math185f09-hw10sol - MATH 185: COMPLEX ANALYSIS FALL...

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MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 10 SOLUTIONS 1. (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 > 0 and g : D ( a,ε ) C analytic, g ( a ) 6 = 0, such that f ( z ) = g ( z ) ( z - a ) m for all z D * ( a,ε ). Let the power series representation of g on D ( a,ε ) be g ( z ) = X n =0 a n ( z - a ) n . Let h ( z ) := X n =0 a n + m ( z - a ) n . Then 1 f ( z ) = h ( z ) + m - 1 X n =0 a n ( z - a ) m - n and so e f ( z ) = e h ( z ) m - 1 Y n =0 e an ( z - a ) m - n =: e h ( z ) F ( z ) . Note that e h ( z ) is analytic and non-zero. If e f ( z ) has a pole or a removable singularity at a , then F ( z ) = e f ( z ) e - h ( z ) will have a pole or a removable singularity at a . So since F ( z ) has an essential singularity at a , e f ( z ) 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 > 0, 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 > 0, Re f ( z ) ≤ - m log | z - a | for all z D * ( a,ε ). Show that a is a removable singularity of f . Solution. Note that the condition implies that | e f ( z ) | = e Re f ( z ) e - m log | z - a | = 1 | z - a | m . Date : December 5, 2009. 1 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. 1
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Hence | ( z - a ) m e f ( z ) | ≤ 1 for all z D * ( a,ε ) and thus lim z a | ( z - a ) m +1 e f ( z ) | ≤ lim z a | z - a | = 0 . 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.
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math185f09-hw10sol - MATH 185: COMPLEX ANALYSIS FALL...

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