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.
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
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
Solution. as follows: f a is analytic in C and its Laurent expansion about = may be obtained e e = + e + 2! e 2 + + n! e n + = + + + ] 2! 2 + + + 2 2! + 2! + + + n n! + 2! ( n ) 2 + ] +. Observe that the coefficient of the term is simply ( ) ] 2 2 + + + 2 2! + 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 + = ( 3 + 3 3! 2 ) ( 5! 4 + + 6 3! 2 ) ] 2 5! 4 + +. 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
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
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
(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