The complex exponential function

Transcription

1 Capter Te complex exponential function Tis is a very important function!. Te series For any z C, we define: exp(z) := n! zn = + z + z2 2 + z3 6 + z On te closed disk D(0,R) := {z C z R}, one as n! zn n! Rn and we know tat te series n! Rn converges for any R > 0. Terefore, exp(z) is a normally convergent series of continuous functions, and z exp(z) is a continuous function from C to C. Teorem.. For any a, b C, one as exp(a + b) = exp(a) exp(b). Proof. - We just sow te calculation, but tis sould be justified by arguments from real analysis (absolute convergence implies commutative convergence): exp(a + b) = (a + b)n n! = = k+l=n = k,l 0 (k + l)! n! a k b l k+l=n k!l! n! (k + l)! a k b l k!l! k!l! ak b l = exp(a) exp(b). 4

2 We now give a list of basic, easily proved properties. First, te effect of complex conjugation: z C, exp(z) = exp(z). Since obviously exp(0) =, on draws from te teorem: z C, exp(z) C andexp( z) = exp(z) Also, z R exp(z) R and ten, writing exp(z) = ( exp(z/2) ) 2, one sees tat exp(z) R +. Last, if z ir (pure imaginary), ten z = z, so putting w := exp(z), one as w = w so tat w =. In oter words, exp sends ir to te unit circle U := {z C z = }. Summarizing, if x := Re(z) and y := Im(z), ten exp sends z to exp(z) = exp(x) exp(iy), were exp(x) R + and exp(iy) U. exp(z) + exp( z) exp(z) exp( z) Exercice..2 For z C, define cos(z) := and sin(z) :=, so 2 2i tat cos is an even function, sin is an odd function and exp(z) = cos(z) + isin(z). Translate te property of teorem.. into properties of cos and sin..2 Te function exp is C-derivable Lemma.2. If z R, ten exp(z) z er 2 z 2. Proof. - exp(z) z = z2 2 ( + ) z6 z and + z 6 + z R 6 + R er. Teorem.2.2 For any fixed z 0 C: lim = exp(z 0). 0 Proof. - exp(z 0 + ) exp(z 0 ) 0. = exp(z 0 ) exp() and, after te lemma, exp() wen Terefore, exp is derivable wit respect to te complex variable: we say tat it is C-derivable (we sall cange terminology later) and tat its C-derivative is itself, wic we write d exp(z) = exp(z) or exp = exp. Corollary.2.3 On R, exp restricts to te usual real exponential function; tat is, for x R, exp(x) = e x. 5

3 Proof. - Te restricted function exp : R R sends 0 to and it is its own derivative, so it is te usual real exponential function. For tis reason, for now on, we sall put e z := exp(z) wen z is an arbitraty complex number. Corollary.2.4 For y R, one as exp(iy) = cos(y) + i sin(y). Proof. - Put f (y) := exp(iy) and g(y) := cos(y)+isin(y). Tese functions satisfy f (0) = g(0) = and f = i f, g = ig. Terefore te function := f /g wic is well defined from R to C satisfies (0) = and = 0, so tat it is constant equal to. Note tat tis implies te famous formula of Euler e iπ =. Corollary.2.5 For x,y R, one as e x+iy = e x (cosy + isiny). Corollary.2.6 Te exponential map exp : C C is surjective. Proof. - Any w C can be written w = r(cosθ + isinθ), so w = exp(ln(r) + iθ). One can find a proof wic does not use trigonometry in te preliminary capter of Rudin. Exercice.2.7 Let a,b > 0 and U := {z C a < Re(z) < aand b < Im(z) < b} (tus, an open rectangle under te identification of C wit R). Assuming b < π, describe te image V := exp(u) C and define an inverse map V U. Te exponential viewed as a map R 2 R 2. It will be useful to consider functions f : C C as functions R 2 R 2, under te usual identification of C wit R 2 : x + iy (x,y). In tis way, f is described by (x,y) F(x,y) := ( A(x,y),B(x,y) ) { A(x,y) := Re ( f (x + iy) ),, were B(x,y) := Im ( f (x + iy) ). In te case were f is te exponential function exp, we compute easily: { A(x,y) = e x cos(y), B(x,y) = e x = F(x,y) = ( e x cos(y),e x sin(y) ). sin(y), We are going to compare te differential of te map F wit te C-derivative of te exponential map. On te one and, te differential df(x,y) is te linear map defined by te relation: F(x + u,y + v) = F(x,y) + df(x,y)(u,v) + o(u,v), were o(u,v) is small compared to te norm of (u,v) wen (u,v) (0,0). Actually, df(x,y) can be expressed using partial derivatives: ( A(x,y) df(x,y)(u,v) = u + A(x,y) v, B(x,y) u + B(x,y) ) v. Terefore, it is described by te Jacobian matrix: JF(x,y) = 6

4 On te side of te complex function f := exp, putting z := x + iy and := u + iv, we write: f (z + ) = f (z) + f (z) + o(), tat is exp(z + ) = exp(z) + exp(z) + o() Here, te linear part is f (z) = exp(z), so we draw te conclusion tat (under our correspondance of C wit R 2 ): f (z) df(x,y)(u,v), tat is, comparing real and imaginary parts: Since it must be true for all u,v, we conclude tat: JF(x,y) = u + A(x,y) v = Re( f (z))u Im( f (z))v, u + B(x,y) v = Im( f (z))u + Re( f (z))v. ( Re( f = (z)) Im( f ) (z)) Im( f (z)) Re( f (z)) As a consequence, te Jacobian determinant det JF(x,y) is equal to f (z) 2 and tus vanises if and only if f (z) = 0: in te case of te exponential function, it vanises nowere. Exercice.2.8 Verify tese formulas wen A(x,y) = e x cos(y), B(x,y) = e x sin(y) and f (z) = exp(x + iy)..3 Te exponential function as a covering map From equation e x+iy = e x (cosy + isiny), one sees tat e z = z 2iπZ, i.e. k Z : z = 2iπk. It follows tat e z = e z 2 z 2 z 2iπZ, i.e. k Z : z 2 = z +2iπk. We sall write tis relation: z 2 z (mod 2iπZ) or more sortly z 2 z (mod 2iπ). Teorem.3. Te map exp : C C is a covering map, tat is: for any w C, tere is a neigborood V C of w suc tat exp (V ) = U k (disjoint union), were eac U k C is an open set and exp : U k V is an omeomorpism (a bicontinuous bijection). Proof. - Coose a particular z 0 C suc tat exp(z 0 ) = w 0. Coose an open neigborood U 0 of z 0 suc tat, for any z,z U 0, one as z z < 2π. Ten exp maps bijectively U 0 to V := exp(u 0 ). Moreover, one as exp (V ) = U k were k runs in Z and te U k = U 0 + 2iπk are open sets. It remains to sow tat V is an open set. Te most generalizable way is to use te local inversion teorem, since te Jacobian determinant vanises nowere. Anoter way is to coose an open set as in exercice.2.7. Te fact tat exp is a covering map is a very important topological property and it as many consequences. 7

5 Corollary.3.2 (Pat lifting property) Let a < b in R and let γ : [a,b] C be a continuous pat wit origin γ(a) = w 0 C. Let z 0 C be suc tat exp(z 0 ) = w 0. Ten, tere exists a unique lifting, a continuous pat γ : [a,b] C suc tat t [a,b], expγ(t) = γ(t) and subject to te initial condition γ(a) = z 0. Exercice.3.3 If one cooses anoter z 0 C suc tat exp(z 0 ) = w 0, one gets anoter lifting γ : [a,b] C suc tat t [a,b], expγ (t) = γ(t) and subject to te initial condition γ (a) = z 0. Sow tat tere is some constant k Z suc tat t [a,b], γ (t) = γ(t) + 2iπk. Corollary.3.4 (Index of a loop wit respect to a point) Let γ : [a,b] C be a continuous loop, tat is γ(a) = γ(b) = w 0 C. Ten, for any lifting γ of γ, one as γ(b) γ(a) = 2iπn for some n Z. Te number n is te same for all te liftings, it depends only on te loop γ: it is te index of γ around 0, written I(0,γ). Actually, anoter property of covering maps (te omotopy lifting property) allows one to conclude tat I(0,γ) does not cange if γ is continuously deformed witin C : it only depends on te omotopy class of γ (see te topology course). Example.3.5 If γ(t) = e nit on [0,2π], ten all liftings of γ ave te form γ(t) = nit + 2iπk for some k Z and one finds I(0,γ) = n..4 Te exponential of a matrix x x n For a complex vector X =. C n, we define X := max ( x i ). Ten, fot a complex square i n matrix A = (a i, j ) i, j n Mat n (C), define te subordinated norm: AX A := sup = max X C n X i n X 0 n j= a i, j. Ten, for te identity matrix, I n = ; and, for a product, AB A B. It follows easily tat k! Ak k! A k for all k N, so tat te series k 0 k! Ak converges absolutely for any A Mat n (C). It actually converges normally on all compacts and terefore define a continuous map exp : Mat n (C) Mat n (C), A k 0 k! Ak. We sall also write for sort e A := exp(a). In te case n =, te notation is consistent. Examples.4. (i) For a diagonal matrix A := Diag(λ,...,λ n ), one as k! Ak = Diag(λ k /k!,...,λk n/k!), so tat exp(a) = Diag(e λ,...,e λ n ). (ii) If A is an upper triangular matrix wit diagonal D := Diag(λ,...,λ n ), ten k! Ak is an upper triangular matrix wit diagonal k! Dk, so tat exp(a) is an upper triangular matrix wit diagonal exp(d) = Diag(e λ,...,e λ n ). Similar relations old for lower triangular matrices. 8

6 0 a b (iii) Take A :=. Ten A 0 2 = I 2, so tat exp(a) = ai 2 + ba =, were a = b a and b = (2k + )!. k 0 k 0 (2k)! Te same kind of calculations as for te exponential map gives te rules exp(0 n ) = I n ; exp(a) = exp(a); and: AB = BA = exp(a + B) = exp(a)exp(b) = exp(b)exp(a). Remark.4.2 Te condition AB = BA is( required ) to use te Newton binomial formula. If we take for instance A := and A :=, ten AB BA. We ave A = B 2 = 0, so tat 0 2 exp(a) = I 2 + A = and exp(b) = I B =, tus exp(a) exp(b) =. On 0 te oter and, A + B = and te previous example gave te value of exp(a + B), wic 0 was different. It follows from te previous rules tat exp( A) = ( exp(a) ) so tat exp actually sends Mat n (C) to GL n (C). Now tere are rules more specific to matrices. For te transpose, using te fact tat t (A k ) = ( t A) k, and also te continuity of A t A (tis is required to go to te limit in te infinite sum), we see tat exp( t A) = t (exp(a)). Last, if P GL n (C), from te relation (PAP ) n = PA n P (and also from te continuity of A PA n P ), we deduce te very useful equality: Pexp(A)P = exp(pap ). Now any complex matrix A is conugate to an upper triangular matrix T aving te eigenvalues of A on te diagonal; using te exaamples above, one concludes tat if A as eigenvalues λ,...,λ n, ten exp(a) as eigenvalues e λ,...,e λ n : Note tat tis implies Tr(e A ) = e deta. Sp(e A ) = e Sp(A). 0 π Example.4.3 Let A :=. Ten A is diagonalisable wit spectrum Sp(A) = {iπ, iπ}. π 0 Tus, exp(a) is diagonalisable wit spectrum {, }. Terefore, exp(a) = I 2. 0 b Exercice.4.4 Compute exp in two ways: by diagonalisation as in te example above; b 0 a b by direct calculation as in a previous example. Deduce from tis te value of exp. b a.5 Application to differential equations Let A Mat n (C) be fixed. Ten, z e za is a C-derivable function from C to te complex linear space Mat n (C); tis simply means tat eac coefficient is a C-derivable function from C to itself. 9

7 Derivating our matrix-valued function coefficientwise, we find: Indeed, e(z+)a e za = e za ea I n d eza = Ae za = e za A. Now consider te vectorial differential equation: = ea I n e za and ea I n = A + 2 A2 + d X(z) = AX(z), were X : C C n is searced as a C-derivable vector-valued function, and again derivation is performed coefficientwise. We solve tis by canging of unknown function: X(z) = e za Y (z). Ten, applying Leibniz rule for derivation: ( f g) = f g+ f g (it works te same for C-derivation), we find: X = AX = e za Y + Ae za Y = Ae za Y = e za Y = 0 = Y = 0. Terefore, Y (z) is a constant function. If now we fix z 0 C, X 0 C n and we adress te Caucy problem: d X(z) = AX(z), X(z 0 ) = X 0, we see tat te unique solution is X(z) := e (z z 0)A X 0. An important teoretical consequence is te following. Call Sol(A) te set of solutions of d X(z) = AX(z). Tis is obviously a complex linear space. Wat we proved is tat te map X X(z 0 ) from Sol(A) to C n, wic is obviously linear, is also bijective. Terefore, it is an isomorpism of Sol(A) wit C n. (Tis is a very particular case of te Caucy teorem for complex differential equations.) Example.5. To solve te second order scalar equation (wit constant ( coefficients) ) f + p f + f (z) q f = 0 (p,q C), we introduce te vector valued function X(z) := f and find tat our (z) scalar equation is actually equivalent to te vector equation: X 0 = AX, were A :=. q p Terefore, te solution will be searced in te form X(z) := e (z z 0)A X 0, were z 0 may be cosen at will or else imposed by initial conditions. Exercice.5.2 Compute e (z z 0)A and solve te problem wit initial conditions f (0) = a, f (0) = b. Tere will be a discussion according to weter p 2 4q = 0 or 0. 0