The hyperbolic functions also have practical use in putting the tangent function into the Cartesian form u + ⅈ v.
Using Definition 5.6, and Equations (5-34) and (5-35), we have
(5-38) tan z = tan (x + ⅈ y) = sin (x+ⅈ y)
cos (x+ⅈ y) = sin x cosh y + ⅈ cos x sinh y cos x cosh y - ⅈ sin x sinh y .
If we multiply each term on the right by the conjugate of the denominator, the simplified result is
tan z = cos x sin x + ⅈ cosh y sinh y
As with sin z, we obtain a graph of the mapping w = tan z parametrically. Consider the vertical line segments in the z plane obtained by successively setting x = -π4 +k π
16 for
k = 0,1, ...,8, and for each z value letting y vary continuously, 3 ≤ y ≤ 3. In the exercises we ask you to show that the images of these vertical segments are circular arcs in the uv plane, as Figure 5.8 shows. In Section 10.4, we give a more detailed investigation of the mapping w = tan z.
-p 4 4 p x
Figure 5.8 Vertical segments mapped onto circular arcs by w = tan z.
Exploration.
How should we define the complex hyperbolic functions? We begin with
8 Chapter05Section04.nb
cosh z = 12 (ⅇz+ ⅇ-z)
sinh z = 1
2 (ⅇz- ⅇ-z)
With these definitions in place, we can now easily create the other complex hyperbolic trigonometric functions, provided the denominators in the following expressions are not zero.
Definition 5.8. Identities for the hyperbolic trigonometric functions are
tanh z = sinh zcosh z ,
coth z = cosh zsinh z ,
sech z = 1
cosh z ,
csch z = 1
sinh z .
As the series for the complex hyperbolic sine and cosine agree with the real hyperbolic sine and cosine when z is real, the remaining complex hyperbolic trigonometric functions likewise agree with their real counterparts. Many other properties are also shared. We state several results with-out proof, as they follow from the definitions we gave using standard operations, such as the quo-tient rule for derivatives. We ask you to establish some of these identities in the exercises.
The derivatives of the hyperbolic functions follow the same rules as in calculus:
d
dz cosh z = sinh z, and d
dz Sinh z = cosh z, d
dz Tanh z = sech2z, and d
dz coth z = - csch2z, d
dz sech z = - sech z Tanh z, and d
dz csch z = - csch z coth z.
The hyperbolic cosine and hyperbolic sine can be expressed as
cosh z = cosh (x + ⅈ y) = cos y cosh x + ⅈ sinh x sin y sinh z = sinh (x + ⅈ y) = sinh x cos y + ⅈ cosh x sin y Some of the important identities involving the hyperbolic functions are
cosh2z - sinh2z = 1,
sinh (z1+z2) = sinh z1cosh z2 + cosh z1sinh z2,
cosh (z1+z2) = cosh z1cosh z2 + sinh z1sinh z2,
cosh (z + 2 π ⅈ) = cosh z,
sinh (z + 2 π ⅈ) = sinh z,
cosh (-z) = cosh z,
sinh (-z) = - sinh z.
The "legacy version" being used to implement the commands "CartesianMap and PolarMap." The following Initialization Cell will load these subroutines.
Initialization Cell
10 Chapter05Section04.nb
John H. Mathews and Russell W. Howell Jones and Bartlett Learning
Prof. John H. Mathews Department of Mathematics California State University Fullerton
Fullerton, CA 92634 [email protected]
Prof. Russell W. Howell
Mathematics & Computer Science Department Westmont College
Santa Barbara, CA 93108 [email protected] 5.5 Inverse Trigonometric and Hyperbolic Functions
We expressed trigonometric and hyperbolic functions in Section 5.4 in terms of the exponen-tial function. In this section we look at their inverses. When we solve equations such as w = sin z for z, we will obtain formulas that involve the logarithm. Because trigonometric and hyperbolic functions are all periodic, they are many-to-one; hence their inverses are necessarily multivalued.
The formulas for the inverse trigonometric functions are
(5 - 45) arcsin z = - ⅈ log ⅈ z + (1 - z2)1/2,
arccos z = - ⅈ log z + ⅈ (1 - z2)1/2,
arctan z = ⅈ2 log ⅈ +ⅈ -zz.
We derive the first equation and leave the others as exercises. If we take a particular branch of the multivalued function, w = arcsin z, we have
z = sin w = ⅇⅈw2 ⅈ- ⅇ-ⅈw
which we can also write as
ⅇⅈw - 2 ⅈ z - ⅇ-ⅈw = 0.
Multiplying both sides of this equation by ⅇⅈw gives
(ⅇⅈw)2 - 2 ⅈ z ⅇⅈw - 1 = 0,
which is a quadratic equation in terms of ⅇⅈw. Using the quadratic equation to solve for ⅇⅈw, we obtain
ⅇⅈw = 2 ⅈ z+-4 z2+41/2
2 = ⅈz + (1 - z2)1/2,
where the square root is a multivalued function. Taking the logarithm of both sides and obtain ⅈw = log ⅈ z + (1 - z2)1/2, then simplify this and obtain the desired result:
w = arcsin z = - ⅈ log ⅈ z + (1 - z2)1/2,
where the multivalued logarithm is used. To construct a specific branch of arcsin z, we must first select a branch of the square root and then select a branch of the logarithm.
Remark. The function arcsin z is known to be an odd function and therefore arcsin (-z) = arcsin z. As an immediate consequence we have
- ⅈlog -ⅈ z + (1 - z2)1/2 = ⅈlog ⅈ z + (1 - z2)1/2.
We can now use the trigonometric identity arccos z = π2 - arcsin z and obtain an alternate expression for arccos z:
arccos z = π
2 -arcsin z
= π
2 + ⅈlog ⅈ z + (1 - z2)1/2
= π
2 - ⅈlog -ⅈ z + (1 - z2)1/2
= - ⅈ π ⅈ
2 - ⅈlog -ⅈ z + (1 - z2)1/2
= - ⅈlog (ⅈ) - ⅈ log -ⅈ z + (1 - z2)1/2
= - ⅈ log (ⅈ) + log -ⅈ z + (1 - z2)1/2
= -log ⅈ (- ⅈ) z + ⅈ (1 - z2)1/2
= -log z + ⅈ (1 - z2)1/2
We mention this derivation because Mathematica uses the second line for determining the principal value, i.e.
arccos z = π2 + ⅈlog ⅈ z + (1 - z2)1/2
an we will see shortly that ArcCos [z] = π2 + ⅈLog[ⅈ z + Sqrt[1 - z2]].
The principal value used by Mathematica for the inverse trigonometric functions are
2 Chapter05Section05.nb
ArcSin z = - ⅈLogⅈ z + (1 - z2)1/2,
"spider web" in the w plane, as Figure 5.10 shows.
-10-8 -6 -4 -2 2 4 6 8 10
Figure 5.10 A rectangular grid is mapped onto a spider web by w = Arcsin z.The derivatives of arcsin(z), arccos(z), arctan(z)
We can find the derivatives of any branch of these functions by using the chain rule:
(5 - 46) dzd arcsin z = 1 differen-tiating both sides, using the chain rule:
d
dzsin w = dzd z
d
dwsin w dw
dz = 1 cos w dwdz = 1
dw
dz = 1
cos w
Then cos2w + sin2w = 1 is used to get the substitution cos w = 1 - sin2w = 1 - z2 for the preceding equation, and we obtain the desired result:
d
dzarcsin z = dw
dz = 1
1 - z2
.
Derivation of the other formulas are left as exercises for the reader.
The Inverse Sine arcsin(z) . Verify that the formula
(i) Arcsin z = - ⅈ Logⅈ z + 1 - z2
is correct. (At least for values of z in the upper half plane Im (z) > 0.)