3 Elementary Functions

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3 Elementary Functions 3 Elementary Functions We already know a great deal about polynomials and rational functions: these are analytic on their entire domains. We have thought a little about the square-root function and seen some difficulties. The remaining elementary functions are the exponential, logarithmic and trigonometric functions. 3.1 The Exponential and Logarithmic Functions (§30–32, 34) We have already defined the exponential function exp : C ! C : z 7! ez using Euler’s formula ez := ex cos y + iex sin y (∗) and seen that its real and imaginary parts satisfy the Cauchy–Riemann equations on C, whence exp C d z = z is entire (analytic on ). Indeed recall that dz e e . We have also seen several of the basic properties of the exponential function, we state these and several others for reference. Lemma 3.1. Throughout let z, w 2 C. 1. ez 6= 0. ez 2. ez+w = ezew and ez−w = ew 3. For all n 2 Z, (ez)n = enz. 4. ez is periodic with period 2pi. Indeed more is true: ez = ew () z − w = 2pin for some n 2 Z Proof. Part 1 follows trivially from (∗). To prove 2, recall the multiple-angle formulae for cosine and sine. Part 3 requires an induction using part 2 with z = w. Part 4 is more interesting: certainly ew+2pin = ew by the periodicity of sine and cosine. Now suppose ez = ew where z = x + iy and w = u + iv. Then, by considering the modulus and argument, ( ex = eu exeiy = eueiv =) y = v + 2pin for some n 2 Z We conclude that x = u and so z − w = i(y − v) = 2pin. Example 3.2. Find all z 2 C such that ez = 5(−1 + i). Following the Lemma, write z = x + iy and take the polar form of 5(−1 + i) to see that ( p p x = ln(5 2) z = (− + ) () x iy = 3pi/4 () e 5 1 i e e 5 2e 3p y = 4 + 2pn for some n 2 Z p 3p () z = ln(5 2) + + 2pn i for some n 2 Z 4 We see that there are infinitely many suitable z! Duplicate Notation Warning! When n 2 N, the expression e1/n can now mean two things. For instance e1/3 can mean: p p p 1. The set of cube roots of e, namely f 3 e, 3 ee2pi/3, 3 ee−2pi/3g; p 2. The real value 3 e 2 R+. Given that ez is such a common function, in both real and complex analysis, we default to the second meaning: if you mean the set of nth roots, say so! The periodicity of the exponential leads to the far more interesting notion of the complex logarithm. Definition 3.3. Let z = reiq be a non-zero complex number with principal argument q = Arg z. The principal logarithm of z is the value Log z := ln r + iq = ln jzj + i Arg z where ln is the usual natural logarithm. The logarithm of z is any (and all!) of the valuesa log z = ln jzj + i arg z = ln r + i(q + 2pn) : n 2 Z aThis is identical to how we use arg z, which, depending on context, means either the set fArg z + 2pnig or some particular value from this set. We’ll more formally discuss such multi-valued functions in the next section. Examples 3.4. 1. Since −4 = 4epi, we see that Log(−4) = ln 4 + pi and log(−4) = ln 4 + (1 + 2n)pi 2. Again write in polar form to compute: p p p p Log( 3 − i) = Log(2e−pi/6) = ln 2 − i and log( 3 − i) = ln 2 − i + 2pni 6 6 These examples involve solving equations of the form ew = z: writing z = reiq = eln r+iq as above, and appealing to part 4 of Lemma 3.1, we instantly see that ew = z () w = log z Read this carefully, remembering that the logarithm is multi-valued and the exponential periodic: elog z = z and log(ew) = w + 2pni where n 2 Z Before moving on, we clear up some of the basic properties of the principal logarithm function. All parts of this should be clear from Definition 3.3. 2 Lemma 3.5. Throughout, z and w are complex numbers with z 6= 0, and n 2 Z. • Log : C n f0g ! fw 2 C : Im w 2 (−p, p]g is a bijection with inverse exp. • Log(ew) = w + 2pni where n 2 Z is chosen such that Im w + 2pn 2 (−p, p]. • If z 2 R+, then Log z = ln z is the usual natural logarithm. The Logarithm Laws Just as the standard rules for exponentiation (Lemma 3.1 parts 2 and 3) apply to the complex expo- nential, something similar works for the log laws. However, the multi-valued nature of the logarithm makes this a little more subtle. Suppose non-zero z, w are given: since jzwj = jzj jwj and arg zw = arg z + arg w, we conclude that log zw = ln jzj jwj + i(arg z + arg w) = ln jzj + i arg z + ln jwj + i arg w = log z + log w Be very careful with this expression, since it is not an identity of functions. Indeed it means two things: • We have set equality: in particular, the following sets are identical: log z + log w = fa + b : a 2 log z, b 2 log wg = fjzj + i Arg z + 2pki + jwj + i Arg w + 2pmi : k, m 2 Zg log zw = fln jzj jwj + i Arg zw + 2pni : n 2 Zg • There exist particular choices of the arguments of z, w and zw so that arg zw = arg z + arg w. Unless you are sure you won’t make a mistake, it is therefore safer to write log zw = log z + log w + 2pni where n 2 Z Given its restricted range, we can be more precise for the principal logarithm: 9n 2 f0, ±1g such that Log zw = Log z + Log w + 2pni p p Example 3.6. Let z = − 3 + i = 2e5p/6 and w = 2(1 + i) = 2epi/4. Then 5p p log z = ln 2 + i + 2pki, log w = ln 2 + i + 2pmi 6 4 5p + p 13p/12 13p log zw = log(4e 6 4 ) = log(4e ) = ln 4 + i + 2pni 12 For the above boxed formula to make sense for particular choices of logarithms, we’d need to select k + m = n + 1. More explicitly, 5p p Log z = ln 2 + i, Log w = ln 2 + i 6 4 11p Log zw = Log(4e−11p/12) = Log(4e−11p/12) = ln 4 − i = Log z + Log w − 2pi 12 3 We can similarly demonstrate the other log law, with the same caveat: z log = log z − log w w You might assume that the final log law (log zn = n log z whenever n 2 N) also makes sense, but you’d be very wrong: an example should explain why you should ignore this law! p Example 3.7. Let z = − 3 + i = 2e5p/6 and compute what is generally meant by the set 2 log z: 5p 5p 2 log z = 2 ln 2 + i + 2pmi = ln 4 + i + 4pmi 6 3 that is, we double the value of everything in the set log z. This is different from the set 5p log z2 = log(4e10p/6) = ln 4 + i + 2pki 3 However, in the language of the previous exercise, there would be no problem if each copy of log z gots its own multiples of 2pi: 5p log z + log z = ln 4 + i + 2p(l + m)i where l, m 2 Z 3 = log z2 Since this is not how 2 log z is generally interpreted, it is safer to state that log z2 6= 2 log z. Since the principal logarithm is a function rather than a set, we can be more precise: for any n 2 N, n Log zn = n Log z + 2pki for some integer k with jkj ≤ 2 Example 3.8. Let z = e−13pi/16 and consider z16. We see that Log z16 = Log e−13pi = Log epi = ip, 16 Log z = −13pi =) Log z16 = 16 Log z + 14pi 4 Exercises. 3.1.1. Compute the following: p 2 (a) exp(3 − 2 i) (b) Log(ie) (c) log(3 − 4i) (d) Log[(−1 + i) ] 3.1.2. (a) If ez is real, show that Im z = np for some integer n. (b) If ez is imaginary, what restriction is placed on z? 3.1.3. Show in two ways that the function f (z) = exp(z2) is entire, and find its derivative. 2 3.1.4. Prove, for any z 2 C, that exp(z2) ≤ exp jzj . What must z satisfy if this is to be equality? 3.1.5. Find Re(e1/z) in terms of x and y. Why is this function harmonic in every domain that does not contain the origin? 3.1.6. Show that Log i3 6= 3 Log i. p p 3.1.7. The square roots of i are i = epi/4 and − i = e−3pi/4. p p p 1 (a) Compute Log i and Log(− i) and check that Log i = 2 Log i. (b) Show that the set of all logarithms of all square roots of i is 1 log i1/2 = n + pi where n 2 Z 4 1/2 1 and therefore deduce that log i = 2 log i as sets. 3.2 Multi-valued functions, Branch Cuts and the Power Function (§33, 35, 36) The complex logarithm is often called a multi-valued function, since each log z represents a set of complex numbers.
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