3 Elementary Functions 3.1 The Exponential Function For z = x + iy we have ez = exeiy where Euler's formula gives diy = cos y + i sin y The note: • When y = 0 we have ex the usual exponential. p • When z = 1=n for n = 1; 2; 3::: we have ez = e1=n = n e, the positive nth root of e. • In polar coordinates, ez = exeiy = reiθ, from which j ez j= ex = r and arg(ez) = y + 2πn = θ + 2πn: • Since ex 6= 0, this means ez 6= 0. • Other properties as expected, ez1 d ez1 ez2 = ez1+z2 ; = ez1−z2 ; ez = ez ez2 dz and ez is entire. • Some properties are unexpected, ez+2πi = eze2πi = ez so the function is periodic with period 2πi. And ez can be negative, since ei(2n+1)π = ei2πn+iπ = ei2nπeiπ = (1)(−1) Examples: 1. Find numbers z = x + iy such that p ez = 1 + 3i: p Write 1 + 3i as ez = exeiy = 2eiπ=3; x π then e = 2 and y = 3 + 2nπ for n = 0; ±1 ± 2; :::. Then, π ln(ex) = x =) x = ln(2) =) z = ln(2) + iπ + 2nπ n 2 3 Z 1 3.2 The Logarithmic Function The log function comes from solving ew = z for w, where z is a non-zero complex number. Note z = reiΘ with modulus j z j= r, and the principle argument −π < Θ ≤ π, and w = u + iv so ew = z =) eueiv = reiΘ =) eu = r and v = Θ + 2nπ =) u = ln r and v = Θ + 2nπ =) w = ln r + i(Θ + 2nπ): Definition: The logarithmic function is the multi-valued complex function given by log(z) = ln r + i(Θ + 2nπ) n 2 Z where Θ = Arg(z). The principle value of log(z) is Log(z) = ln r + iΘ; and so log(z) = Log(z) + 2nπi n 2 Z: Some notes: • log(z1z2) = log(z1) + log(z2) • One usual composition holds, for z = reiΘ, elog(z) = eln r+i(Θ+2nπ) = eln reiΘe2nπi = reiΘ · 1 = z • The other composition doesn't necessarily hold, since for z = x + iy log(ez) = log ex+iy = log exeiy = ln(ex) + i(y + 2nπi) = x + iy + 2nπi takes multiple values. 2 Examples: p 1. Let z = −1 − 3i. Then r = 2 and Θ = −2π=3. Hence Log(z) = ln(2) − 2πi=3 and 1 log(z) = ln(2) − 2πi=3 + 2nπi = ln(2) + 2π n − i n 2 : 3 Z 2. Since the complex number 1 has r = 1 and θ = 0 we have log(1) = ln(1) + 2nπi = 2nπi and Log(1) = 0 as we would want. 3. The complex number −1 has r = 1 and θ = π so log(−1) = ln(1) + i(π + 2nπ) = (1 + 2n)πi and so Log(−1) = πi. 4. Some properties carry over from calculus, for example p 2 (1 + i)2 = 2eπi=4 = 2eπi=2 so p Log(1 + i) = ln( 2) + πi=4 and h i Log (1 + i)2 = ln(2) + πi=2 imply h i Log (1 + i)2 = 2Log(1 + i): p But this isn't always true, for example −1 + i = 2e3πi=4 gives p 2 (−1 + i)2 = 2e3πi=4 = 2e3πi=2 −π so r = 2 while Arg(−1 + i) = 2 . So we have p 3πi Log(−1 + i) = ln( 2) + 4 and h i πi Log (−1 + i)2 = ln(2) − 2 and so h i Log (−1 + i)2 6= 2Log(−1 + i) 3 3.3 Branches and Derivatives of Logarithm Functions For z = reiθ a non-zero complex number, if Arg(z) = Θ and arg(z) = θ = Θ + 2nπ for n 2 Z, and hence log(z) = ln r + i(Θ + 2nπ) = ln r + iθ for n 2 Z. Let α 2 R and restrict θ so that α < θ < α + 2π then we have a complex function log(z) = ln r + iθ (r > 0; α < θ < α + 2π) with components u(r; θ) = ln r and v(r; θ) = θ defined on the domain C n rayα where iα rayα = fre : 0 < r < 1g: On this domain, log(z) is single-valued, continuous, and analytic on its domain, since ur = 1=r vr = 0 uθ = 0 vθ = 1 =) rur = vθ and uθ = −rvr and from C-R it follows d log(z) = e−iθ(u + iv ) = 1=z (j z j> 0; α < arg(z) < α + 2π) dz r r and d Log(z) = e−iΘ(u + iv ) = 1=z (j z j> 0; −π < Arg(z) < π) dz r r Away from this domain, i,e, on rayα the function is not continuous, since [IMAGE] Definition: For a multiple valued function, f, a branch of f is any single-valued function F that is analytic in some domain at each point z for which F (z) is a value of f. • The portion of line or curve being removed to define the branch is called the branch cut. • The branch cut for the example above is the origin and rayα. 4 • Log(z) = ln r + iΘ(r > 0; −π < Θ < π) is the principle branch of log(z). • The branch cut for the principle branch is the origin and rayπ = ray−π. Examples: 1. A non-prinicple branch for the logarithm function is π 9π log(z) = ln r + iθ r > 0; < θ < 4 4 We will show that for this branch, log(i2) = 2 log(i): Write, 4πi log(i2) = log(−1) = ln1 + = πi 4 and notice 2πi 2 log(i) = 2 ln 1 + = πi: 4 2. Another non-principle branch for the logarithm function is 3π 11π log(z) = ln r + iθ r > 0; < θ < : 4 4 In this case, we write 5πi 5πi log(i2) = log(−1) = ln 1 + = 4 4 while 10πi 20πi 2 log(i) = 2 ln 1 + = 4 4 3. What does it mean when we say that log(z1z2) = log(z1) + log(z2) (1) holds? Let z1 = z2 = −1. Recall, log(1) = ln 1 + i (0 + 2nπ) = 2nπi and log(−1) = ln 1 + i (π + 2nπ) = (1 + 2n)πi for n 2 Z. Then, since z1z2 = 1, we have log(z1z2) = 0 for n = 0 5 and log(z1) = πi for n = 0 and so log(z2) = −πi for n = −1 makes the statement true. It really means log(z1) + log(z1) = (1 + 2n)πi + (1 + 2m)πi = 2(1 + n + m)πi = log(z1z2): 4. Now we will show that sometimes (but not always) Log(z1z2) = Log(z1) + Log(z2) holds. Let z1 and z2 be non-zero complex numbers with Re z1 > 0 and Re z2 > 0: Then, iΘ1 iΘ2 z1 = r1e and z2 = r2e π π where − 2 < Θi < 2 , and thus −π < Θ1 + Θ2 < π: Then i(Θ1+Θ2) z1z2 = r1r2e and hence Arg (z1z2) = Θ1 + Θ2 is principal. Therefore, Log(z1z2) = ln j z1z2 j +iArg (z1z2) = ln r1r2 + i(Θ1 + Θ2) = ln r1 + ln r2 + i(Θ1) + i(Θ2) = Log(z1) + Log(z2) 5. Just as we would expect from calculus, zn = en log(z) To see this, just notice z = reiθ =) zn = rneinθ and en log(z) = en(ln r+iθ) = en ln reinθ = rneinθ: And we also see that for nonzero z the following holds, 1=n 1 log(z) z = e n : 6 To see this, iΘ z = re =) log(z) = ln r + i(Θ + 2kπ) for k 2 Z 1 ln r i(Θ + 2kπ) =) log(z) = + for k 2 n n n Z 1 log(z) ln r + i(Θ+2kπ) =) e n = e n n for k 2 Z 1 log(z) ln r i(Θ+2kπ) =) e n = e n e n for k 2 Z 1 p Θ 2kπ log(z) n i( + ) =) e n = r · e n n for k 2 Z 1 log(z) 1=n =) e n = z 3.4 The Power Function Definition: For z 6= 0 and complex constant c, the power function is defined by zc = ec log z: Some properties: • We already know that zn = en log z and z1=n = e1=n log(z). 1 e0 0−z −z • Recalling that ez = ez = e = e , we have 1 1 = = e−c log z = z−c: zc ec log z • For a fixed branch, (r > 0; α < θ < α + 2π) the log function is analytic, so we define zc there. Then, d d c zc = ec log z = ec log z: dz dz z then recalling elog z = z, we have d c ec log z zc = ec log z = c = ce(c−1) log z = czc−1 dz elog z elog z on the branch (r > 0; α < θ < α + 2π). • The principal value of zc is zc = ecLogz; and the principle branch is this function on the domain (r > 0; −π < Arg z < π). Examples: 1. Consider the function ii = ei log i. Then, π π log(i) = ln j i j +i + 2nπ = i + 2nπ 2 2 and hence i i·i π +2nπ −π 1 +2n i = e ( 2 ) = e ( 2 ) where all values are in R, and with principle value i − π i = e 2 : 7 2. Consider log(−1) = ln 1 + i(π + 2nπ) = (1 + 2n)πi from which it's easy to see 1/π 1 log(−1) 1 ((1+2n)πi) (1+2n)i (−1) = e π = e π = e : 3. The principle branch of z2=3 is p 2=3 2 Logz 2 r+ 2 iΘ 3 i 2Θ z = e 3 = e 3 3 = r2e 3 : with −π < Θ < π, and so the principle value of z2=3 is p3 2Θ p3 2Θ z2=3 = r2 cos + i r2 sin : 3 3 this is analytic in domain r > 0 − π < Θ < π.