Math 120 A Midterm 2 Solutions

Transcription

1 Math 2 A Midterm 2 Solutions Jim Agler. Find all solutions to the equations tan z = and tan z = i. Solution. Let α be a complex number. Since the equation tan z = α becomes tan z = sin z eiz e iz cos z = 2i e iz +e iz 2 = i e iz e iz e iz + e iz, or equivalently, Multiplying by e iz gives or equivalently, Hence, tan z = α if and only if When α = we have from () that i e iz e iz = α, e iz + e iz e iz e iz = iα(e iz + e iz ). e 2iz = iα(e 2iz + ), ( iα)e 2iz = + iα. e 2iz = + iα iα. () e 2iz = + i i = i which leads to 2iz = (π/2)i + 2nπi. Therefore, tan z = has the solutions z = π 4 + nπ, n =, ±, ±2,.... When α = i, () implies that e 2iz = + i i i i =.

2 Therefore, tan z = i does not have any solutions. 2. Let be the boundary of the triangle with vertices,, and i, oriented in the counterclockwise direction. (i) ompute ez dz directly from the definition. (ii) arefully state a theorem that gives the answer to part (i) without having to do any computation. Solution. (i) For z, z 2 a pair of points in, let [z, z 2 ] denote the contour parametrized by z(t) = z + t(z 2 z ), t. We have that e iz dz = e iz dz + e iz dz + e iz dz. [,] [,i] [i, ] Furthermore, e z dz = [,] e ( +2t) 2 dt = 2e e 2t dt = 2e ( 2 ) e2t = e (e 2 ) = (e e ), 2

3 e z dz = [,i] e (+(i )t) (i ) dt = (i )e e (i )t dt = (i )e ( i ) e(i )t = e (e (i ) ) = e i e, and e z dz = [i, ] e (i (+i)t) ( ( + i)) dt = ( + i)e i e (+i)t dt = ( + i)e i ( + i ) e (+i)t = e i (e (+i) ) Therefore, = e e i. e iz dz = (e e ) + (e i e) + (e e i ) =. Remark: Note that instead of computing the above 3 integrals separately, one could save a bit of time by fixing a pair of complex numbers z and z 2 and then deriving the formula, e iz dz = e iz 2 e iz, [z,z 2 ] which implies each of the above 3 integrals quickly (cf. Midterm 2). Problem #3 on Practice 3

4 (ii) auchy s Theorem asserts that if is a simple closed contour and f is a function that is analytic on and inside, then f(z) dz =. onsequently, since f(z) = ez is analytic on and inside the triangle with vertices,, and i (indeed, f is analytic on the entire plane), it follows from auchy s Theorem that eiz dz =. 3. Let R denote the upper half of the circle z = R, parametrized in the counterclockwise direction. Show that lim dz =. R R z 3 + Solution. R is parametrized by z(t) = Re it, t π. Therefore, π R z 3 + dz = Re it Log(Re it ) ire it dt. R 3 e it + Now, if t π Re it Log(Re it ) = R ln R + it R(ln R + π) and R 3 e it + R 3. Therefore, R z 3 + dz = π Re it Log(Re it ) R 3 e it + ire it dt π π Reit Log(Re it ) R 3 e it + R(ln R + π) R dt R 3 ire it dt But = π R2 (ln R + π). R 3 π R2 (ln R + π) R 3 = π ln R R + π R R 3 4

5 as R. Therefore, lim R R z 3 + dz =. does not have an antiderivative on the punc- 4. Give two different proofs that z tured plane D = {z z }. Proof. We argue by contradiction. Accordingly, assume that /z has an antiderivative on D. By the Antiderivative Theorem, it follows that if is a closed contour in D, then /z dz =. But if is the closed contour in D defined by z(t) = eit, t 2π, /z dz = 2π e it ieit dt 2π = i dt = 2πi. This contradiction implies that /z does not have an antiderivative on D. Proof 2. We again argue by contradiction. Suppose that F is an antiderivative for /z on D, i.e., F is analytic on D and F (z) = /z for all z D. Since F is differentiable on D, in particular, F is continuous on D. (2) Now let D = \ {x R x }, the set of complex numbers with the negative real axis removed. Both f(z) and Log z are analytic on D and as f (z) = /z = d dz Log z for all z D, it follows (cf. Theorem on pg. 73 of the text) that there exists a constant c such that Log z = F (z) + c for all z D. (3) Now, (2) implies that lim F θ π (eiθ ) = lim F θ π+ (eiθ ). 5

6 Hence, (3) implies that But while lim Log θ π (eiθ ) = lim Log θ π+ (eiθ ). lim Log θ π (eiθ ) = π, lim Log θ π+ (eiθ ) = π. This contradiction implies that /z cannot have an antiderivative on D. Proof 3. We again argue by contradiction. Suppose that F is an antiderivative for /z on D, i.e., F is analytic on D and F (z) = /z for all z D. If we let F (z) = u(r, θ) + iv(r, θ), then u and v are well defined real valued functions on r >, θ R, that for each fixed r are periodic in θ with period 2π. Furthermore (cf. Theorem on page 69 of the text), holds as well, for all r and θ, the auchy-riemann equations, ru r = v θ and u θ = rv r hold, and in addition, the formula F (z) = e iθ (u r + iv r ) (4) holds as well. Since F (z) = /z = /(re iθ ), we see that (4) implies that u r + iv r = r. Therefore, u r = /r and v r =, and there exist functions a(θ) and b(θ) such that But since ru r = v θ, we have that u(r, θ) = ln r + a(θ) and v(r, θ) = b(θ). b (θ) = v θ (r, θ) = ru r (r, θ) = r(/r) = for all θ. This implies that there exists a constant b such that b(θ) = θ + b for all θ. onsequently, v(r, θ) = b(θ) = θ + b, contradicting the fact that v is periodic. This contradiction implies that /z cannot have an antiderivative on D. 6