CHAPTER 4. Elementary Functions. Dr. Pulak Sahoo
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1 CHAPTER 4 Elementary Functions BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India sahoopulak1@gmail.com 1
2 Module-4: Multivalued Functions-II Theorem 1. The most general solution of e w = z, z 0 is given by w = log z = ln z +i(arg z + nπi) = ln z +i arg z, n I. (1) Proof. Setting z = re iθ (r > 0, θ = Arg z), we have e w = z e u e iv = re iθ. Comparing we get e u = r and e iv = e iθ. This gives u = ln r and v = θ + nπ, 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 + nπi), n I which has infinitely many values at every z 0. Remark 1. Since log z is not a uniquely defined function of z, it is appropriate to introduce the principal value of log z for z 0. For z 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 + nπi, n I. We note that the function w = log z, z 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
3 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 e log 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 e z. 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 πi. Thus, if log z is a fixed branch, then log z = Logz + nπi () 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 (n 1)π < arg z (n + 1)π, n I. Next we examine whether the identity log(z 1 z ) = log z 1 + log z (z 1, z 0) is valid in the complex plane. Now for a fixed branch, we have log(z 1 z ) = ln z 1 z +i arg(z 1 z ) and log z 1 + log z = ln z 1 + ln z +i(arg z 1 + arg z ) = ln z 1 z +i(arg z 1 + arg z ). Since arg(z 1 z ) and arg z 1 + arg z differ by an integer multiple of π, we can write log(z 1 z ) = log z 1 + log z + nπi, n I. 3
4 Again, the property Log(z 1 z ) = Logz 1 + Logz does not hold true always. The choice z 1 = z = 1 serves a simple example for the job. Here, as z 1 z = 1 we have Log z 1 z = Log 1 = 0 but Logz 1 + Logz = πi. 1 Complex Exponents We recall that there are n distinct complex values associated with w = z 1/n, z 0. If z = re iθ then these values are z k = r 1/n e i[(θ+kπ)/n], k = 0, 1,..., (n 1). The values of z 1/n changes as the argument of z takes the values θ, θ + π,..., θ + (n 1)π. By restricting arg z to a particular branch we can get a unique value of z 1/n. This seems to indicate a link between the roots of z 1/n and logarithm of a complex number. We may define the function z 1/n as w = z 1/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 + kπ, 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 (z 1/n ) m = e (m/n) log z, z 0. This, too, has n distinct values. We now define z α for complex values of α by z α = e α log z. (5) In view of this definition we see that z α is a multivalued function, provided α is not an integer. Replacing log z by Logz in (5) we obtain the principal value of z α as z α = e αlogz. The principal branch and any other branch of the multivalued function z α are obtained from Logz and log z respectively. 4
5 Inverse Trigonometric Functions as The inverse of sine and cosine functions of a complex variable z = x + iy are defined Result 1 w = sin 1 z, when z = sin w; and w = cos 1 z, when z = cos w. The inverse trigonometric functions are multivalued functions. Proof By definition of sin 1 z we have Similarly, it can be shown that Also w = sin 1 z z = sin w z = eiw e iw i (e iw ) ize iw 1 = 0 e iw = iz + (1 z ) 1/ iw = log[iz + (1 z ) 1/ ] sin 1 z = i log[iz + (1 z ) 1/ ]. (6) cos 1 z = i log[z + i(1 z ) 1/ ]. (7) w = tan 1 z z = tan w z = (eiw e iw )/i (e iw + e iw )/ iz = eiw 1 e iw + 1 e iw = 1 + iz 1 iz w = 1 ( ) i z i log i + z tan 1 z = i log ( i + z i z ). (8) In view of (6)-(8), we see that sin 1 z, cos 1 z and tan 1 z are multivalued as the logarithmic 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
6 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 dz (sin 1 z) = 1 (1 z ) 1/, d 1 dz (cos 1 z) = (1 z ). 1/ The derivative of tan 1 z does not depend upon the way it is made single valued and d dz (tan 1 z) = z. 3 Inverse Hyperbolic Functions Inverse of hyperbolic trigonometric functions are defined in an analogous manner as sinh 1 z = log[z + (z + 1) 1/ ]; cosh 1 z = log[z + (z 1) 1/ ]; tanh 1 z = 1 log 1 + z 1 z. Replacing z by 1/z in the above three relations, the corresponding expressions for csc h 1 z, sec h 1 z 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 (1+i) log(1+i) = e = e (1+i)[ln +i(nπ+π/4)] = e ( 1 ln nπ π/4)+i( 1 ln +nπ+π/4) = e 1 ln nπ π/4 e i( 1 ln +nπ+π/4), n I. This gives the general value. The principal value is obtained from the general value by putting n = 0 and is e 1 ln π/4 e i( 1 ln +π/4). Example. Find all possible values of z 1+i =. 6
7 Solution. We have z 1+i = e (1+i) log z = = e ln +nπi, n I (1 + i) log z = ln + nπi log z = 1 (ln + nπ) i (ln nπ) z = e 1 (ln +nπ) i (ln nπ), n I. Example 3. Solve the equation cot z = i. Solution. The solution set is given by z = cot 1 (i). We have tan 1 z = i log ( ) i+z i z. ( ) Hence, cot 1 z = i log i+1/z = i log ( iz+1 i 1/z iz 1). Therefore z = cot 1 (i) = i log 1 3 = nπ i ln 3, n I. Example 4. Prove that tan [ ] i log a ib a+ib =, (a, b) (0, 0). a b Solution. We know that tan 1 z = i log ( ) i+z i z. Now replacing z by we get a b ( ) ( ) tan 1 = i i + a b log a b i a b = i ( ) (a log b )i + (a b )i = i ( ) a log b i a b + i = i ( ) (a ib) log (a + ib) ( ) a ib = i log. a + ib From this we get our desired result. 7
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