Math 460: Complex Analysis MWF am, Fulton Hall 45 Homework 8 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book:.4.,.4.0,.4.-.4.6, 4.., 4.., 4..6 Problem K. Show that there exists no non-constant analytic function f : C \ {0} D 0). Problem L. Suppose that f is a holomorphic function with Laurent development f) 0 ) + a +... 0 on the annulus D r 0 ) \ { 0 }, and suppose that g) is analytic in the disk D r 0 ). Show that Resf g, 0 ) Resf, 0 ) g 0 ) + g 0 ). Problem M. Let γ be the circle of radius around 0, oriented counterclockwise. Compute + 4 d. γ Solution.4.. No: Suppose f was entire, and f/n) 0 for all n N. Since /n 0 as n and f is continuous at 0, we must have f0) f/n) 0. In that case, f is ero on the non-discrete subset {/n} {0} C. By the Identity Theorem, f would be identically ero. On the other hand, sinπ/) is a non-constant analytic function on C \ {0} vanishing on {/n : n N}. Solution.4.0. For n N, consider the deleted neighborhood D /n 0 ) \ { 0 }. By assumption, f takes the value c in every deleted neighborhood of 0, so there is n D /n 0 ) such that f n ) c.
Suppose that f has a removable singularity at 0. In that case, f has an analytic extension ˆf to U, which is therefore continuous at 0. By Exercise..6 on homework ), 0 ˆf) n ˆfn ), so ˆf 0 ) 0 ˆf) n ˆfn ) n f n ) c. In that case, ˆf is equal to c on the set {n : n N} { 0 }, which is non-discrete since n 0. By the Identity Theorem, ˆf must be identically c, so that f is constant, a contradiction. Thus c is not a removable singularity of f. On the other hand, if f has a pole of order k at c, then we must have 0 ) k f) n 0 ) k f n ) 0 n ) ) n 0 ) k f n) n n 0 c 0. This violates the definition of a pole of order k at c. Thus c is neither removable or a pole, so that f has an essential singularity at c. Solution.4.. We have f) ) ) + ). Let f ) /, f ) / ), and f ) / + ). Note that f is analytic on C \ {0}, f is analytic on C \ {}, and f is analytic on C \ { }. Thus f )f ) is analytic in a neighborhood of 0, f )f ) is analytic in a neighborhood of, and f )f ) is analytic in a neighborhood of. Since f) f )f ) f )f ) f )f ), + the function f has simple poles at each of 0,, and. Solution.4.. The function f) sin/) is analytic on C \ {0}. Suppose that there were a non-negative integer k so that 0 k sin/) existed and was finite. In that case, this it is equal to the it obtained by approaching 0 through the positive imaginary axis. In other words, 0 k sin ) ) sin. y 0 +iy)k iy
Since sin e i e i )/i, we have ) sin e/y e /y, iy i so we have ) 0 k sin iy) k y 0 + i e /y e /y ) ik yk e /y y k e /y. y 0 + Since we have y 0 + y k e /y 0, for any k, we would have ) 0 k sin ik yk e /y. y 0 + As we discussed in class, the latter it is not finite. We conclude that 0 k sin/) doesn t exist either, for any integer k. Thus 0 can be neither removable nor a pole, so that 0 is an essential singularity of f. Solution.4.4. Using the power series expansion of e, we have e + + +! +... +! +... +! +..., which converges for all C. The latter can be shown using the ratio test, as with showing that the power series for e converges for all C). Solution.4.5. Let f). Using the power series expansion for e, we may rewrite e f): e + + + +... + + +...!! ) + + +... + + ) +...)! + + +...!! )...! + +...! +... + + +....!! This last expression is the quotient of two power series, each convergent on all of C. Since the denominator is non-ero at 0, it is non-ero in a disk neighborhood of 0, and the quotient is analytic at 0. Thus the singularity of f) at 0 is removable.
4 There are several other singularities of f, namely πin, for any non-ero integer n. At these singularities, / is analytic. Moreover, we have e πin) + πin) + +..., so that e + πin) + πin) +... πin) + πin) +... πin + πin +..., which shows that /e ) has a simple pole at πin. Thus f has a simple pole at πin for each non-ero integer n, and a removable singularity at 0. Solution.4.6. Recall the power series expansion for the principal branch of the logarithm log, centered at : log ) ) + )..., which converges in D ). For any D ) \ {}, we have Since ) log ) +.... ) log ) ) +... 0, ) the function log / ) has a simple pole at. Solution 4... The function f) / ) is given by f) ) +. Since / ) is analytic on { : < } and / is analytic on { : 0 < } and vanishes at, the decomposition from Theorem 4.. is given by g) / ) and h) /. The function g)
5 has the power series expansion g) n, and the desired Laurent development of h in the annulus { : 0 < } is given by h) /. Thus the Laurent development of f is given by f) + n n n. Solution 4... For any 0, we have e / n! n, so that e / n! n n!. n Re-indexing, we have e / n n n)!. Solution 4..6. The function f) / + ) can be decomposed using the method of partial fractions: + + i) i) / + i + / i. Since +i) is analytic on { : i < }, and i) is analytic on { : 0 < i } and vanishes at, the decomposition from Theorem 4.. is given by g) + i) and h) i). Evidently, a power series for h in the variable w i) is already clear, so it
6 only remains to compute the power series for g. We have + i) i) + i) i i) + i i i i) i i)) n i i n i)n. Thus, we obtain the Laurent development f) i) + i n i)n. Solution K. Suppose f : C \ {0} D 0) is a non-constant analytic function. Since f) for all C \ {0}, f is bounded in absolute value in a neighborhood of 0. By the Removable Singularity Theorem, the isolated singularity 0 is removable, and f extends to a function ˆf which is analytic on the entire complex plane, i.e. ˆf is entire. Note that ˆf) max{, ˆf0) } for all C one could use continuity to say that ˆf) for all, but this unnecessary), so that ˆf is bounded, and by Liouville s Theorem it is constant. Since f) ˆf) for all 0, we conclude f is constant, contradicting the stated assumptions. Solution L. Note that the residue of any function at an isolated singularity 0 is given by the coefficient of 0 ) in its Laurent development in the annulus D r 0 ) \ { 0 }, for r small enough so that f is analytic on D r 0 ) \ { 0 }. The function g is holomorphic in D r 0 ), so it has a power series expansion g) c n 0 ) n that converges in D r 0 ). In that case, using the Laurent development for f in D r 0 ) \ { 0 }, for D r 0 ) we have f) g) 0 ) + a ) + a 0 +... c 0 + c 0 ) + c 0 )... ) 0 c 0 0 ) + a c 0 + c + c + a c + a 0 c 0 ) +..., 0
so that Resf g, 0 ) a c 0 + c. Since a Resf, 0 ), c 0 g 0 ), and c g 0 ), we have as desired. Resf g, 0 ) Resf, 0 ) g 0 ) + g 0 ), Solution M. Let h) + )/ 4 ). Since for 0 or for 4 ) 0 ± ±, the function h is analytic for any C\{0, +, }. Evidently, ind γ 0) ind γ ) and ind γ + ) 0, since <, while + >. By the Residue Theorem we have + 4 d πi Resf, 0) + Resf, ) ). γ Factoring the denominator of h, we have h) + ) + ). Thus Resh, + ) Res ) + ), ) ) + ) ) + ) ) 5 4 ) ) 5 8 4 5 + + 4. In order to compute Resh, 0), we use Problem L, with f) / and g) + )/ ). The Laurent development of f in the annulus C \ {0} is evidently /, so that the coefficient of is 0, and Resf, 0) 0. Problem L now says that Resh, 0) g 0), so we compute: ) g ) + ) ) 0) ). 0 7
Putting these computations together, we find that + 4 ) d πi + πi 4 6 4 ). γ