CHAPTER 4

Elementary Functions

BY

Dr. Pulak Sahoo

Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India E-mail : [email protected]

1 Module-4: Multivalued Functions-II

Theorem 1. The most general solution of ew = z, z 6= 0 is given by

w = log z = ln | z | +i(Arg z + 2nπi) = ln | z | +i arg z, n ∈ I. (1)

Proof. Setting z = reiθ (r > 0, θ = Arg z), we have

ew = z ⇒ eueiv = reiθ.

Comparing we get eu = r and eiv = eiθ. This gives

u = ln r and v = θ + 2nπ, n ∈ I where ln r is the natural logarithm, to the base e, of a positive real number. Therefore, we have the expression

w = log z = ln | z | +i(Arg z + 2nπi), n ∈ I which has infinitely many values at every z 6= 0.

Remark 1. Since log z is not a uniquely defined function of z, it is appropriate to intro- duce the principal value of log z for z 6= 0. For z 6= 0, we call

ln | z | +iArgz as the principal value of log z and we denote it by Logz. Therefore, (1) can be rewritten in the form

log z = Logz + 2nπi, n ∈ I.

We note that the function w = log z, z 6= 0 is multivalued and this multivaluedness is due to the fact that many values connected with the argument of a complex number. The method for investigating continuity and other properties for single valued functions

2 cannot be used for multivalued functions. Since a multivalued function can be considered as a collection of single valued functions, the nature of multivalued function may then be examined from the point of view of its single valued counterparts. We define a branch of log z to be any single valued function log∗ z satisfying the identity elog∗ z = z for all nonzero complex values of z. There are infinitely many branches associated with the multivalued function log z. Each branch is an inverse of the function ez. Among all the branches of log z, there is exactly one branch whose imaginary part e.g. arg z is defined in the interval (−π, π]. This branch is called the principal branch of log z and is denoted by Logz. Each branch of log z differs from the principal branch by a multiple of 2πi. Thus, if log∗ z is a fixed branch, then

log∗ z = Logz + 2nπi (2) for some integer n. We note that

Logz = ln | z | +iArgz (−π < Argz ≤ π). (3)

The restriction in (3) may be viewed geometrically as a cut of the z-plane along the negative real axis. This ray is then called the branch cut for the function Logz. Other branches of log z may be defined by restricting arg z to

(2n − 1)π < arg z ≤ (2n + 1)π, n ∈ I.

Next we examine whether the identity

log(z1z2) = log z1 + log z2 (z1, z2 6= 0) is valid in the complex plane. Now for a fixed branch, we have

log(z1z2) = ln | z1z2 | +i arg(z1z2) and

log z1 + log z2 = ln | z1 | + ln | z2 | +i(arg z1 + arg z2)

= ln | z1z2 | +i(arg z1 + arg z2).

Since arg(z1z2) and arg z1 + arg z2 differ by an integer multiple of 2π, we can write

log(z1z2) = log z1 + log z2 + 2nπi, n ∈ I.

3 Again, the property

Log(z1z2) = Logz1 + Logz2

does not hold true always. The choice z1 = z2 = −1 serves a simple example for the job. Here, as z1z2 = 1 we have Log z1z2 = Log 1 = 0 but Logz1 + Logz2 = 2πi.

1 Complex Exponents

We recall that there are n distinct complex values associated with w = z1/n, z 6= 0.

iθ 1/n i[(θ+2kπ)/n] If z = re then these values are zk = r e , k = 0, 1, ... , (n − 1). The values of z1/n changes as the argument of z takes the values θ, θ + 2π, ... , θ + 2(n − 1)π. By restricting arg z to a particular branch we can get a unique value of z1/n. This seems to indicate a link between the roots of z1/n and logarithm of a complex number. We may define the function z1/n as

w = z1/n = e(1/n) log z = e(1/n)(ln|z|+i arg z). (4)

Since the logarithm function is multivalued, the function defined in (4) is also multivalued. Setting arg z = Argz + 2kπ, k is an integer, we see that (4) assumes different values for k = 0, 1, ... , (n − 1). More generally, if m and n are positive integers with no common factors, we define

(z1/n)m = e(m/n) log z, z 6= 0.

This, too, has n distinct values. We now define zα for complex values of α by

zα = eα log z. (5)

In view of this definition we see that zα is a multivalued function, provided α is not an integer. Replacing log z by Logz in (5) we obtain the principal value of zα as

zα = eαLogz.

The principal branch and any other branch of the multivalued function zα are obtained from Logz and log z respectively.

4 2 Inverse Trigonometric Functions

The inverse of sine and cosine functions of a complex variable z = x + iy are defined as

w = sin−1 z, when z = sin w; and w = cos−1 z, when z = cos w.

Result 1 The inverse trigonometric functions are multivalued functions. Proof By definition of sin−1 z we have

w = sin−1 z ⇒ z = sin w eiw − e−iw ⇒ z = 2i ⇒ (eiw)2 − 2izeiw − 1 = 0

⇒ eiw = iz + (1 − z2)1/2

⇒ iw = log[iz + (1 − z2)1/2]

⇒ sin−1 z = −i log[iz + (1 − z2)1/2]. (6)

Similarly, it can be shown that

cos−1 z = −i log[z + i(1 − z2)1/2]. (7)

Also

w = tan−1 z ⇒ z = tan w (eiw − e−iw)/2i ⇒ z = (eiw + e−iw)/2 e2iw − 1 ⇒ iz = e2iw + 1 1 + iz ⇒ e2iw = 1 − iz 1 i − z  ⇒ w = log 2i i + z i i + z  ⇒ tan−1 z = log . (8) 2 i − z In view of (6)-(8), we see that sin−1 z, cos−1 z and tan−1 z are multivalued as the logarith- mic function is so. Replacing z by 1/z in (6), (7) and (8) we can get the corresponding expression of csc−1 z, sec−1 z and cot−1 z respectively.

5 The derivatives of sin−1 z and cos−1 z depend on the values chosen for square roots. It happens due to being multivalued. Therefore,

d 1 d 1 (sin−1 z) = , (cos−1 z) = − . dz (1 − z2)1/2 dz (1 − z2)1/2

The derivative of tan−1 z does not depend upon the way it is made single valued and

d 1 (tan−1 z) = . dz 1 + z2

3 Inverse Hyperbolic Functions

Inverse of hyperbolic trigonometric functions are defined in an analogous manner as

sinh−1 z = log[z + (z2 + 1)1/2];

cosh−1 z = log[z + (z2 − 1)1/2]; 1 1 + z tanh−1 z = log . 2 1 − z Replacing z by 1/z in the above three relations, the corresponding expressions for csc h−1z, sec h−1z and coth−1 z are available, respectively. Note Inverse hyperbolic functions are multivalued.

Example 1. Find all the values of (1 + i)1+i. Also specify the principal value.

Solution. We have

(1 + i)1+i = e(1+i) log(1+i) √ = e(1+i)[ln 2+i(2nπ+π/4)]

( 1 ln 2−2nπ−π/4)+i( 1 ln 2+2nπ+π/4) = e 2 2

1 ln 2−2nπ−π/4 i( 1 ln 2+2nπ+π/4) = e 2 e 2 , n ∈ I.

This gives the general value. The principal value is obtained from the general value by putting n = 0 and is

1 ln 2−π/4 i( 1 ln 2+π/4) e 2 e 2 .

Example 2. Find all possible values of z1+i = 2.

6 Solution. We have

1+i (1+i) log z ln 2+2nπi z = 2 ⇒ e = 2 = e , n ∈ I ⇒ (1 + i) log z = ln 2 + 2nπi 1 i ⇒ log z = (ln 2 + 2nπ) − (ln 2 − 2nπ) 2 2 1 (ln 2+2nπ)− i (ln 2−2nπ) ⇒ z = e 2 2 , n ∈ I.

Example 3. Solve the equation cot z = 2i.

−1 −1 i i+z
Solution. The solution set is given by z = cot (2i). We have tan z = 2 log i−z . −1 i  i+1/z  i iz+1
Hence, cot z = 2 log i−1/z = 2 log iz−1 . Therefore

i 1 z = cot−1(2i) = log 2 3 √ = nπ − i ln 3, n ∈ I.

 a−ib  2ab Example 4. Prove that tan i log a+ib = a2−b2 , (a, b) 6= (0, 0).

−1 i i+z
2ab Solution. We know that tan z = 2 log i−z . Now replacing z by a2−b2 we get

  2ab ! 2ab i i + 2 2 tan−1 = log a −b a2 − b2 2 2ab i − a2−b2 i (a2 − b2)i + 2ab = log 2 (a2 − b2)i − 2ab i a2 − b2 − 2abi = log 2 a2 − b2 + 2abi i (a − ib)2  = log 2 (a + ib)2 a − ib = i log . a + ib

From this we get our desired result.

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