17.7 Trigonometric and Hyperbolic Functions
INTRODUCTION
In this section we define the complex trigonometric and hyperbolic functions. Analogous to the complex functions ez and Ln z defined in the previous section, these functions will agree with their real counterparts for real values of z. In addition, we will show that the complex trigonometric and hyperbolic functions have the same derivatives and satisfy many of the same identities as the real trigonometric and hyperbolic functions.
Trigonometric Functions
If x is a real variable, then Euler’s formula gives
eix = cos x + i sin x and e−ix = cos x − i sin x.
By subtracting and then adding these equations, we see that the real functions sin x and cos x can be expressed as a combination of exponential functions:
(1)
Using (1) as a model, we now define the sine and cosine of a complex variable:
DEFINITION 17.7.1 Trigonometric Sine and Cosine
For any complex number z = x + iy,
(2)
As in trigonometry, we define four additional trigonometric functions in terms of sin z and cos z:
(3)
When y = 0, each function in (2) and (3) reduces to its real counterpart.
Analyticity
Since the exponential functions eiz and e−iz are entire functions, it follows that sin z and cos z are entire functions. Now, as we shall see shortly, sin z = 0 only for the real numbers z = nπ, n an integer, and cos z = 0 only for the real numbers z = (2n + 1)π/2, n an integer. Thus, tan z and sec z are analytic except at the points z = (2n + 1)π/2, and cot z and csc z are analytic except at the points z = nπ.
Derivatives
Since (d/dz)ez = ez, it follows from the Chain Rule that (d/dz)eiz = ieiz and (d/dz)e−iz = −ie−iz. Hence,
In fact, it is readily shown that the forms of the derivatives of the complex trigonometric functions are the same as the real functions. We summarize the results:
(4)
Identities
The familiar trigonometric identities are also the same in the complex case:
Zeros
To find the zeros of sin z and cos z we need to express both functions in the form u + iv. Before proceeding, recall from calculus that if y is real, then the hyperbolic sine and hyperbolic cosine are defined in terms of the real exponential functions ey and e−y:
(5)
Now from Definition 17.7.1 and Euler’s formula we find, after simplifying,
Thus from (5) we have
(6)
It is left as an exercise to show that
(7)
From (6), (7), and cosh2y = 1 + sinh2y, we find
|sin z|2 = sin2x + sinh2y (8)
and |cos z|2 = cos2x + sinh2y. (9)
Now a complex number z is zero if and only if |z|2 = 0. Thus, if sin z = 0, then from (8) we must have sin2x + sinh2y = 0. This implies that sin x = 0 and sinh y = 0, and so x = nπ and y = 0. Thus the only zeros of sin z are the real numbers z = nπ + 0i = nπ, n = 0, ±1, ±2, . . . . Similarly, it follows from (9) that cos z = 0 only when z = (2n + 1)π/2, n = 0, ±1, ±2, . . . .
EXAMPLE 1 Complex Value of the Sine Function
From (6) we have, with the aid of a calculator,
sin(2 + i) = sin 2 cosh 1 + i cos 2 sinh 1 = 1.4031 − 0.4891i. ≡
In ordinary trigonometry we are accustomed to the fact that |sin x| ≤ 1 and |cos x| ≤ 1. Inspection of (8) and (9) shows that these inequalities do not hold for the complex sine and cosine, since sinh y can range from −∞ to ∞. In other words, it is perfectly feasible to have solutions for equations such as cos z = 10.
EXAMPLE 2 Solving a Trigonometric Equation
Solve the equation cos z = 10.
SOLUTION
From (2), cos z = 10 is equivalent to (eiz + e−iz)/2 = 10. Multiplying the last equation by eiz then gives the quadratic equation in eiz:
e2iz − 20eiz + 1 = 0.
From the quadratic formula we find eiz = 10 ± 3. Thus, for n = 0, ±1, ±2, … , we have iz = loge(10 ± 3) + 2nπi. Dividing by i and utilizing loge(10 − 3) = −loge(10 + 3), we can express the solutions of the given equation as . ≡
Hyperbolic Functions
We define the complex hyperbolic sine and cosine in a manner analogous to the real definitions given in (5).
DEFINITION 17.7.2 Hyperbolic Sine and Cosine
For any complex number z = x + iy,
(10)
The hyperbolic tangent, cotangent, secant, and cosecant functions are defined in terms of sinh z and cosh z:
(11)
The hyperbolic sine and cosine are entire functions, and the functions defined in (11) are analytic except at points where the denominators are zero. It is also easy to see from (10) that
(12)
It is interesting to observe that, in contrast to real calculus, the trigonometric and hyperbolic functions are related in complex calculus. If we replace z by iz everywhere in (10) and compare the results with (2), we see that sinh(iz) = i sin z and cosh(iz) = cos z. These equations enable us to express sin z and cos z in terms of sinh(iz) and cosh(iz), respectively. Similarly, by replacing z by iz in (2) we can express, in turn, sinh z and cosh z in terms of sin(iz) and cos(iz). We summarize the results:
sin z = −i sinh(iz), cos z = cosh(iz) (13)
sinh z = −i sin(iz), cosh z = cos(iz). (14)
Zeros
The relationships given in (14) enable us to derive identities for the hyperbolic functions utilizing results for the trigonometric functions. For example, to express sinh z in the form u + iv we write sinh z = −i sin(iz) in the form sinh z = −i sin(−y + ix) and use (6):
sinh z = −i [sin(−y) cosh x + i cos(−y) sinh x].
Since sin(−y) = −sin y and cos(−y) = cos y, the foregoing expression simplifies to
(15)
Similarly, (16)
It also follows directly from (14) that the zeros of sinh z and cosh z are pure imaginary and are, respectively,
Periodicity
Since sin x and cos x are 2π-periodic, we can easily demonstrate that sin z and cos z are also periodic with the same real period 2π. For example, from (6), note that
that is, sin(z + 2π) = sin z. In exactly the same manner, it follows from (7) that cos(z + 2π) = cos z. In addition, the hyperbolic functions sinh z and cosh z have the imaginary period 2πi. This last result follows from either Definition 17.7.2 and the fact that ez is periodic with period 2πi, or from (15) and (16) and replacing z by z + 2πi.
17.7 Exercises Answers to selected odd-numbered problems begin on page ANS-46.
In Problems 1–12, express the given quantity in the form a + ib.
- cos(3i)
- sin(−2i)
- cos(2 − 4i)
- tan(i)
- sec(π + i)
- csc(1 + i)
- cosh(πi)
- cosh(2 + 3i)
In Problems 13 and 14, verify the given result.
In Problems 15–20, find all values of z satisfying the given equation.
- sin z = 2
- cos z = −3i
- sinh z = −i
- sinh z = −1
- cos z = sin z
- cos z = i sin z
In Problems 21 and 22, use the definition of equality of complex numbers to find all values of z satisfying the given equation.
- cos z = cosh 2
- sin z = i sinh 2
- Prove that cos z = cos x cosh y − i sin x sinh y.
- Prove that sinh z = sinh x cos y + i cosh x sin y.
- Prove that cosh z = cosh x cos y + i sinh x sin y.
- Prove that |sinh z|2 = sin2y + sinh2x.
- Prove that |cosh z|2 = cos2y + sinh2x.
- Prove that cos2z + sin2z = 1.
- Prove that cosh2z − sinh2z = 1.
- Show that tan z = u + iv, where
- Prove that tanh z is periodic with period πi.
- Prove that (a) = sin and (b) .