3 Elementary Functions

Transcription

1 3 Elementary Functions 3.1 The Exponential Function For z = x + iy we have where Euler s formula gives The note: e z = e x e iy iy = cos y + i sin y When y = 0 we have e x the usual exponential. When z = 1/n for n = 1,, 3... we have e z = e 1/n = n e, the positive n th root of e. In polar coorinates, e z = e x e iy = re iθ, from which e z = e x = r arg(e z ) = y + πn = θ + πn. Since e x 0, this means e z 0. Other properties as expecte, e z is entire. e z 1 e z = e z 1+z, e z 1 e z = ez 1 z, z ez = e z Some properties are unexpecte, e z+πi = e z e πi = e z so the function is perioic with perio πi. An e z can be negative, since e i(n+1)π = e iπn+iπ = e inπ e iπ = (1)( 1) Examples: 1. Fin numbers z = x + iy such that e z = 1 + 3i. Write 1 + 3i as e z = e x e iy = e iπ/3, then e x = y = π 3 + nπ for n = 0, ±1 ±,... Then, ( π ) ln(e x ) = x = x = ln() = z = ln() + iπ 3 + nπ n Z 1

2 3. The Logarithmic Function The log function comes from solving e w = z for w, where z is a non-zero complex number. Note z = re iθ with moulus z = r, the principle argument π < Θ π, w = u + iv so e w = z = e u e iv = re iθ = e u = r v = Θ + nπ = u = ln r v = Θ + nπ = w = ln r + i(θ + nπ). Definition: The logarithmic function is the multi-value complex function given by log(z) = ln r + i(θ + nπ) n Z where Θ = Arg(z). The principle value of log(z) is Log(z) = ln r + iθ, so log(z) = Log(z) + nπi n Z. Some notes: log(z 1 z ) = log(z 1 ) + log(z ) One usual composition hols, for z = re iθ, e log(z) = ln r+i(θ+nπ) e = e ln r e iθ e nπi = re iθ 1 = z The other composition oesn t necessarily hol, since for z = x + iy takes multiple values. log(e z ) = log ( e x+iy) = log ( e x e iy) = ln(e x ) + i(y + nπi) = x + iy + nπi

3 Examples: 1. Let z = 1 3i. Then r = Θ = π/3. Hence Log(z) = ln() πi/3 ( log(z) = ln() πi/3 + nπi = ln() + π n 1 ) i n Z. 3. Since the complex number 1 has r = 1 θ = 0 we have log(1) = ln(1) + nπi = nπi as we woul want. Log(1) = 0 3. The complex number 1 has r = 1 θ = π so log( 1) = ln(1) + i(π + nπ) = (1 + n)πi so Log( 1) = πi. 4. Some properties carry over from calculus, for example ( ) (1 + i) = e πi/4 = e πi/ so imply Log(1 + i) = ln( ) + πi/4 [ Log (1 + i) ] = ln() + πi/ [ Log (1 + i) ] = Log(1 + i). But this isn t always true, for example 1 + i = e 3πi/4 gives ( ) ( 1 + i) = e 3πi/4 = e 3πi/ so r = while Arg( 1 + i) = π. So we have so Log( 1 + i) = ln( ) + 3πi 4 [ Log ( 1 + i) ] = ln() πi [ Log ( 1 + i) ] Log( 1 + i) 3

4 3.3 Branches Derivatives of Logarithm Functions For z = re iθ a non-zero complex number, if for n Z, hence for n Z. Let α R restrict θ so that then we have a complex function with components efine on the omain C \ ray α where Arg(z) = Θ arg(z) = θ = Θ + nπ log(z) = ln r + i(θ + nπ) = ln r + iθ α < θ < α + π log(z) = ln r + iθ (r > 0, α < θ < α + π) u(r, θ) = ln r v(r, θ) = θ ray α = {re iα : 0 < r < }. On this omain, log(z) is single-value, continuous, analytic on its omain, since from C-R it follows u r = 1/r v r = 0 u θ = 0 v θ = 1 = ru r = v θ u θ = rv r z log(z) = e iθ (u r + iv r ) = 1/z ( z > 0, α < arg(z) < α + π) z Log(z) = e iθ (u r + iv r ) = 1/z ( z > 0, π < Arg(z) < π) Away from this omain, i,e, on ray α the function is not continuous, since [IMAGE] Definition: For a multiple value function, f, a branch of f is any single-value function F that is analytic in some omain at each point z for which F (z) is a value of f. The portion of line or curve being remove to efine the branch is calle the branch cut. The branch cut for the example above is the origin ray α. 4

5 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 ray π = ray π. Examples: 1. A non-prinicple branch for the logarithm function is ( π log(z) = ln r + iθ r > 0, 4 < θ < 9π 4 We will show that for this branch, log(i ) = log(i). ) Write, notice log(i ) = log( 1) = ln1 + 4πi 4 = πi ( log(i) = ln 1 + πi ) = πi. 4. Another non-principle branch for the logarithm function is ( log(z) = ln r + iθ r > 0, In this case, we write 3π 4 < θ < 11π 4 ). while log(i ) = log( 1) = ln 1 + 5πi 4 = 5πi 4 ( log(i) = ln πi ) = 0πi What oes it mean when we say that log(z 1 z ) = log(z 1 ) + log(z ) (1) hols? Let z 1 = z = 1. Recall, log(1) = ln 1 + i (0 + nπ) = nπi log( 1) = ln 1 + i (π + nπ) = (1 + n)πi for n Z. Then, since z 1 z = 1, we have log(z 1 z ) = 0 for n = 0 5

6 so log(z 1 ) = πi for n = 0 log(z ) = πi for n = 1 makes the statement true. It really means log(z 1 ) + log(z 1 ) = (1 + n)πi + (1 + m)πi = (1 + n + m)πi = log(z 1 z ). 4. Now we will show that sometimes (but not always) Log(z 1 z ) = Log(z 1 ) + Log(z ) hols. Let z 1 z be non-zero complex numbers with Re z 1 > 0 Re z > 0. Then, z 1 = r 1 e iθ 1 z = r e iθ where π < Θ i < π, thus π < Θ 1 + Θ < π. Then z 1 z = r 1 r e i(θ 1+Θ ) hence Arg (z 1 z ) = Θ 1 + Θ is principal. Therefore, Log(z 1 z ) = ln z 1 z +iarg (z 1 z ) = ln r 1 r + i(θ 1 + Θ ) = ln r 1 + ln r + i(θ 1 ) + i(θ ) = Log(z 1 ) + Log(z ) 5. Just as we woul expect from calculus, z n = e n log(z) To see this, just notice z = re iθ = z n = r n e inθ e n log(z) = e n(ln r+iθ) = e n ln r e inθ = r n e inθ. An we also see that for nonzero z the following hols, z 1/n = e 1 n log(z). 6

7 To see this, 3.4 The Power Function z = re iθ = log(z) = ln r + i(θ + kπ) for k Z = 1 ln r i(θ + kπ) log(z) = + for k Z n n n = e 1 n log(z) = e ln r n + i(θ+kπ) n for k Z = e 1 n log(z) = e ln r i(θ+kπ) n e n for k Z = e 1 n log(z) = n r e i( Θ n + kπ n ) for k Z = e 1 n log(z) = z 1/n Definition: For z 0 complex constant c, the power function is efine by Some properties: z c = e c log z. We alreay know that z n = e n log z z 1/n = e 1/n log(z). Recalling that 1 e z = e0 e z = e 0 z = e z, we have 1 z c = 1 e c log z = e c log z = z c. For a fixe branch, (r > 0, α < θ < α + π) the log function is analytic, so we efine z c there. Then, z zc = z ec log z = c z ec log z. then recalling e log z = z, we have z zc = c e log z ec log z = c ec log z e log z on the branch (r > 0, α < θ < α + π). The principal value of z c is z c = e clogz, = ce(c 1) log z = cz c 1 the principle branch is this function on the omain (r > 0, π < Arg z < π). Examples: 1. Consier the function i i = e i log i. Then, ( π ) log(i) = ln i +i + nπ hence i i = e i i( π +nπ) = e π( 1 +n) where all values are in R, with principle value i i = e π. 7 ( π ) = i + nπ

8 . Consier from which it s easy to see 3. The principle branch of z /3 is log( 1) = ln 1 + i(π + nπ) = (1 + n)πi ( 1) 1/π = e 1 π log( 1) = e 1 π ((1+n)πi) = e (1+n)i. z /3 = e 3 Logz = e 3 r+ 3 iθ = 3 r e i Θ 3. with π < Θ < π, so the principle value of z /3 is z /3 = 3 r cos Θ 3 + i 3 r sin Θ 3. this is analytic in omain r > 0 π < Θ < π. 4. Consier the non-zero complex numbers, Taking principle values of powers, thus On the other h, z 1 = 1 + i z = 1 i z 3 = 1 i. (z 1 z ) i = i = e ilog = e i ln z1 i = (1 + i) i = e ilog (1+i) = e i(ln +i π 4 ) = e π/4 i(ln )/ e z i = (1 i) i = e ilog (1 i) = e i(ln i π 4 ) = e π/4 i(ln )/ e (z 1 z ) i = z i 1z i. (z z 3 ) i = e ilog (z z 3 ) = e i(ln +iπ) = e π e i ln but so hence z3 i = ( 1 i) i = e ilog (1 i) = e i(ln i 3π 4 ) = e 3π/4 i(ln )/ e z i z i 3 = e π/4 e i(ln )/ e 3π/4 e i(ln )/ = e π e i ln (z z 3 ) i = z i z i 3e π 8

9 3.5 Trig Functions From Euler s formula, we know that { e ix = cos x + i sin x e ix = cos x i sin x = Therefore, it s not a stretch to efine sin z = eiz e iz i cos z = eiz + e iz. other trig functions are as expecte Some properties: tan z = sin z cos z sec z = 1 cos z { e ix e ix = i sin x e ix + e ix = cos(x) cot z = cos z sin z csc = 1 sin z sin z cos z are linear combinations of entire functions, are thus entire. Since we get z eiz = ie iz z e iz = ie iz z sin z = cos z cos z = sin z z Definition: A zero of a function is a number z 0 such that f(z 0 ) = 0. Extening a real value function to the complex plane can a zeroes, for example, f(x) = x + 1 has no zeroes in R but zeroes ±i in C. Extening the omain to C, trig functions gain no zeroes, that is zeroes of sin = {nπ : n Z} zeroes of cos = {nπ + π } : n Z Other trig functions are analytic except at singularities, those are {nπ : n Z} for tan z csc z, {nπ + π : n Z } for cot z sec z. 9

10 Definition: The hyperbolic sine cosine functions are efine by sinh z = ez e z cosh z = ez + e z 10