Second Midterm Exam Name: Practice Problems March 10, 2015

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1 Math Treibergs Second Midterm Exam Name: Practice Problems March 10, Determine the singular points of the function and state why the function is analytic everywhere else: z 1 fz) = z + 1)z + z + 7) Points z 0 where the function fails to be analytic, but is analytic at some point in every neighborhood of z 0 are called singular points. The function is analytic at a point z 0 if it is analytic in some neighborhood, an open set containing z 0. The given function blows up where the denominator is zero and the numerator is not zero.) For this function, this occurs at the points z 0 = 1 and z 0 = ± = ± 19 i. Away from these isolated points the denominator is nonzero so the complex derivative exists because it is a rational function of z. Thus every other point ζ is some distance r > 0 from these points, thus is analytic because the function is differentiable in the open disk of radius r about ζ. Also these points are singular because the there is no derivative at these points and every open set containing them contains other points where the function is analytic.. Let A be an open subset of and A = {z : z A} be its reflection across the real axis. Suppose that fz) is analytic on A. Show that f z) is analytic on A. First observe that A is open. That is because every point Z A has an open neighborhood U = B, namely, the reflection of the little open disk about z B A. Second, if fz) = ux, y) + ivx, y), then gz) = f z) = ux, y) ivx, y) and we show that the auchy Riemann equations hold for g. By the auchy Riemann Equations for f, Re g) = ux, y) = x x x u = x, y) y x, y) v = y [ vx, y)] = Im g) y and y Re g) = y ux, y) = y u = x, y) x v = x, y) x [ vx, y)] = Im g). x For every point of A, both real partial derivatives exist and are continuous for g because they are for f on A. Since the auchy Riemann equations hold as well, g is analytic on A.. Suppose that fz) = ux, y) + ivx, y) is an analytic function defined on a domain D. If aux, y) + bvx, y) = c, where a, b, c are real constants not all zero, then fz) is constant on A. Is the result still valid if a, b, c were allowed to be complex constants? Observe that we can t have both a = b = 0 because that implies that c = 0 too. Differentiating we find 0 = au x + bv x a b 0 = au y + bv y = bu x av x. b a = a b Since the determinant of the coefficient matrix is strictly negative if a and b are real and not both zero, we must have u x = v x = 0 everywhere in D. Hence f z) = u x + iv x = 0 and since D is connected, f is constant. 1

2 The result continues to hold if a, b.c are complex. If a + b 0 then the same argument shows f = 0 on D. But if b = a then b = ±ia, both nonzero, and the system has only one independent equation, 0 = au x ± aiv x = au x ± iv x ) = 0 = u x ± iv x. But since both u x and v x are real, the real and imaginary parts of this equation say u x = v x = 0 so f is again constant on D. 4. Let f be an entire function that equals a polynomial on [0, 1] on the real exis. Show that f is a polynomial. Let fx) = a 0 + a 1 x + + a n x n on [0, 1]. Then fz) and a 0 + a 1 z + + a n z n agree on [0, 1], and both are analytic on that is, both are entire). By the Identity Theorem, if two analytic functions agree on a line segment contained in the domain of both, then they agree over the whole domain. 5. Prove the identity: sinh e z + sinh e z = sinhcosh z) coshsinh z) Identities are proved starting from one side of the equality and deducing the other side. sinh e z + sinh e z = expez ) exp e z ) + expe z ) exp e z ) { = 1 e z + e z exp + ez e z ) exp ez + e z e z + e z + exp ez e z ) exp ez + e z ez e z ) + ez e z = 1 { exp cosh z + sinh z) exp cosh z sinh z) } + exp cosh z sinh z) exp cosh z + sinh z) = 1 { exp cosh z) exp sinh z) exp cosh z )exp sinh z) } + exp cosh z) exp sinh z) exp cosh z )exp sinh z) = 1 { }{ } exp cosh z) exp cosh z) exp sinh z )+ exp sinh z) = sinhcosh z) coshsinh z). ) } 6. Find all values z such that We write the equation as e z+i =. e x e y+1)i = e z+i = = e πi so that and so x = ln and y + 1 = π + nπ, n Z); z = 1 ln + 1 π 1 + nπ) i, n Z).

3 7. Show that logi 1/ ) = 1 log i. The logarithm is a multiple valued function. First, the three cube roots of i = e πi/ are e π/+πn)i/ n Z) so that the numbers corresponding to n = 0, 1, are the distinct roots e πi/6, e 5πi/6, e πi/, Then log e πi/6) ) 1 = ln n πi, log e 5πi/6) ) 5 = ln n πi, log e πi/) ) = ln n πi, n Z), n Z), n Z). Since 5 6 = and = 1 6 +, their union is log i 1/) = n ) πi, n Z). On the other hand, so that which is the same as log i 1/). ) 1 log i = ln n πi, n Z) 1 1 log i = 6 + n ) πi, n Z) 8. Verify the following relation for the inverse function. arccoth z are the solutions w of the equation Solving for e w yields or Finally, taking logarithm gives the solution. arccoth z = 1 log z + 1 z 1. z = coth w = cosh w sinh w = ew + e w e w e w = ew + 1 e w 1. e w + 1 = ze w 1) e w = z + 1 z 1

4 9. Show that for different choices of z 1 and z the following expression may or may not be valid: Log z 1 z = Log z 1 Log z. The issue is whether or not the differnce of arguments remains in the domain of the principal value. Thus if we have z k = r k e iθ k then Log z k = log z k in the branch π < Θ k π. Subtracting arguments in this range implies π < Θ 1 Θ < π which may put them out of range for the principal value if Θ 1 Θ π or Θ 1 Θ > π. For example if z 1 = 1 + i = e πi/4 and z = 1 + i = e πi/4 we have Log z 1 = Log i) = π [ z i = ln + π ] 4 i [ln + π4 ] i = Log z 1 Log z. On the other hand, if z 1 = 1 i = e πi/4 and z = 1 + i = e πi/4 we have Log z 1 = Log 1) = πi πi = [ln π ] z 4 i [ln + π4 ] i = Log z 1 Log z. 10. Find 1 + i) i and its principal value. 1 + i = e 1/4+n)πi for n N. Its powers are thus { 1 + i) i = e i log1+i) = exp i ln ) }) n πi, n Z) = exp π4 ) ) ln + 6nπ exp i, n Z) Its principal value corresponds to N = 0 or P. V. 1 + i) i = exp 11. Find the roots of the equation cos z =. Expressing the real and imaginary parts we find which reduces to two real equations π ) ) ln exp 4 i cos z = cos x cosh y i sin x sinh y = cos x cosh y = and sin x sinh y = 0. The second says either sinh y = 0 which implies y = 0 or sin x = 0 which implies x = nπ, n N). If y = 0 then the first equation becomes cos x cosh y = cos x cosh 0 = cos x = which has no real solution because cos x 1. Thus we have to solve the first equation in the second case x = nπ, cos x cosh y = cos nπ cosh y = 1) n cosh y =. 4

5 But cosh y 1 so that this equation only has solutions if n is even, in which case y = cosh 1, namely, real s that solves Equivalently, or = cosh s = 1 es + e s ). 0 = e s 6e s + 1 e s = 6 ± = ±. Noting that + and are reciprocals, we may write all solutions as z = nπ ± i ln + ), n Z). 1. Show that tanh z = i taniz) and d dz tanh z = sech z. We use the identities sinh z = i sin iz and cosh z = cos iz. Then tanh z = sinh z i sin iz = = i tan iz. cosh z cos iz Also, by the definitions of sinh z and cosh z and the quotient rule d dz tanh z = d sinh z dz cosh z = d e z e z dz e z + e z = ez + e z ) e z + e z ) e z + e z ) e z e z ) e z + e z ) e z + + e z) e z + e z) 4 = e z + e z ) = e z + e z ) = 1 cosh z = sech z. 1. Evaluate the following integral, assuming Re z < 0. 0 e zt dt This is an improper integral. Observing that d e zt dt z T [ e e zt dt = lim e zt zt dt = lim 0 T 0 T z For z = x + iy, the latter limit depends on the estimate ] T e zt = e xt e iyt = e xt which tends to zero as T because x = Re z < 0. = ezt we obtain 0 [ e zt = lim T z 1 ] = 1 z z. 5

6 14. let denote the right half of the circle z = in the counterclockwise direction. onsider the two given parametric representations z 1 θ) and z y) of. Find the reparameterization θ = φy) such that z y) = z 1 [φy)] and check that φ has positive derivative. z 1 θ) = e iθ, π θ π ) z y) = + iy, y ). Putting the first representation in rectangular coordinates gives Equating the ratio we find x = cos θ and y = sin θ. y = y x = sin θ cos θ = tan θ or ) y θ = φy) = Atn, π θ π ). To see that z y) = z 1 [φy)], we note that cosatn t) = 1/ 1 + t and sinatn t) = t/ 1 + t, so )) y Re z 1 φy)) = cos Atn = = = Re z y) y and )) y Im z 1 φy)) = sin Atn = y 1 + y = y 4 = Im z y). For < y < we find φ 1 y) = ) y 1 + y + = ) > 0 6

7 15. Using the contour shown in the figure, evaluate the integral Ree iz ) dz The function is the derivative in F z) = d dz e iz i = e iz Thus the integral is independent of path, and its value may be computed using the end points z 1 = 1 and z = to give ) I = Ree iz ) dz = Re fz) dz = Re F z ) F z 1 )). But so [ e iz F z ) F z 1 ) = i ] 1 = e i i ei i = i[ cos ) + i sin ) cos 1 i sin 1 ] I = sin sin Let denote the positively oriented circle z = 5 about the origin. For the branch at angle α R, with its corresponding log z = ln r + iθ where α < θ < α + π, find z / dz. We have zθ) = 5e iθ for α < θ < α + π so dz = 5ie iθ dθ. Also z / z θ) = exp ) logzθ)) 5ie iθ = 5i exp ln ) iθ As this extends continuously to the interval α θ α + π, we may integrate using real antiderivatives as usual [ ] α+π α+π z / dz = z / z θ) dθ = i e iθ/ 5 α = 5 [ e iα+π)/ e iα/] = 5 i α [ e πi/ 1 ] e iα/. 7

8 17. Let N denote the boundary of the square formed by the lines x = ± N + 1 ) π and y = ± N + 1 where N is a positive integer and N is given the counterclockwise orientation. Show that the integral tends to zero as N. dz z sin z N We bound the absolute value. The length of the square is L N = 4N + 1)π. To bound the integrand, we observe that It follows that ) π sin z = sinx + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y. sin z = sin x cosh y + cos x sinh y = sin x1 + sinh y) + cos x sinh y = sin x + sinh y sin x so that on the vertical lines, [ sin z sin x = ± sin N + 1 ] π) = 1. The estimate also showed that sin z sinh y so that on the horizontal lines, since sinh y is increasing in y and N 0, [ sin z sinh y = ± sinh N + 1 ] [ π) = sinh N + 1 ] ) π ) π sinh 1. So on all four sides of the square sin z 1. Thus we may estimate the integrand 1 z sin z = 1 z sin z 1 ) = M N + 1 N π because z x on the verticals and z y on the horizontals, which are both equal N + 1 ) π. Then the contour integral may be estimated N dz z sin z L 4N + 1)π NM N = ) = 16 N + 1 π N + 1) π. Thus the value of the integral tends to zero as N. 8

9 18. Find the integral, where the integrand is computed using the principal branch and is any contour from z = to z = that, except for its end points, lies above the real axis. z i dz The restriction on means that in polar coordinates, for z we have r > 0 and 0 θ π. In this range z i = exp i Log z), 0 θ π). For this range of θ s, Log z = log z where the second logarithm has the branch cut along the negative imaginary axis R π/ so π < θ < π. The function has an antiderivative on the domain D = R π/, namely F z) = z i+1 i + 1 = 1 + i exp i + 1) log z). 5 Hence the integral is found by evaluating the antiderivative function at the end points z 1 = = e 0 i and z = = e πi of the contour D z i dz = F ) F ) = 1 + i ) exp[ i + 1)ln + πi)] exp[ i + 1) ln ] 5 = 1 + i e π e i ln + e i ln ) Using the contour which is a unit square with corners 0, 1, 1 + i and i shown in the figure, evaluate the integral cos + 1 ) dz z Observe that the integrand function fz) = cos + 1 ) z is the composition of an entire function with a rational function which has only one singularity at z =. So fz) is analytic in the punctured plane, D = {}. Since is a simple, closed contour in D, by the auchy-goursat Theorem, the integral is zero. 9

10 0. Show that the integral on the real axis I = x + 4 i x + i) 4 dx = 0. Let fz) = z + 4 i z + i) 4. Since f decays quadratically in z, the function is integrable on R. For R > 0, letting R denote the line segment from R to R, we have z + 4 i I = lim R R z + i) 4 dz. Observe that the only singularity of the rational function fz) is at z = i so it is analytic in D = { i}. Let G R denote the arc of the circle z = R in the upper half plane. Thus R + G R is a simple closed contour in D so that, by the auchy Goursat Theorem, 0 = fz) dz = R +G R fz) dz + R fz) dz. G R We show that the limit of the second integral vanishes as R, thus so does the first, so I = 0. We estimate the second integral. The length of the contour is L R = πr. Also for z Γ R and R >, by the triangle inequalty Hence fz) = z + 4 i z + i 4 z + 4 i z i 4 = R + 5 R ) 4 = M R fz) dz L R M R = πrr + 5) G R R ) 4 which tends to zero as R, completing the argument. 10