MATH 45 SAMPLE 3 SOLUTIONS May 3, 06. (0 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. Because f is holomorphic, u and v satisfy the Cauchy-Riemann equations: u x = v y, u y = v x. Furthermore, we know that a holomorphic function is infinitely differentiable. In particular, the second partial derivatives of u and v exist and are continuous. By the above, these satisfy u xx = v yx = v xy = u yy. The middle equality is Clairaut s theorem from Calc III. It follows that, u xx + u yy = u yy + u yy = 0, so u is harmonic.. (0 pts) Find a harmonic conjugate of u(x, y) = y x +y. We need to find a function v(x, y) satisfying the C-R equations with u: By the quotient rule, Therefore v(x, y) = x u x = v y, u y = v x. v y = u x = y dy = x (x + y ) xy (x + y ). u du = xu + C(x) = where C(x) is constant relative to y, but may depend on x. Now On the other hand, v x = (x + y ) x(x) (x + y ) + C (x) = y x (x + y ) + C (x). x x + y + C(x), u y = y x (x + y ). Comparing, we see that C (x) = 0, so C(x) = C is an arbitrary constant, and x v(x, y) = x + y + C. (Note: Taking C = 0, we find that u(x, y) + iv(x, y) = z.)
3. (5 pts) In each integral below, the integrand has singularities. Classify each singularity as: removable, pole (find the order), or essential. Then compute the integral. All closed contours are taken counterclockwise. e z (a) dz. Here the only singularity inside the contour is 0. Because the function z = z(z ) φ(z) = is nonzero when z = 0, we see that 0 is a simple pole. So by (??), this (b) ez (z ) e integral is = (πi) 0 (0 ) e z = πi. dz. Now the singularities 0 and are both inside the contour. 0 is a z =3 z(z ) simple pole as shown above, and is a pole of order (since the function ϕ(z) = ez is z nonzero when z = ). By the residue theorem, the value of the integral is πi (residues at 0 and ) = πi( 4 + ϕ ()! ) = πi( 4 + zez e z z z= ) (c) = πi( 4 + e e 4 ) = πi ( + e ). z dz. Note that the numerator and denominator both vanish at 0. This tells z = sin z us nothing about the nature of the singularity there (we need to examine the Laurent series for that). In fact, the integrand has a removable singularity at z = 0 because lim z 0 z sin z = lim z 0 z z z 3 /3! + z 5 /5! = lim z 0 z /3! + z 4 /5! =. Therefore by Cauchy s theorem the integral equals 0. e z g(z) (d) dz. The integrand has the form h(z) z = sin z = ez, where g(0) 0, h(0) = 0, and sin z h (0) = cos(0) 0. Therefore it has a simple pole at z = 0. The denominator has other (e) zeroes, namely πk for nonzero k Z, but these lie outside the contour. By the residue theorem, the value of the integral is πi g(0) = πi. h (0) ze /z dz. The integrand has an essential singularity at 0. This is seen from the z = fact that the Laurent expansion has infinitely many negative power terms: ze /z = z( + z +!z 4 + 3!z 6 + ) = z + z +!z 3 + 3!z 5 + From this we also see that the residue equals. So the value of the integral is πi = πi.
4. (0 pts) Expand the function f(z) = as a Laurent series in the annulus < z. (z ) Let F (z) = (z ) = (z ). Note that F (z) = f(z), so if we can get the Laurent series for F (z), we can differentiate it to get the one for f(z). Note that the center of the annulus is 0, so this will be a Laurent series in powers of (z 0) = z. On the given annulus, /z <. So we write: F (z) = z ( z ) = z n z = n z n. n This series converges whenever /z <, i.e. z >. Therefore on this annulus, f(z) = F (z) = ( n ) n z n = 5. (0 pts) Expand the function f(z) = < z <. By partial fractions, f(z) = separately. First, z = + (z ) = 4 z(z )(z ) (n + ) n z n+. as a Laurent series in the annulus z(z )(z ) = / z z + /. We treat each term z + z provided z <, i.e. z <. Second, = 4 ( z ) = 4 ( z )n = z = (z ) + = (z ) + = z ( ) = (z ) (z ) z = ( ) n+ (z ) n = provided <, i.e. < z. Lastly, z n= z = (z ) ( ) n (z ) n, ( ) n n+ (z )n, ( z )n is already a Laurent series about, valid for all z. It can be combined with the n = term from the previous case, namely, (z ), to give (z ). Putting everything together, the Laurent series f(z) = n= ( ) n (z ) n (z ) + ( ) n (z )n n+
is valid on the annulus < z <. 6. (5 pts) Find and classify the singularities of tan z = sin z cos z. Find the residue at each isolated singularity. We know from a problem earlier in the semester that the only zeros of cos(z) are the real ones, namely numbers of the form π + πk for k Z (prove it). Because the numerator sin( π + πk) = ± is nonzero, we see that these are not removable singularities of tan z. In fact, cos ( π + πk) = sin(π + πk) 0, so such a point is a simple zero of cos(z), and hence a simple pole of tan z. By our formula g(w) h (w) for the residue at a simple pole, we have Res z= π +πk sin z cos z = sin( π + πk) sin( π =. + πk) 7. (5 pts) (From MT review) Suppose f = u + iv is an entire function, and that u(x, y) is bounded. Prove that f is constant. (Hint: Apply Liouville s Theorem to the function g(z) = e f(z).) By hypothesis, there exists M > 0 such that u(x, y) M for all x, y R. With g(z) = e f(z), we see that for all z = x + iy C, g(z) = e u(x,y)+iv(x,y) = e u(x,y) e M, since e iv(x,y) = (it is an instance of e iθ = ). Thus g is bounded on C. Since f is entire, and e z is entire, it follows from the chain rule that g(z) = e f(z) is entire. By Liouville s theorem, g(z) is constant, and hence f(z) is also constant. x 8. (5 pts) Evaluate 0 + x dx. 4 (Turn it into an integral from to, then use residues like we did for a similar example in class.) Let f(z) = z. Because f(z) = P (z) +z 4 Q(z) and deg Q deg P +, the integral 0 f(x)dx = is a ratio of polynomials, with Q(x) 0 for real x, f(x)dx = R lim f(x)dx R is convergent. For the first equality above, we used the fact that f(x) is an even function.
For any large number R > 0, let C R be the closed curve going along the line segment on the real axis from R to R, and then following the semicircular arc centered at 0 and traveling from R counterclockwise back to R. By the residue theorem, CR z + z 4 dz = πi (Residues of f(z) inside C R.) Hence x + x dx = 4 πi (Residues of f(z) in the upper half-plane) lim f(z)dz. 0 We first use the ML-inequality to show that lim f(z)dz = 0. For z on, we have z = R. So z + z 4 R R 4 by the reverse triangle inequality. So by the ML-inequality, f(z)dz R R 4 πr = πr3 R 4. It follows that lim f(z)dz = 0, as claimed. Hence x + x dx = 4 πi (Residues of f(z) in the upper half-plane). 0 The singularities of f(z) are the zeros of + z 4. As done in class, we solve for these roots by setting z 4 =. Taking the norm of both sides, we see that z =. So writing z = e iθ, we have e 4iθ = e iπ = 4θ = π + πk = θ = π 4 + π k. This gives four distinct roots: z = e iπ/4, e i3π/4, e i5π/4, e i7π/4.
The first two are in the upper half-plane; the last two are in the lower half-plane. Since d ( + dz z4 ) = 4z 3 is nonzero at each of these points, they are all simple poles of f(z), so using our formula for the residue at a simple pole, the integral is ( ) (e iπ/4 ) = πi 4(e iπ/4 ) + (ei3π/4 ) = πi 3 4(e i3π/4 ) 3 4 [ e πi/4 + e 3πi/4] = πi i ( ) = 4 π. To evaluate the sum in brackets, I drew a picture and used some trig. (Note that a real integral must have a real value! This can serve as a check for your work.) 9. (5 pts) Find the maximum value of z on the unit disk z. The function z is holomorphic, so by the maximum modulus principle, the maximum of z occurs on the boundary of the disk. (We don t actually need this fact to solve this particular problem, but it is nice to know what to expect.) Note that z = (z )(z ) = (z )(z ) = z z (z + z ) + = z 4 Re(z ) +. From this calculation, we see that the above quantity is maximized when Re(z ) is as small as possible. Because z also belongs to the unit disk, its real part is minimal when z =, so z = ±i. Thus the maximum occurs at ±i, and the maximum value is i =.