 log��i3i

�) i

2 log�� 13

�= i

2 log�i2 � 13

�= i

2

�log i2 + log

�13

��= i

2 (1:364� 0:477)= i

2 (0:8868)= 0:443i:Solution b:tan�1 (1 + i) = i

2 logh1�(1+i)1+(1+i)

i= i

2 [log (�i)� log (2 + i)]Solution c:We have cosh�1 (z) = log

�z +

pz2 � 1

�cosh�1 (�1) = log

��1 +

q(�1)2 � 1

�= log

��1 +

p1� 1

�= log (�1)= 1:364Solution d:We have tanh�1 z = 1

2 logh(1+z)(1�z)

i13

) tanh�1 (0) = 12 log

h(1+0)(1�0)

i= 1

2 log�11

�= 1

2 log (1)= 1

2 (0)= 0:

Solution:Let z = x+iy. We recall that log z = log

px2 + y2 + i (2n� + �) ; where

n 2 Z; � = tan�1 (y=x) and �0 � � � 2�:(a) We have z = tan�1 (2i)) tan z = 2i

) eiz�e�izi(eiz+e�iz) = 2i

) eiz � e�iz = �2�eiz + e�iz

�) 3eiz = �e�iz) e2iz = � 1

3) 2iz = log

�� 13

�= � log (�3)

) z = i2 [log (�3)]

) z = i2 [log 3 + (2n+ 1)�i] :

(b) We have z = tan�1 (1 + i)) tan z = 1 + i

) eiz�e�izi(eiz+e�iz) = 1 + i

) eiz � e�iz = (i� 1)�eiz + e�iz

�) (2� i) eiz = ie�iz) e2iz = i

(2�i) = �15 + i

25

) 2iz = log�� 15 + i

25

�) 2iz = log

hq125 +

425

i+ i

h2n� + tan�1

�2=5�1=5

�i) 2iz = log

�1p5

�+ i

�2n� + tan�1 (�2)

�) 2iz = � 1

2 log 5 + i�2n� � tan�1 (2)

�) z = n� � 1

2 tan�1 (2) + i

4 log 5:

(c) We have z = cosh�1 (�1)) cosh z = �1) ez+e�z

2 = �1) e2z + 2ez + 1 = 0) (ez + 1)

2= 0

) ez = �1) z = (2n+ 1)�i for all n 2 Z(d) We have z = tanh�1 0) tanh z = 0

) ez�e�z(ez+e�z) = 0

) ez � e�z = 0) e2z = 1) ez = 1 or ez = �1) z = 2n�i or z = (2n+ 1)�i for all n 2 Z) z = n�i for all n 2 Z:

14

2. (a) Solve the equation sin z = 2 for z by equating real parts andimaginary parts in that equation;(b) Solve the equation sin z = 2 for z by using the expression

sin�1 z = �i loghiz +

�1� z2

�1=2i:

Solution a:

Set sin z = sinx cosh y + i cosx sinh y = 2) z = x+ iy; sin z = sin (x+ iy) = sinx cosh y + i cosx sinh y) sinx cosh y + i cosx sinh y = 2 + 0 � i) sinx cosh y = 2cosx sinh y = 0The equation cosx sinh y = 0 holds when x =

�n+ 1

2

�� or when y = 0.

When y = 0, the equation sinx cosh y = 2 becomes sin x = 2, which has nosolution. So we must have x =

�n+ 1

2

��; which has no solution. So we

must have x =�n+ 1

2

��: In that case the equation sinx cosh y = 2 becomes

(�1)n cosh y = 2: However cosh y is always positive. So n must be an even

integer. We then solve for cosh y = (ey+e�y)2 = 2; take ey = !:

e�y = 1ey =

1!

(!+ 1! )

2 = 2 = !2+12! = 2 = !2 + 1 = 6!

) !2 � 6! + 1 = 0) ! = 6�

p16�42 = 4�

p12

2 = 2�p3:

) ! = ey = 2�p3

cosh y = ey+e�y

2 = 2; take ey = !e�y = 1

ey =1!

(!+ 1! )

2 = 2 =(!2+1)2! = 2 = !2 + 1 = 6!

) !2 � 6! + 1 = 0) ! = 6�

p16�42 = 4�

p12

2 = 2�p3:

) ! = ey = 2�p3

y = ln�2�

p3�

Hence the solution set isz =

�2n+ 1

2

�� + i ln

�2�

p3�:

Solution b:If we put z = 2 into the formula

zin�1z = �i loghiz +

�1� z2

�1=2i; then

sin�1 2 = �i log�i2 +

p�3�

= �i log�2i+

p�3�

= �i log�2i+ i

p3�

= �i log�2i+ i

p3�

sin�1 2 = �i log�i�2�

p3��

= �i log��2�

p3�ei(�=2)+2n�

�t= �i

�ln�2�

p3�+ i

��2 + 2n�

��15

=��2 + 2n�

�+ i ln

�2�

p3�:

Solution:Let z = x+iy. We recall that log z = log

px2 + y2 + i (2n� + �) ; where

n 2 Z; � = tan�1 (y=x) and o� � � 2�:(a) We have sin z = 2) sin (x+ iy) = 2) sinx cos (iy) + cosx sin (iy) = 2) sinx cosh y + i cosx sinh y = 2:Equating the real and imaginary parts on both sides, we have sinx cosh y = 2

and cosx sinh y = 0: Now cosx sinh y = 0) cosx = 0 or sinh y = 0: If sinh y =0, then we have y = 0 + 2n� for n 2 Z: Then sinx cos (2n�) = 2 ) sinx = 2which is impossible because �1 � sinx �� 1 for all x 2 R: Therefore wemust have cosx = 0; and it implies that x = 2n� + �

2 for n 2 Z: Now forx = 2n� + �=2; we have sin (2n� + �=2) cosh y = 2) cosh y = 2

) ey+e�y

2 = 2) e2y � 4ey + 1 = 0) ey = 4�

p16�42

) ey = 2�p3

) y = log�2�

p3�:

Thus the required solution is z = x+ iy = (2n� + �=2) + i log�2�

p3�for

all n 2 Z:(b) We havesin z = 2)z = sin�1 (2)

= �i logh2i+ (1� 4)1=2

i= �i log

��2�

p3�i�

= �i�log

q02 +

�2�

p3�2+ i

h2n� + tan�1

�2�p3

0

�i�= �i

�log�2�

p3�+ i

�2n� + tan�11

��= �i

�log�2�

p3�+ i (2n� + �=2)

�= (2n� + �=2)� i log

�2�

p3�:

Thus the required solution is z = (2n� + �=2)� i log�2�

p3�for all n 2 Z:

From part (a) and part (b), we see that the solution of sin z = 2 is given byz = (2n� + �=2)� i log

�2�

p3�for all n 2 Z:

3. Solve the equation cos z =p2 for z.

Solution (3): From cos (z) =p2 we have eiz + e�iz = 2

p2: Let X = eiz:

Then X+1=X = 2p2: Thus X2�2

p2X+1 = 0: We solve X =

p2�1: Thus

z = 1i log

�p2� 1

�= �i

�ln��p2� 1��+ i2n�� = 2n� � i ln

�p2� 1

�, where n

can be any integer.7. Derive expression (9), Section 35, for cosh�1 z:

Solution: This is rather similar to what we did above for real values. Forz given, we want to solve (

ew+e�w)2 = coshw = z:

This is equivalent to e2w � 2zew + 1 = 0:

16

The solutions to this quadratic equation are ew =(2z�

p4z2�4)2 = z +�

z2 � 1�1=2

:Recall that we do not need to write � since the second root of a complex

number is a set containing two elements. The above says that cosh�1 z =

loghz +

�z2 � 1

�1=2i: Here cosh�1 takes on in�nitely many values (not we are

not using, e.g., the principal branch of the logarithm). This, however, shouldbe expected for complex numbers due, e.g., to the periodicity of cosh and sinh.

17