Math 1260 Solutions for Final Exam Page 1

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1 Math 6 Solutions for Final Exam Page Problem. ( points) For each of the following functions, circle what kind of singularity it has at z =, and compute the residue at z =. (a) f(z) = z sin(z). Removable Pole Essential (b) f(z) = sin(/z). Removable Pole Essential Solution. (a) f(z) has an pole at z =. To see this, we note that z lim z z f(z) = lim z z sin(z) =, so in fact, f(z) has a pole of order. The residue can be computed using the derivative formula that we often use, or alternatively using the power series for sin z. Thus Hence z sin(z) = ) z (z z3 6 + = z z 6 + = ) ( z + z 6 + = z [ ] Res z sin(z), =. But the easiest way to compute the residue is to note that f(z) = f( z), i.e., the function f(z) is even, so in its Laurent expansion c k z k, k= we have c k = for every odd k. In particular, we have c =, so the residue is. Math 6 Final Exam Fri, Dec, 3 5pm

2 Math 6 Solutions for Final Exam Page (b) f(z) has an essential singularity at z =. We can see this by observing that lim z z m f(z) does not exist for any integer m. To ease notation, let w = /z, so we re looking at f(w) = sin(w). If we write f(w) as a Laurent series in w, then the residue is the coefficient of w. Thus f(w) = ) (w w3 6 + = w w 6 + = ) ( w + w 6 + = w + w 6 + Thus the Laurent series of f(z) looks like f(z) = z + 6z + a z 3 + b z 5 + so [ ] Res sin(/z), = 6. Problem. (5 points) Compute the values of each of the following integrals. e 3z (a) z = z dz. 3 (b) z = z(e z ) dz. (c) z dz, where γ is the line segment from to + i. γ Math 6 Final Exam Fri, Dec, 3 5pm

3 Math 6 Solutions for Final Exam Page 3 Solution. (a) There is a one pole at z =, and it is a pole of order 3. The residue theorem gives [ ] e 3z e 3z dz = Res πi z = z z, [ ) ] = Res ( + 3z + (3z) + (3z)3 +, z 3! 3! [ = Res z z + 9 z + 9 ] +, So = 9. z = e 3z dz = 9πi. z (b) Again there is one pole at z =, but this time it is a pole of order. Letting f(z) =, we have z(e z ) e z ze z (e z ) = Res [ f(z), ] d ( = lim z f(z) ) z dz d = lim z dz ( z ) e z = lim z (e z ) ze z (e z ). One way to proceed now is to use L Hopital s rule a couple of times. An easier(?) way is to use the Taylor series expansion of e z. Thus ( ) ( ) + z + z + z3 + z + z + z + z = ( z + z + z3 + ) 6 ( ) ( ) z + z3 + z + z3 + z = z + z +. z + z 3 + So the limit as z is. Therefore [ ] ( dz = πi Res z = z(e z ) z(e z ), = πi ) = πi. Math 6 Final Exam Fri, Dec, 3 5pm

4 Math 6 Solutions for Final Exam Page 4 (c) This line integral needs to be done directly from the definition. The curve γ is parametrized by z(t) = ( + i)t for t. So γ z dz = = = = =. ( + i)t d ( ( + i)t ) ( i)t ( + i) dt t dt Problem 3. ( points) Use residue theory to compute the value of the definite integral x (x + ) dx. (I expect you to use complex analysis for this problem, although it can also be done using continued fractions as you learned in first-year calculus.) Solution. Let f(z) = z (z + ), and let D R = { re iθ : R < r < R and < θ < π }. (This is the usual half-disk in the upper half-plane.) The function f(z) has one pole in D R, namely z = i, and it s a double pole, since z f(z) = (z + ) = z (z + i) (z i). Math 6 Final Exam Fri, Dec, 3 5pm

5 Math 6 Solutions for Final Exam Page 5 The residue at z = i is Res [ f(z), i ] d ( = lim (z i) f(z) ) z i dz ( ) d z = lim z i dz (z + i) ( ) ( ) z d z = lim z i z + i dz z + i ( ) ( ) z i = lim z i z + i (z + i) i = i i 4 = i 4. So the residue theorem says that f(z) dz = πi Res [ f(z), i ] = π D R. Let L R be the line segment R x R, and let Γ R be the semicircle of radius R in the upper halfplane. Then R x f(z) dz = L R R (x + ) dx. Next we estimate f(z) dz sup f(z) Length(Γ R ) (ML estimate), Γ R z Γ R = sup z z Γ R (z + ) πr R (R ) πr (note it s R, not R +,) R. So letting R and combining these calculations gives ( x ) dx = lim f(z) dz f(z) dz = π (x + ) R D R Γ R. Problem 4. ( points) For each of the following functions, describe the Taylor series expansion about the indicated point, and compute the radius of convergence. Math 6 Final Exam Fri, Dec, 3 5pm

6 Math 6 Solutions for Final Exam Page 6 (a) f(z) = log(z) centered at z =. (b) f(z) = ( z) centered at z =. Solution. In general the Taylor series expansion of f(z) centered at a is f (k) (a) (z a) k. k! k= Each part of this problem can be done by computing the derivatives at the indicated point. Alternatively, one can use related series and differentiate or integrate them. (a) We have f (z) = z, so f (z) = z = + (z ) = + z Now we can expand using the geometric series to get f (z) = ( ) k z ( ) k ( ) k = (z k+ )k. Integrating gives k= f(z) = k= k= ( ) k (k + ) k+ (z )k+ + C. The constant is obtained by setting z =, so log() = f() = C. Finally, relabeling, we get ( ) k+ f(z) = log(z) = log() + (z ) k. k k The radius of convergence ρ may be computed using the ratio test or the root test. The latter gives ρ = lim ( ) k+ /k k k k = lim k k = /k, so ρ =. (b) Again, it s not very hard to compute the derivatives. But even easier to note that f(z) is the derivative of, which is just a geometric z series. So f(z) = d ( ) = d z k = kz k = (k )z k. dz z dz k= k= k= Math 6 Final Exam Fri, Dec, 3 5pm k=.

7 Math 6 Solutions for Final Exam Page 7 The radius of convergence is ρ = lim =. k (k ) /k Problem 5. ( points) Let f(z) = z z. (a) Find the Laurent series of f(z) centered at in the domain z <. (b) Find the Laurent series of f(z) centered at in the domain z >. Solution. (a) We note that f(z) has a pole at z =, but that s okay. The partial fraction expansion of f(z) is f(z) = / z / z. We leave the second term alone and expand the first using the geometric series / z = 4 z = ( z n = 4 ) n+ zn. This converges on z <. Further, if we also include n =, we get the other term, so the Laurent series of f on the domain z < is f(z) = n+ zn. n= (b) For z >, we want an expansion in the variable /z, so / z = z = ( ) n n = z z z = n z n+ z. n This gives n= n= f(z) = z + n= n= n z n. We can simplify by noting that the n = term cancels the /z, so n f(z) = z. n n= Problem 6. (5 points) Let D be a bounded domain with nice boundary. Math 6 Final Exam Fri, Dec, 3 5pm n= n=

8 Math 6 Solutions for Final Exam Page 8 (a) Suppose that f(z) is analytic on D, continuous on D D, and does not vanish on D D. Let m = inf f(z) z D be the smallest value of f(z) on the boundary of D. Prove that f(z) m for all z D. (This is a minimum principle that complements the maximum principle.) Problem 6. (continued) (b) Let D be the unit disk. Find a function that is analytic on D D and satisfies f() =, and such that f does not satisfy the minimum principle. (c) Suppose that f(z) is analytic and non-constant on D and continuous on D D. Assume further that f(z) is constant for z D. Prove that f(z) must have a zero in D. Solution. (a) For any function h, we write M(h) = sup h(z) and m(h) = inf h(z). z D z D Since f(z) does not vanish on D, we know that g(z) = /f(z) is analytic on D. So we can apply the maximum principle to g(z) to conclude that g(z) M(g) for all z D. Since g = /f, this implies that f(z) M(/f) for all z D. But if T is any set of positive real numbers, we have { } sup t : t T = inf{t : t T }. This implies that M(/f) = /m(f). Substituting this in above gives m(f) f(z) for all z D. (b) The simplest example is f(z) = z. Then m(f) =, but f(z) is not larger than m(f). In fact, we have f(z) < m(f) for all z in the unit circle. (c) The maximum principle says that f(z) M(f) for all z D. Math 6 Final Exam Fri, Dec, 3 5pm

9 Math 6 Solutions for Final Exam Page 9 Suppose that f does not vanish. Then the minimum principle (proven in (a)) says that f(z) m(f) for all z D. However, we re given that f(z) is constant for z D, so m(f) = M(f) directly from the definitions of m(f) and M(f). So our two inequalities imply that f(z) = M(f) = m(f) for all z D. In particular, there are points z D for which f(z) = M(f), so the other half of the maximum principle tells us that f is constant. Problem 7. ( points) Let f(z) be the polynomial f(z) = z 4 + 5z +. (a) Prove that f(z) has exactly one root inside the disk z <. (b) How many roots does f(z) have inside the annulus < z <? Prove that your answer is correct. Solution. (a) For z = we have 5z = 5 = z 4 + z 4 +. So from Rouché s theorem, the polynomial f(z) and the polynomial 5z have the same number of zeros in the disk z <. Since 5z clearly has one zero in the disk, so does f(z). (b) On the circle z = we have z 4 = 6 6 = 5z + 5z +, so f(z) and z 4 have the same number of zeros in the disk z <. Since z 4 has four zeros (counted with mulitplicity), so does f(z). That s the number of zeros in the disk z <, and we know from (a) that there is one zero in the disk z <, so f(z) has three zeros in the annulus < z <. Problem 8. ( points) Let f(z) be analytic in a domain D, and suppose that f satisfies Re ( f(z) ) = Im ( f(z) ) for all z D. Prove that f is constant in D. Math 6 Final Exam Fri, Dec, 3 5pm

10 Math 6 Solutions for Final Exam Page Solution. There are probably lots of ways to do this problem. Here s one. Write f(z) = u(x, y) + iv(x, y) as usual. The assumption that Re ( f(z) ) = Im ( f(z) ) says that u(x, y) = v(x, y). Now the Cauchy Riemann equations yield It follows that u x = v y = u y = v x = u x A similar calculation gives (Cauchy Riemann equation) (since v = u) (Cauchy Riemann equation) (since v = u). u x =. u y =. Alternatively, we can use u x = and compute = u x = v y = u y. Thus u x = and u y =, which implies that u is a constant. And since v = u, we find that f = u + iv is also a constant. Problem 9. ( points) Proe that there exists a function f(z) with the following properties: f(z) is meromorphic on C. f(z) has simple poles at the points {,, 3, 4,...} and no other poles. For k {,, 3,...}, the residue of f(z) at k is equal to k. Be sure to prove that the function that you define is meromorphic, as well as having the indicated poles and residues. Solution. We d like to use k z k, k= Math 6 Final Exam Fri, Dec, 3 5pm

11 Math 6 Solutions for Final Exam Page but it doesn t converge. The function k/(z k) looks like when k is large, so we might try summing k z k + = z z k. But z/(z k) looks like z/k when k is large, so its sum won t converge, either. So we add on z/k to compensate, k z k + + z k = z z k + z k = z (z k)k. Note that this calculation shows that z /((z k)k has a simple pole at z = k with residue k. Then we define f(z) = k= ( k z k z ) = k k= z (z k)k. The usual argument shows that f is meromorphic with the correct poles. We briefly indicate. Choose any R (not an integer) and break up f as f(z) = f (z) + f (z) = z (z k)k + z (z k)k. k<r k>r Let D R = { z < R} be a disk of radius R. Then f (z) is meromorphic on D R with simples poles at the integers k < R and residue k at k. On the other hand, for z D R and k > R we have z (z k)k R (k R)k, so z (z k)k k>r k>r R (k R)k <. The Weierstrass M-test implies that the series defining f (z) converges to an analytic function on D R. Hence f(z) is meromorphic on D R with the desired poles and residues at {k < R}. Since R is arbitrary, this shows that f(z) is entire with the desired poles and residues. Problem. ( points) Let f(z) be an analytic function that maps the unit disk conformally to a domain D. In other words, if we denote the unit disk by D = { z < }, then Also let f : D D is analytic, one-to-one, and onto. m = inf f() w w D Math 6 Final Exam Fri, Dec, 3 5pm

12 Math 6 Solutions for Final Exam Page be the distance from f() to the boundary of D. Prove that f () m. (Hint. Consider the inverse function f : D D, note that the disk around f() of radius m is contained in D, and use Schwarz s lemma.) Solution. We consider the inverse function f : D D. Our choice of m tells us that the disk B = { w C : f() w < m } is contained in D, so f is analytic on B; and since the image of f is in D, we know that f (w) for all w D. We want to shift B to be the unit disk. The map z mz + f() send the unit disk to B, so we should look at the function g(z) = f ( mz + f()) ). Then g : D D with g() =, so the derivative version of Schwarz s lemma says that g (). Note that g () = (f ) ( f() ) m. So we find that (f ) ( f() ) m. Okay, now we differentiate the identity f ( f(z) ) = z to get (f ) ( f(z) ) f (z) =. Evaluating at z = gives (f ) ( f() ) f () =. So (f ) ( f() ) = f (), and substituting this above gives f () m. Cross-multipying gives m f (), which is the desired result. Math 6 Final Exam Fri, Dec, 3 5pm

13 Math 6 Solutions for Final Exam Page 3 Problem. ( points) Compute the value of the integrals cos(x ) dx and sin(x ) dx Hint #. Integrate the function f(z) = e iz around the boundary of the region { D R = re iθ : < r < R and < θ < π }. 4 Hint #. The following integral from 3rd semester calculus may be useful: π e t dt =. Warning. Don t spend too much time on this problem until you ve worked on the other problems. Solution. The function f(z) = e iz is entire, so Cauchy s theorem tells us that D R f(z) dz =. The boundary of D R consists of 3 pieces: L = {x : x R}, L = {t i : t R} Γ R = {Re iθ : θ π/4}. (in reverse direction), Here i is the square root in the first quadrant, i.e., i = +i. The integral along L gives the integrals that we re trying to compute, L f(z) dz = For L, we have R e ix dx = L f(z) dz = R R = i R cos(x ) dx + i sin(x ) dx. ( ) e i( it) d it R = + i R e t dt e t dt. Math 6 Final Exam Fri, Dec, 3 5pm

14 Math 6 Solutions for Final Exam Page 4 Hence lim f(z) dz = + i R L e t dt = + i π. Finally, for Γ R we use Jordan s lemma, which says that if C R is the semicircle of radius R in the upper halfplane, then C R e iz dz < π. Our curve Γ R is not equal to C R. By making the change of variables w = z 4, we could map Γ R to C R, but then we wouldn t get the integral in Jordan s Lemma. So instead we use the change of variables w = z, which maps Γ R to the quarter-circle Then Hence B R = {Re iθ : θ π/}. f(z) dz = f(w / ) d(w / ) Γ R B R = dw w. / B R e iw f(z) dz = e iw dw Γ R B R w / e iw dw B R w / = e iw dw. R / B R In order to use Jordan s lemma, we note that C R is the union of the quarter-circle B R and the quarter-circle B R = { z : z B R}. In other words, the map z z maps B R to B R. We also note that if z = x + iy B R, then z = x + iy, so and similarly e zi = e ( x+iy)i = e y ix = e y = e (x+iy)i = e zi, d( z) = dz, so e iz dz = e iz dz. B R B R Math 6 Final Exam Fri, Dec, 3 5pm