ELEMENTARY FUNCTIONS
34. TRIGONOMETRIC FUNCTIONS Euler’s formula (Sec. 6) tells us that
104 Elementary Functions chap. 3
EXERCISES
1. Show that (a)(1+i)i =exp
−π 4 +2nπ
exp
iln 2
2
(n=0,±1,±2, . . .);
(b)(−1)1/π =e(2n+1)i (n=0,±1,±2, . . .).
2. Find the principal value of (a)ii; (b)
e
2(−1−√ 3i)
3π i
; (c)(1−i)4i.
Ans. (a) exp(−π/2); (b)−exp(2π2); (c)eπ[cos(2 ln 2)+isin(2 ln 2)].
3. Use definition (1), Sec. 33, ofzcto show that(−1+√
3i)3/2= ±2√ 2.
4. Show that the result in Exercise 3 could have been obtained by writing (a) (−1+√
3i)3/2=[(−1+√
3i)1/2]3and first finding the square roots of−1+√ 3i; (b) (−1+√
3i)3/2=[(−1+√
3i)3]1/2 and first cubing−1+√ 3i.
5. Show that the principalnth root of a nonzero complex numberz0that was defined in Sec. 9 is the same as the principal value ofz1/n0 defined by equation (5), Sec. 33.
6. Show that ifz=0 andais a real number, then|za| =exp(aln|z|)= |z|a, where the principal value of|z|a is to be taken.
7. Letc=a+bi be a fixed complex number, wherec=0,±1,±2, . . . ,and note that icis multiple-valued. What additional restriction must be placed on the constantcso that the values of|ic|are all the same?
Ans.c is real.
8. Let c, c1, c2,and z denote complex numbers, where z=0. Prove that if all of the powers involved are principal values, then
(a)zc1zc2 =zc1+c2; (b) zc1
zc2 =zc1−c2; (c)(zc)n=zc n (n=1,2, . . .).
9. Assuming thatf(z)exists, state the formula for the derivative of cf (z).
34. TRIGONOMETRIC FUNCTIONS
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sec. 34 Trigonometric Functions 105
It is, therefore, natural to define the sine and cosine functions of a complex variable zas follows:
sinz= eiz−e−iz
2i and cosz= eiz+e−iz
2 .
(1)
These functions are entire since they are linear combinations (Exercise 3, Sec. 25) of the entire functionseiz ande−iz.Knowing the derivatives
d
dzeiz=ieiz and d
dze−iz = −ie−iz of those exponential functions, we find from equations (1) that
d
dzsinz=cosz and d
dzcosz= −sinz.
(2)
It is easy to see from definitions (1) that the sine and cosine functions remain odd and even, respectively:
sin(−z)= −sinz, cos(−z)=cosz.
(3) Also,
eiz=cosz+isinz.
(4)
This is, of course, Euler’s formula (Sec. 6) whenzis real.
A variety of identities carry over from trigonometry. For instance (see Exercises 2 and 3),
sin(z1+z2)=sinz1cosz2+cosz1sinz2, (5)
cos(z1+z2)=cosz1cosz2−sinz1sinz2. (6)
From these, it follows readily that
sin 2z=2 sinzcosz, cos 2z=cos2z−sin2z, (7)
sin
z+π 2
=cosz, sin
z− π 2
= −cosz, (8)
and [Exercise 4(a)]
sin2z+cos2z=1.
(9)
The periodic character of sinzand coszis also evident:
sin(z+2π )=sinz, sin(z+π )= −sinz, (10)
cos(z+2π )=cosz, cos(z+π )= −cosz.
(11)
106 Elementary Functions chap. 3
Wheny is any real number, definitions (1) and the hyperbolic functions sinhy= ey−e−y
2 and coshy= ey+e−y 2 from calculus can be used to write
sin(iy)=isinhy and cos(iy)=coshy.
(12)
Also, the real and imaginary components of sin z and cos z can be displayed in terms of those hyperbolic functions:
sinz=sinx coshy+icosxsinhy, (13)
cosz=cosxcoshy−isinx sinhy, (14)
wherez=x+iy.To obtain expressions (13) and (14), we write z1=x and z2=iy
in identities (5) and (6) and then refer to relations (12). Observe that once expres- sion (13) is obtained, relation (14) also follows from the fact (Sec. 21) that if the derivative of a function
f (z)=u(x, y)+iv(x, y) exists at a pointz=(x, y),then
f(z)=ux(x, y)+ivx(x, y).
Expressions (13) and (14) can be used (Exercise 7) to show that
|sinz|2=sin2x+sinh2y, (15)
|cosz|2=cos2x+sinh2y.
(16)
Inasmuch as sinhy tends to infinity asy tends to infinity, it is clear from these two equations that sinz and coszare not bounded on the complex plane, whereas the absolute values of sinxand cosxare less than or equal to unity for all values ofx.
(See the definition of a bounded function at the end of Sec. 18.)
A zero of a given functionf(z) is a numberz0such thatf (z0)=0. Since sinz becomes the usual sine function in calculus whenzis real, we know that the real numbersz=nπ (n=0,±1,±2, . . .)are all zeros of sinz. To show that there are no other zeros, we assume that sinz=0 and note how it follows from equation (15) that
sin2x+sinh2y=0.
This sum of two squares reveals that
sinx=0 and sinhy=0.
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sec. 34 Trigonometric Functions 107
Evidently, then,x=nπ (n=0,±1,±2, . . .)andy=0 ; that is, sinz=0 if and only if z=nπ (n=0,±1,±2, . . .).
(17) Since
cosz= −sin
z−π 2
, according to the second of identities (8),
cosz=0 if and only if z= π
2 +nπ (n=0,±1,±2, . . .).
(18)
So, as was the case with sinz, the zeros of coszare all real.
The other four trigonometric functions are defined in terms of the sine and cosine functions by the expected relations:
tanz= sinz
cosz, cotz= cosz sinz, (19)
secz= 1
cosz, cscz= 1 sinz. (20)
Observe that the quotients tanz and secz are analytic everywhere except at the singularities (Sec. 24)
z= π
2 +nπ (n=0,±1,±2, . . .),
which are the zeros of cosz. Likewise, cotzand csczhave singularities at the zeros of sinz, namely
z=nπ (n=0,±1,±2, . . .).
By differentiating the right-hand sides of equations (19) and (20), we obtain the anticipated differentiation formulas
d
dztanz=sec2z, d
dzcotz= −csc2z, (21)
d
dzsecz=secztanz, d
dzcscz= −csczcotz.
(22)
The periodicity of each of the trigonometric functions defined by equations (19) and (20) follows readily from equations (10) and (11). For example,
tan(z+π )=tanz.
(23)
Mapping properties of the transformation w=sinz are especially important in the applications later on. A reader who wishes at this time to learn some of those properties is sufficiently prepared to read Sec. 96 (Chap. 8), where they are discussed.
108 Elementary Functions chap. 3
EXERCISES
1. Give details in the derivation of expressions (2), Sec. 34, for the derivatives of sinz and cosz.
2. (a) With the aid of expression (4), Sec. 34, show that
eiz1eiz2 =cosz1cosz2−sinz1sinz2+i(sinz1cosz2+cosz1sinz2).
Then use relations (3), Sec. 34, to show how it follows that
e−iz1e−iz2 =cosz1cosz2−sinz1sinz2−i(sinz1cosz2+cosz1sinz2).
(b) Use the results in part (a) and the fact that sin(z1+z2)= 1
2i
ei(z1+z2)−e−i(z1+z2)
= 1 2i
eiz1eiz2−e−iz1e−iz2
to obtain the identity
sin(z1+z2)=sinz1cosz2+cosz1sinz2
in Sec. 34.
3. According to the final result in Exercise 2(b),
sin(z+z2)=sinzcosz2+coszsinz2.
By differentiating each side here with respect tozand then settingz=z1,derive the expression
cos(z1+z2)=cosz1cosz2−sinz1sinz2
that was stated in Sec. 34.
4. Verify identity (9) in Sec. 34 using
(a) identity (6) and relations (3) in that section;
(b) the lemma in Sec. 27 and the fact that the entire function f (z)=sin2z+cos2z−1 has zero values along thex axis.
5. Use identity (9) in Sec. 34 to show that
(a) 1+tan2z=sec2z; (b) 1+cot2z=csc2z.
6. Establish differentiation formulas (21) and (22) in Sec. 34.
7. In Sec. 34, use expressions (13) and (14) to derive expressions (15) and (16) for|sinz|2 and|cosz|2.
Suggestion: Recall the identities sin2x+cos2x=1 and cosh2y−sinh2y=1.
8. Point out how it follows from expressions (15) and (16) in Sec. 34 for|sinz|2 and
|cosz|2that
(a)|sinz| ≥ |sinx|; (b)|cosz| ≥ |cosx|.
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sec. 35 Hyperbolic Functions 109
9. With the aid of expressions (15) and (16) in Sec. 34 for|sinz|2and|cosz|2, show that (a)|sinhy| ≤ |sinz| ≤coshy; (b)|sinhy| ≤ |cosz| ≤coshy.
10. (a) Use definitions (1), Sec. 34, of sinz and coszto show 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 cosz1=cosz2, then at least one of the numbersz1+z2andz1−z2 is an integral multiple of 2π.
11. Use the Cauchy–Riemann equations and the theorem in Sec. 21 to show that neither sinznor coszis an analytic function ofzanywhere.
12. Use the reflection principle (Sec. 28) to show that for allz, (a) sinz=sinz; (b) cosz=cosz.
13. With the aid of expressions (13) and (14) in Sec. 34, give direct verifications of the relations obtained in Exercise 12.
14. Show that
(a) cos(iz)=cos(iz) for allz;
(b) sin(iz)=sin(iz) if and only if z=nπ i (n=0,±1,±2, . . .).
15. Find all roots of the equation sinz=cosh 4 by equating the real parts and then the imaginary parts of sinzand cosh 4.
Ans.
π 2 +2nπ
±4i (n=0,±1,±2, . . .).
16. With the aid of expression (14), Sec. 34, show that the roots of the equaion cosz=2 are
z=2nπ+icosh−12 (n=0,±1,±2, . . .).
Then express them in the form
z=2nπ±iln(2+√
3) (n=0,±1,±2, . . .).