MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 10 SOLUTIONS. f(z) = a n. h(z) := a n+m (z a) n. f(z) = h(z) + (z a) m n. =: e h(z) F (z).

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1 MATH 85: COMPLEX ANALYSIS FALL 29/ PROBLEM SET SOLUTIONS. (a) Show that if f has a pole or an essential singularity at a, then e f has an essential singularity at a. Solution. If f has a pole of order m at a, then there exists ε > and g : D(a, ε) C analytic, g(a), such that f() = g() ( a) m for all D (a, ε). Let the power series representation of g on D(a, ε) be g() = a n ( a) n. Let Then and so h() := n= a n+m ( a) n. n= m a n f() = h() + ( a) m n m e f() = e h() n= e n= an ( a) m n =: e h() F (). Note that e h() is analytic and non-ero. If e f() has a pole or a removable singularity at a, then F () = e f() e h() will have a pole or a removable singularity at a. So since F () has an essential singularity at a, e f() must have an essential singularity at a. If f has an essential singularity at a, then f(d (a, ε)) is dense in C for all ε >, ie. f(d (a, ε)) = C (here S denotes the closure of the set S). Recall that if g is any continuous function, then g(s) g(s). Since exp : C C is a continuous function, let S = f(d (a, ε)) and we have exp(s) exp(s) = exp(c) = C. Hence, (exp f)(d (a, ε)) is dense in C. By part (a), a is an essential singularity of exp f. (b) Let Ω C be a region. Let a Ω and f : Ω\{a} C be a function with an isolated singularity at a. Suppose for some m N and ε >, Re f() m log a for all D (a, ε). Show that a is a removable singularity of f. Solution. Note that the condition implies that e f() = e Re f() e m log a = a m. Date: December 5, 29. Now that you have learnt about Laurent series and its relation with poles, you could of course just write this down without going through the preceding arguments.

2 Hence for all D (a, ε) and thus ( a) m e f() ( a a)m+ e f() a =. a So a must either be a pole (of order k m) or a removable singularity of e f. By part (a), a cannot be a pole nor an essential singularity of f (otherwise a will be an essential singularity of e f, contradicting the previous statement). Hence a must be a removable singularity of f. 2. (a) Let f : D (, ) C be analytic. Show that if f() log for all D (, ), then f. Solution. Clearly f has an isolated singularity (which may be removable) at. Let the Laurent expansion of f in D (, ) be f() = a n n. n= Then by the integral formula for Laurent coefficients, a n = f() d i n+ where r is given by :, ] C, (t) = re it, and < r <. Hence a n = r r n log re it f(re it )ire it re it n+ f(re it ) dt r n = r n log r. Since this true for all < r <, for n =, 2, 3,..., we may take it as r + to get a n r r n log + r = (note that this only works when n is positive). Hence a n = for all n < and so f() = a n n n= must have a removable singularity at. Upon defining f() = a, we may assume that f : D(, ) C is an analytic function. Now we may apply maximum modulus theorem to see that for all < r <, Hence max r f() = max f() max log =r =r = log r. dt max f() log < r r =, dt 2

3 and so f on D(, ). (b) Let f : C C be analytic on C with a pole of order at. Show that if f() R for all =, then for some α C and β R, f() = α + α + β for all C. Solution. Since f has a pole of order at and is analytic otherwise, the Laurent expansion of f takes the form f() = n= a n n. (2.) Let be the closed curve :, ] C, (θ) = e iθ. Since f(e iθ ) R for all θ, ], we must have f(e iθ ) = f(e iθ ). The integral formula for Laurent coefficients yields, for any n Z, a n = f() d i n+ Note that and so we get = = = f(e iθ )e inθ d f(e iθ )e inθ d f(e iθ )e inθ d. a n = f(e iθ )e inθ d a n = a n for all n Z. By (2.), a n = for all n >. So a n = for all n >. Let a = α and so a = a = α. Let a = β. Then (2.) becomes f() = α + α + β. We know that α since f has a simple pole at. β = a = a = β. We also know that β R since 3. Evaluate the integral for i = a, b. (a) f a : C C is given by i f i f a () = e e and a is the boundary D(, 2) traversed once counter-clockwise. 3

4 Solution. as follows: f a is analytic in C and its Laurent expansion about = may be obtained e e = + e + 2! e n! e n + = ] 2! ! + 2! n n! + 2! ( n ) 2 + ] +. Observe that the coefficient of the term is simply ( ) ] ! + 3 3! + + n n! + = +! + 2! + + (n )! + By the residue theorem (b) f b : D (, π) C is given by = e. a f a = i Res(f a ; ) Ind( a ; ) = ei. f b () = (sin ) 3 and b is the boundary D(, ) traversed once counter-clockwise. Solution. f b is analytic in D (, π) and its Laurant expansion about = may be obtained as follows: (sin ) 3 = 3! 3 + ] 3 5! 5 = ( 3 3! 2 )] 3 5! 4 + = ( ! 2 ) ( 5! ! 2 ) ] 2 5! Now observe that the terms enclosed in the second parentheses onwards would all have powers at least 4 and so will not contribute to the term. The only term in the first parentheses that contribute to the term is the 2 term, which has coefficient 3/3! = /2. By the residue theorem, b f b = i Res(f b ; ) Ind( b ; ) = πi. 4. (a) Let the Laurent expansion of cot(π) on A(;, 2) be cot(π) = a n n. n= Compute a n for n <. Solution. Let be the circle D(, r) traversed once counter-clockwise and < r < 2. Note that Ind(; ) = for all D(, r), ie. the bounded component of. By the integral formula for Laurent coefficients, a k = i cot(π) d = k+ i k cot(π) d 4

5 for k N. For k =, k cot(π) = cot(π) = cos(π) sin(π) has three isolated (non-removable) singularities in the bounded component of, namely,,,. So by the residue theorem 2, a = cot(π) d i = Res(cot(π); ) + Res(cot(π); ) + Res(cot(π); ) = cos(π) π cos(π) + cos(π) = π cos(π) + cos(π) = π cos(π) = 3 π. = For k 2, k cot(π)] = sin(π) k cos(π)] ] π sin(π) = π = ] k cos(π) In other words, is a removable singularity of k cot(π) for k 2. Note that the residue about any removable singularity is. So by the residue theorem, a k = k cot(π) d i = Res( k cot(π); ) + Res( k cot(π); ) + Res( k cot(π); ) = k cos(π) d d sin(π) + + k cos(π) d = d cos(π) = ( )k + π π if k is odd, = 2 if k is even, π for k 2. (b) For n =,, 2,..., compute i n d 3 sin where n is the circle D(, r n ) traversed once counter-clockwise and r n = (n + 2 )π. Solution. Note that for m, ( mπ) mπ 3 sin = mπ = m 3 π 3 = m 3 π 3 = mπ 3 sin( mπ) mπ mπ sin( mπ) 2 We use result that Res(ϕ/ψ; a) = ϕ(a)/ψ (a) if ϕ(a), ψ(a) = and ψ (a). 5

6 while for m = 4 3 sin = sin =. So the integrand 3 csc has a pole of order 3 at and simple poles at mπ for all m Z, m. Since D(, r n ), the bounded component of n, contains {mπ m = n,,,,,..., n}, the residue theorem yields d n i 3 sin = n m= n ( ) Res 3 sin ; mπ. Now for m, ( ) Res 3 sin ; mπ = /3 d d sin = /m3 π 3 cos(mπ) = ( )m m 3 π 3, =mπ and for m =, ( ) Res 3 sin ; =. For the latter, observe that the Laurent expansion of 3 csc contains only even powers, and so a =. Hence d n i 3 sin = ( ) m n m 3 π 3 ( ) m m 3 π 3 =. n m= m= 5. (a) Does the following function have an antiderivative on A(; 4, )? ( )( 2)( 3) Solution. Let f : A(; 4, ) C be f() := ( )( 2)( 3). Let A(; 4, ) be the boundary of a rectangle R traversed once counter-clockwise. Since A(; 4, ), we must have, 2, 3 R, the bounded component of. So the winding numbers Ind(; n) = for n =, 2, 3. By the residue theorem f() d = Res(f; ) + Res(f; 2) + Res(f; 3). i Since f has simple poles at, 2, 3, the required residues may be evaluated by Res(f; ) = ( )f() = Res(f; 2) = 2 ( 2)f() = 2 ( 2)( 3) = 2, ( )( 3) = 2, Res(f; 3) = ( 3)f() = 3 3 ( )( 2) = 3 2. Hence, f() d =. i Note that this holds for arbitrary and thus all A(; 4, ), = R for some R. Hence by Problem Set 9, Problem 3(a) (which is really Morera s Theorem), f has an antiderivative on A(; 4, ). 6

7 (b) Does the following function have an antiderivative on A(; 4, )? Solution. Let g : A(; 4, ) C be g() := 2 ( )( 2)( 3) 2 ( )( 2)( 3). The same argument above applies but this time 2 Res(g; ) = ( )g() = ( 2)( 3) = 2, 2 Res(g; 2) = ( 2)g() = 2 2 ( )( 3) = 4, 2 Res(g; 3) = ( 3)g() = 3 3 ( )( 2) = 9 2. So g() d =. i Recall that if g has an antiderivative on A(; 4, ), then the integral about any closed curve in A(; 4, ) is necessarily 3. Hence g does not have an antiderivative on A(; 4, ). 3 This doesn t depend on having a simply connected region. So it works for annulus too. 7