MATH 381 HOMEWORK 2 SOLUTIONS

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1 MATH 38 HOMEWORK SOLUTIONS Question (p.86 #8). If g(x)[e y e y ] is harmonic, g() =,g () =, find g(x). Let f(x, y) = g(x)[e y e y ].Then Since f(x, y) is harmonic, f + f = and we require x y f x = g (x)[e y e y ] f y = 4g(x)[ey e y ]. g (x) + 4g(x) =. Thus g(x) has the form A sin(x) + B cos(x) and by the initial conditions, A = andb =. Therefore, g(x) = sin(x). Question (p.86 #). Find the harmonic conjugate of tan x y where π < tan x y π. Write u(x, y) = tan x y.thenbythecauchy-riemannequations, () () By (), x = y x + y y = y x + y = v y y = y x x + y y = x x + y = v x. v = log(x + y ) + C(x), and by () v x = x x + y + x C (x) = x + y so C (x) = andc(x) is a constant, call it D. Therefore, v(x, y) = log(x + y ) + D. Question 3. (p.86 #3) Show, if u(x, y) and v(x, y) are harmonic functions, that u + v must be a harmonic function but that uv need not be a harmonic function. Is e u e v a harmonic function? If u and v are harmonic, then u + v is harmonic since (u + v) + (u + v) = u x y x + v x + u y + v y = u x + u y + v x + v y =. To show that uv is not necessarily harmonic, it suffices to show that there exists u, v harmonic such that (uv) x + (uv) y = v x x + v y y. Any u = v harmonic where, willsuffice. Forinstance,takingu = v = x will work, since it s harmonic (both x y of its second-order partials vanish) but v x x + v y y =. Date: October3,.

2 MATH 38 HOMEWORK SOLUTIONS Now, in order for e u e v to be harmonic, we need (e u e v ) + (e u e v ) = e u+v x y x + v x + y + v y =. Thus, the existence of any u, v harmonic such that + v x x + + v y y willshowthate u e v is not harmonic. Again, taking u = v = x gives us what we want as e x is easily seen to be non-harmonic. Question 4 (p.6 #4). State the domain of analyticity of f(z) = e iz. Find the real and imaginary parts u(x, y) and v(x, y) of the function, show that these satisfy the Cauchy-Riemann equations, and find f (z) in terms of z. By definition, Therefore, These are continuous functions at all (x, y) R.Now, f(z) = e iz = e ix e y = e y [cos x + i sin x]. u(x, y) = e y cos x v(x, y) = e y sin x. x = e y sin x = v y y = e y cos x = v x so u, v satisfy the C-R equations, and these derivatives are continuous for allx, y. Therefore, f(z) is entire. Furthermore, f (z) = e y sin x + i(e y cos x) = i(e y [cos x + i sin x]) = ie iz. Question 5 (p.6 #6). State the domain of analyticity of f(z) = e ez. Find the real and imaginary parts u(x, y) and v(x, y) of the function, show that these satisfy the Cauchy-Riemann equations, and find f (z) in terms of z. First, observe that f is an entire function of an entire function, so it is analytic everywhere. Now, so e ez = e ex (cos y+i sin y) = e ex cos y cos(e x sin y) + i sin(e x sin y), u(x, y) = e ex cos y cos(e x sin y) v(x, y) = e ex cos y sin(e x sin y) and f satisfies the C-R equations. Furthermore, x = x cos y ee (e x cos y) cos(e x sin y) e ex cos y (e x sin y) sin(e x sin y) = e ex cos y+x (cos y cos(e x sin y) sin y sin(e x sin y)) v y = x cos y ee ( e x sin y) sin(e x sin y) + e ex cos y cos(e x sin y)(e x cos y) = e ex cos y+x (cos y cos(e x sin y) sin y sin(e x sin y)) y = x cos y ee ( e x sin y) cos(e x sin y) + e ex cos y (e x cos y)( sin(e x sin y)) = e ex cos y+x (cos y sin(e x sin y) + sin y cos(e x sin y)) v x = x cos y ee (e x cos y) sin(e x sin y) + e ex cos y (e x sin y) cos(e x sin y) = e ex cos y+x (cos y sin(e x sin y) + sin y cos(e x sin y)) f (z) = e ex cos y e x (cos y cos(e x sin y) sin y sin(e x sin y)) + i(cos y sin(e x sin y) + sin y cos(e x sin y) = e ex cos y e x cos y cos(e x sin y) + i sin(e x sin y) + e x sin yi cos(e x sin y) sin(e x sin y) = e ex cos y cos(e x sin y) + i sin(e x sin y)e x (cos y + i sin y) = e ez e z.

3 MATH 38 HOMEWORK SOLUTIONS 3 Question 6 (p.6 #3). (a) Prove the expression given in the text for the n th derivative of f(t) = t t + = Re. (Note: t R). t i (b) Find similar expressions for the n th derivative of f(t) = t + = Im.(Note: t R ). t i (a) By the Lemma, for n, Now, observe that f (n) (t) = Re dn dt n t i = Re ( )n n! (t i) n+ = t+i, so by the binomial theorem t i t + ( ) n n! (t i) = ( )n n!(t + i) n+ = ( )n n+ n!(n + )! i k t n+ k n+ (t + ) n+ (t + ) n+ k= (n + k)!k!. But notice that we only get contributions to the real part of this expression when k is even; i.e. when i k R. Summing over the even integers, k = m, we get for n odd that and for n even that (b) In this case we want f (n) (t) = ( )n n!(n + )! (t + ) n+ = f (n) (t) = ( )n!(n + )! (t + ) n+ n!(n + )! (t + ) n+ n+ m= n+ m= n m= i m t n+ m (n + m)!(m)! ( ) m t n+ m (n + m)!(m)! i m t n+ m (n + m)!(m)! f (n) (t) = Im dn dt n t i = Im ( )n n! (t i) n+. By the work above, we want the imaginary part of ( ) n n!(n + )! (t + ) n+ n+ k= i k t n+ k (n + k)!k!. In this case we get contributions when k is odd, so we take the the sum over k = m + for m. Note that i m+ = ( ) m i. It follows that when n is odd, and when n is even, f (n) (t) = f (n) (t) = ( )n!(n + )! (t + ) n+ n m= n n!(n + )! (t + ) n+ m= ( ) m t n m (n m)!(m + )! ( ) m t n m (n m)!(m + )!. Question 7 (p.6 #5). Let P (ψ) = N n= e inψ. (a) Show that (b) Find lim ψ P (ψ). (c) Plot P (ψ) for ψ and N = 3. (a) Note that P (ψ) = sin(nψ) sin(ψ). P (ψ) = einψ e iψ.

4 z = π 4 + kπ, k Z. 4 MATH 38 HOMEWORK SOLUTIONS Thus, Thus, P (ψ) = einψ e iψ = einψ e iψ = einψ e iψ = einψ e iψ (b) By l Hopital s rule we get einψ e inψ e iψ e iψ cos(nψ) + i sin(nψ) cos( Nψ) i sin( Nψ) cos(ψ) + i sin(ψ) cos( ψ) i sin( ψ) i sin(nψ) i sin ψ. P (ψ) = einψ e iψ sin(nψ) sin ψ = sin(nψ) sin ψ. sin(nψ) Nsin(Nψ) lim = lim = N. ψ sin ψ ψ sinψ (c) If you have nothing else, just plug it in Wolfram Alpha. Question 8 (p. #7). Show that sin z cos z = has solutions only for real values of z. What are the solutions? In other words, for z = x + iy we want Equating the real parts and imaginary parts we require (3) (4) sin x cosh y + i cos x sinh y = cos x cosh y i sin x sinh y. sin x cosh y = cos x cosh y cos x sinh y = sin x sinh y. Suppose y andhencesinhy andcoshy. Then in order to have solutions, by (3), weneedcosx = sin x and by (4) we need cos x = sin x. These equations are only satisfied for sin x = cos x =, but no solutions for x exists. Therefore, if there are solutions to the original equation, we must have that y =. Suppose y =. Then since cosh = andsinh= wesimplyneedsolutionstosinx = cos x. Thus we have solutions if and only if Question 9 (p. #). Where does the function f(z) = 3sinz cos z fail to be analytic? Since sin z and cos z are both analytic, f(z) will fail to be analytic when 3sinz cos z =. In other words, when we have solutions to ( sin x cosh y + i cos x sinh y) = cos x cosh y i sin x sinh y. Equating the real parts and imaginary parts we require (5) 3sinx cosh y = cos x cosh y (6) 3cosx sinh y = sin x sinh y. By the same argument as the previous question, there are no solutions when y. Suppose y =. Then since cosh = andsinh= wesimplyneedsolutionsto 3sinx = cos x, thatistotanx = 3. So f(z) is not analytic when z = π 6 + kπ, k Z. Question (p. #). Let f(z) = sin z. (a) Express this function in the form u(x, y) + iv(x, y). Where in the complex plane is this function analytic? (b) What is the derivative of f(z)? Where in the complex plane is f (z) analytic?

5 MATH 38 HOMEWORK SOLUTIONS 5 (a) Since sin z is entire, and z (b) For z, is analytic for z, it follows that f(z) is analytic for z. sin x iy = sin z x + y x = sin x + y cosh y x + y + i cos x x + y sinh y x + y x = sin x + y cosh y x + y i cos x x + y sinh y x + y. which is analytic for all z. d dz sin z = cos z z Question (p. #5). Show that cos z = sinh y + cos x. cos z = cos x cosh y i sin x sinh y = cos x cosh y + sin x sinh y = cos x( + sinh y) + sin x sinh y = cos x + sinh y(cos x + sin x) = cos x + sinh y Question (p.9 #6). Use logarithms to find solutions to e z = e iz. We want solutions to e z( i) =, so taking logs on both sides we get for any k Z, z( i) = πik, so z = πik (i + )πik = = (i )kπ. i Question 3 (p.9 #8). Use logarithms to find solutions to e z = (e z ). In other words, we want solutions to e z 3e z + =. By the quadratic formula, we get that Taking logs gives that for k Z. e z = 3 ± 4 9 = 3 ± 5. z = log 3 ± 5 + πik Question 4 (p.9 #). Use logarithms to find solutions to e ez =. First, taking logs we get e z = πik for k Z. Now for k >, the argument of πik is π + πm where m Z, and for k <, the argument of πik is 3π + πm (again m Z). Thus, for k >, z = log(πk) + i π + mπ and for k <, Question 5 (p.9 #3). Show that where θ R and e iθ. z = log(πk) + i π + mπ. Re log( + e iθ ) = log cos θ

6 6 MATH 38 HOMEWORK SOLUTIONS Re log( + e iθ ) = log + e iθ = log ( + cos θ) + sin θ = log( + cosθ) = log cos θ + sin θ + cos θ sin θ = log cos θ Question 6 (p.7 #9). Intergrate along z =, in the lower half plane. z dz Let z = e it, then we are integrating along the interval t [, π]. Now,dz = ie it dt so z dz = π e i t iei tdt = iπ. Question 7 (p.7 #). Show that x = cos t, y = sin t, wheret ranges from to π, yields a parametric representation of the ellipse x 4 + y =. Use this representation to evaluate i zdz along the portion of the ellipse in the first quadrant. Note that (cost) + sin t = cos t + sin t = 4 and furthermore cos = cosπ = andsin= sin π =. To see that we get all of the ellipse, note that x = cost has solutions t [, π] for all x [, ] and y = sin t has solutions t [, π] for all y [, ]. Furthermore, the parametrization is : except for when x =,y =. Setting z = x + iy = cost + i sin t, wegetdz = (i cos t sint)dt, and i zdz = = π π = 3 + iπ. (cost i sin t)(i cos t sint)dt (i 3sint cos t)dt Question 8 (p.7 #4). Consider I = +i e z dz taken along the line x = y. Without actually doing the integration, show that I 5e 3. Let M be the maximal value attained by e z along the path of integration. Now, for x = y, e z = e x y +ixy = e 3y which attains its maximum when y attains a maximum that is, when z = +i. Therefore M = e 3. By the pythagorean theorem, the length of the path is + = 5, so by the ML inequality, I ML = 5e 3. Question 9 (p.7 #6). Consider I = i e i log z dz taken along the parabola y = x. Without doing the integration, show that I.479e π. Letting θ = arg z e i log z = e i(logz iθ) = e i logz e θ = e θ. Along the given path, this attains a maximum when θ = π, so let M = e π.

7 MATH 38 HOMEWORK SOLUTIONS 7 Now, we need to find the length of the path of integration. So since dy = xdx, L = + dy dx dx = + 4x dx The ML inequality then gives the desired result. <.479. Question (p.8 #). Is the Cauchy-Goursat theorem directly applicable to z= sin z z+i dz? Since sin z z+i directly applicable. is analytic everywhere except for z = i which is not in the unit circle, the C-G theorem is Question (p.8 #6). Is the Cauchy-Goursat theorem directly applicable to z i = log zdz? Since is not in the unit circle about i +, log z is analytic in the desired region so the C-G theorem is directly applicable. Question (p.8 #7). Is the Cauchy-Goursat theorem directly applicable to z= (z ) 4 + dz? Observe that we have a singularity when z is a primitive 8 th root of unity that is, when (z ) 4 =. These roots of unity lie on the unit circle, so shifting over by, we need to determine if the roots closest to the origin, at z = e iπ34 + andz = e iπ54 + have absolute value greater than. By geometry (right angle triangles), it can be seen that at these points, so the C-G theorem directly applies. z = + > Question 3 (p.8 #9). Is the Cauchy-Goursat theorem directly applicable to z=b z + bz + dz where < b <? In this case the singularities are at the roots of the equation x + bx +, that is, when z = b ± b 4 = b 4 b ± i. Here, b z = b = > b 4 therefore C-G applies directly. Question 4 (p.8 #3). Prove that π e cos θ sin(sin θ + θ)dθ =. Begin with e z dz performed around z =. Use the parametric representation z = e iθ, θ π. Separate your equation into real and imaginary parts. Let z = e iθ = cos θ + i sin θ, sodz = e iθ idθ. Sincee z is analytic, But then π z= e z dz = e cos θ+i(sin θ+θ) dθ = π e cos θ+i sin θ e iθ idθ =. so by equating the imaginary part with zero we get the desired result. π e cos θ (cos(θ + sin θ) + i sin(θ + sin θ))dθ =