Chapter 3
Elementary Functions
In this chapter, we will consider elementary functions of a complex variable. We will introduce
complex exponential, trigonometric, hyperbolic, and logarithmic functions.
23. The exponential function
We want to define an exponential function of a complex variable z = x + iy. We require that
• f (x) = ex for all x ∈ R.
• f 0 (z) = f (z).
Definition. (The exponential function)
We define the exponential function of a complex variable z = x + iy to be
ez = ex (cos y + i sin y).
Remark.
1. The domain of the exponential function is the entire complex plane.
2. We sometimes use exp(z) to denote ez .
3. When z = iy, we have the Euler’s formula eiy = cos y + i sin y
π
Example 1. Express e−1+ 2 i in the form a + ib.
Proposition 1. The exponential function is an entire function and
d z
e = ez .
dz
Proof. Exercise.
1
Remark. When z = n1 , n = 2, 3, . . . , then e n =
√
n
1
e. That is e n is not the set of all n−th roots of e.
Proposition 2. Let z = x + iy. Then we have the following properties of ez :
1. |ez | = ex .
2. arg(ez ) = y + 2nπ, n ∈ Z.
3. ez 6= 0 for all z ∈ C.
4. ez1 ez2 = ez1 +z2 .
5.
1
ez
6.
ez1
ez2
= e−z .
= ez1 −z2 .
7. (ez )n = enz for all n ∈ Z.
8. ez is a periodic function with period of 2πi; that is ez+2πi = ez .
Proof. Exercise.
Example 1. Solve the equation ez = 1.
Example 2. Find all the value of z such that eiz = −2.
Exercises (page 67): 1 − 11, 14 − 15
2
24. Trigonometric functions
Euler’s formula gives eix = cos x + i sin x and e−ix = cos x − i sin x. Thus we have cos x = 12 (eix + e−ix )
and sin x = 2i1 (eix − e−ix ). Generalizing these formulas we have the following definition.
Definition. (The sine and cosine functions)
We define the sine and cosine functions of a complex variable z to be
sin z =
and
1 iz
(e − e−iz )
2i
1
cos z = (eiz + e−iz ).
2
Example 1. Find sin(iπ).
Proposition 3.
1. The sine and cosine functions are entire functions.
2.
d
d
sin z = cos z and
cos z = − sin z.
dz
dz
3. sin(−z) = − sin z and cos(−z) = cos z.
Proof. Exercise.
Proposition 4. (Trigonometric identities)
(1) sin2 z + cos2 z = 1.
(2) sin(z1 ± z2 ) = sin z1 cos z2 ± sin z2 cos z1 .
(3) cos(z1 ± z2 ) = cos z1 cos z2 ∓ sin z1 sin z2 .
Proof.
Proposition 5. We have the following:
(1) If z = iy is pure imaginary, then sin(iy) = i sinh y and cos(iy) = cosh y.
(2) If z = x + iy, then cos z = cos x cosh y − i sin x sinh y.
(3) If z = x + iy, then sin z = sin x cosh y + i cos x sinh y.
Proof.
Example 1. Find all the roots of sin z = 1.
Solution: z = (4k + 1) π2 ,
k ∈ Z.
Example 2. Show that cos(iz) = cos(iz).
Example 3. Find all the roots of sin z = cos z.
Solution: z =
π
4
+ 2kπ,
k ∈ Z.
Proposition 6. We have the following:
3
(1) | sin z|2 = sin2 x + sinh2 y.
(2) | cos z|2 = cos2 x + sinh2 y.
(3) sin z = 0 if and only if z = nπ, n ∈ Z
(4) cos z = 0 if and only if z = (2n + 1) π2 , n ∈ Z
Proof. Exercises.
Example 1. Show that | cos z| ≥ | cos x|.
The tangent, cotangent, secant, and cosecant functions
Definition. We define the tangent, cotangent, secant, and cosecant functions of a complex variable
z to be
cos z
1
1
sin z
, cot z =
, sec z =
, csc z =
.
tan z =
cos z
sin z
cos z
sin z
Proposition 7.
(1) tan z and sec z are analytic everywhere except at the singular points z = (2n + 1) π2 , n ∈ Z.
(2) cot z and csc z are analytic everywhere except at the singular points z = nπ, n ∈ Z.
(3)
d
tan z = sec2 z,
dz
d
cot z = − csc2 z,
dz
d
sec z = sec z csc z,
dz
Proof. Exercises.
Example 1. Show that tan(z + π) = tan z.
Exercises(page 70): 1 − 11, 13 − 14, 16 − 17
4
d
csc z = − csc z cot z.
dz
25. Hyperbolic functions
Definition. We define the hyperbolic functions of a complex variable z as follows:
1
sinh z = (ez − e−z ),
2
1
cosh z = (ez + e−z ),
2
and
tanh z =
sinh z
,
cosh z
coth z =
cosh z
,
sinh z
sechz =
1
,
cosh z
Proposition 8.
1. cosh z = cos(iz) and sinh z = −i sin(iz).
2. cosh z and sinh z are entire functions.
3.
d
d
sinh z = cosh z and
cosh z = sinh z.
dz
dz
4. sinh(−z) = − sinh z and cosh(−z) = cosh z.
5. cosh2 z − sinh2 z = 1.
6. sinh(z1 + z2 ) = sinh z1 cosh z2 + sinh z2 cosh z1 .
7. cosh(z1 + z2 ) = cosh z1 cosh z2 + sinh z1 sinh z2
8. | sinh z|2 = sinh2 x + sin2 y.
9. | cosh z|2 = sinh2 x + cos2 y
10. sinh z = 0 if and only if z = nπi, n ∈ Z
11. cosh z = 0 if and only if z = (2n + 1) πi
, n∈Z
2
Proof.
Example 1. Show that sinh(z + iπ) = − sinh z.
Example 2. Find all roots of the equation cosh z = i.
z = ln |1 ±
√
2| + (±
π
+ 2nπ)i,
2
Exercises (page 72): 1 − 10, 14
5
n ∈ Z.
cschz =
1
.
sinh z
26. The logarithmic function and its branches
In this section, we will define the logarithmic function of a complex variable. As a first step we solve
the equation ew = z, where z is a nonzero complex number. Writing w = u + iv and z = |z|eiArgz ,
we have the solutions
w = ln |z| + i(Argz + 2nπ), n ∈ Z.
Definition. We define the logarithm of a nonzero complex number z to be
log z = ln |z| + i(Argz + 2nπ),
n∈Z
or
log z = ln |z| + i arg z.
Example 1. Find log(−1) and log(1 + i).
Example 2. Find log(ez ).
Remark.
1. logz is not a function. We will call it a multiple-valued function.
2. elog z = z for all z 6= 0.
3. log ez = z + 2nπi,
n ∈ Z, for all z ∈ C.
Definition. The principal value of the logarithm of a nonzero complex number z , log z, is defined
to be
Logz = ln |z| + iArgz.
√
Example 1. Find Log(1 + 3i).
Remark.
1. Logz is a function.
2. log z = Logz + 2nπi,
n ∈ Z.
Branches of log z
As we mentioned above, log z is a multiple-valued function. Thus to define a single-valued function
we restrict the domain and define functions as
log z = ln |z| + i arg z,
α < arg z ≤ α + 2π,
where α is any real number. However these functions are not continuous in their domains. To see
this let us consider Logz. If we approach −1 along the curve z = e(π−t)i , 0 ≤ t ≤ π, then
lim Logz = lim (π − t)i = πi.
z−→−1
t−→0
But if we approach −1 along the curve z = e−(π−t)i , 0 ≤ t ≤ π, then
lim Logz = lim −(π − t)i = −πi.
z−→−1
t−→0
Thus lim Logz does not exist and hence Logz is not continuous at −1. The same arguments can
z−→−1
be used to show that Log z is not continuous at any negative real number. Deleting these points
from the domain of Logz, we obtain an analytic function
Logz = ln |z| + iArgz,
|z| > 0, −π < Argz < π.
6
Definition. Let f be a multiple-valued function. A single-valued function F is called a branch of f
if there is a domain D such that
1. F (z) ∈ f (z) for all z ∈ D.
2. F is analytic in D.
Example 2.
Logz = ln |z| + iArgz,
|z| > 0, −π < Argz < π.
is a branch of log z. It is called the principal branch.
Proposition 9. For any real number α, log z = ln |z| + i arg z, |z| > 0, α < arg z < α + 2π, is a
branch of log z and
1
d
log z = , |z| > 0, α < arg z < α + 2π.
dz
z
Proof.
Definition.
1. A branch cut is a portion of a line or curve that is introduced to define a branch of a multiplevalued function.
2. A point that is common to all branch cuts of a function f is called a branch point of f .
Example 1. The branch cut for the principal branch Logz is the origin and the ray θ = π and the
branch point is the origin.
Example 2. Find the roots of the equation log z = 1.
Solution: z = e.
Example 3. Show that Log(z + 1) is analytic everywhere except on the half line y = 0, x ≤ −1.
7
27. Some identities involving logarithms
Many identities involving logarithms of positive real numbers remain valid, with certain modification,
for logarithms of nonzero complex numbers.
Proposition 10. Let z1 and z2 be two nonzero complex numbers. Then
log(z1 z2 ) = log z1 + log z2
and
log
z 1
z2
= log z1 − log z2 .
Proof.
log(z1 z2 ) = ln |z1 z2 | + i arg(z1 z1 )
= ln |z1 | + ln |z2 | + i(arg z1 + arg z2 )
= log z1 + log z2 .
Remark. The statements in the above proposition have the same meaning as the statements for
arg(z1 z2 ) and arg(z1 /z2 ). That is it means that if two of the three logarithms are specified, then
there is a third logarithm such that they are true.
Example 1. Let z1 = −1 and z2 = i. Verify that log(z1 z2 ) = log z1 + log z2 .
Example 2. Let z1 = −1 and z2 = i. Show that Log(z1 z2 ) 6= Logz1 + Logz2 .
Proposition 11. Let z be a nonzero complex number. Then
z n = en log z ,
and
1
z 1/n = e n log z ,
n∈Z
n ∈ N.
Proof.
Example 1. Show that Log(i3 ) 6= 3Log i.
Exercises (page 79): 1 − 13, 15 − 17
8
28. Complex Exponents
In this section we will use the logarithms to define complex exponents, z c and cz .
Definition. Let z be a nonzero complex number and let c be any complex number. We define z c as
z c = ec log z ,
where log z = ln |z| + i arg z is the multiple-valued function.
Example 1. Find the values of 1i .
Example 2. Find the values of i2+i .
Example 3. Find the values of (1 +
√ 3/2
3i) .
Remark. In general, z c is a multiple-valued function.
Proposition 12. If z 6= 0 and c are complex numbers, the
1
= z −c .
zc
Proof. Exercise.
Proposition 13. Let α be a real number and let log z = ln |z| + i arg z, |z| > 0, α < arg z < α + 2π,
be a branch of log z. Then z c = ec log z is analytic in D = {(r, θ) : r > 0, α < θ < α + 2π}. Moreover
d c
z = cz c−1 .
dz
Proof.
Definition. The principal value of z c is z c = ecLog z .
Example 1. Find the principal value of (1 + i)2+3i .
The square root function
As a special case of the function z c we will study z 1/2 . By definition
p
1
i
z 1/2 = e 2 log z = ± |z| e 2 Arg z .
1
This is a two-valued function. The principal value of z 2 is
p
i
z 1/2 = |z| e 2 Arg z
1
and the principal branch of z 2 is
z 1/2 =
p
i
|z| e 2 Arg z , r > 0, −π < Arg z < π.
1
The principal branch of z 2 is analytic in D = {(r, θ) : r > 0, −π < θ < π} and
d 1/2
1
z = 1/2 .
dz
2z
1
Example 1. Find values of (1 + i) 2 .
9
The function cz
Definition. Let c be a nonzero complex number and let z be any complex number. We define
cz = ez log c ,
where log c = ln |c| + i arg c is the multiple-valued function.
Example 1. Consider 2z and find 2i and 2(i+1) .
Proposition 14. If a value of log c is specified, then cz is an entire function and
d z
c = cz log c.
dz
Proof. Exercise.
π
Example 1. If log i = Logi = π2 i, then iz = e 2 iz is an entire function.
10
29. Inverse trigonometric and hyperbolic functions
Inverses of the trigonometric and hyperbolic functions can be defined in terms of logarithms.
Consider the equation
sin w = z.
Substituting sin w =
1
(eiz
2i
− e−iz ) into this equation, we get
eiz − e−iz = 2iz.
Multiplying both sides by eiz and rearranging the equation, we have
e2wi − 2izeiw = 1.
Completing the square, we obtain
(eiw − iz)2 = 1 − z 2 .
Thus we have
1
eiw = iz + (1 − z 2 ) 2 ,
1
where (1 − z 2 ) 2 is double-valued function. Taking logarithms of both side, we have
1
w = −i log[iz + (1 − z 2 ) 2 ].
Definition. ( Inverse of the sine function)
We define sin−1 z as
1
sin−1 z = −i log[iz + (1 − z 2 ) 2 ],
p
1
1
2
where (1 − z 2 ) 2 = ± |1 − z 2 |e 2 Arg(1−z ) .
Remark. sin−1 z is a multiple-valued function.
Example 1. Find the values of sin−1 2
Solution:
√
1
sin−1 2 = (2n + )π ∓ i ln |2 + 3|, n ∈ Z.
2
Definition. ( Inverse of the cosine and tangent functions)
We define
1
cos−1 z = −i log[z + i(1 − z 2 ) 2 ]
and
tan−1 z =
i + z i
log
,
2
i−z
p
1
1
2
where (1 − z 2 ) 2 = ± |1 − z 2 |e 2 Arg(1−z ) .
Remark. When specific branches of the square root and logarithmic functions are used, the functions
sin−1 z, cos−1 z, and tan−1 z become single-valued and analytic in some domain because they are
compositions of analytic functions. Moreover we have
1
d
sin−1 z =
1 ,
dz
(1 − z 2 ) 2
d
−1
cos−1 z =
1 ,
dz
(1 − z 2 ) 2
d
1
tan−1 z =
.
dz
1 + z2
11
Definition. ( Inverse hyperbolic functions)
We define
1 + z 1
tanh−1 z = log
,
2
1−z
1
sinh−1 z = log[z + (z 2 + 1) 2 ],
and
1
cosh−1 z = log[z + (z 2 − 1) 2 ],
p
1
1
2
where (1 − z 2 ) 2 = ± |1 − z 2 |e 2 Arg(1−z ) .
Example 1. Solve the equation sin z = 1.
Example 2. Solve the equation cosh z = i.
Exercises (page 85): 1 − 16
12