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Complex Analysis - Maths KU

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6.5 Residues and Residue Theorem 351<br />

29.<br />

30.<br />

31.<br />

32.<br />

33.<br />

34.<br />

�<br />

�<br />

�<br />

�<br />

�<br />

�<br />

C<br />

cot πz dz, C is the rectangle defined by x = 1<br />

, x = π, y = −1, y =1<br />

2<br />

2z − 1<br />

C z2 (z3 +1)<br />

C<br />

C<br />

C<br />

C<br />

�<br />

z 2 e 1/πz + zez<br />

z4 − π4 �<br />

1<br />

dz, C is the rectangle defined by x = −2, x =1,y = − , y =1<br />

2<br />

dz, C: 4x 2 + y 2 =16<br />

cos z<br />

(z − 1) 2 (z2 dz, C: |z − 1| =1<br />

+9)<br />

1<br />

z 6 +1 dz, C is the semicircle defined by y =0,y = √ 4 − x 2<br />

e 4/(z−2) dz, C: |z − 1| =3<br />

Focus on Concepts<br />

35. (a) Use series to show that z = 0 is a zero of order 2 of 1 − cos z.<br />

(b) In view of part (a), z = 0 is a pole of order two of the function<br />

f(z) =e z /(1 − cos z) and hence has a Laurent series<br />

f(z) =<br />

e z<br />

1 − cos z<br />

a−2 a−1<br />

= +<br />

z2 z + a0 + a1z + a2z2 + ···<br />

valid for 0 < |z| < 2π. Use series for e z and 1−cos z and equate coefficients<br />

in the product<br />

e z � �<br />

a−2 a−1<br />

=(1−cos z) + + a0 + ···<br />

z2 z<br />

to determine a−2, a−1, and a0.<br />

�<br />

e<br />

(c) Evaluate<br />

C<br />

z<br />

dz, where C is |z| =1.<br />

1 − cos z<br />

�<br />

36. Discuss how to evaluate e<br />

C<br />

1/z � �<br />

1<br />

sin dz, where C is |z| = 1. Carry out your<br />

z<br />

ideas.<br />

37. Consider the function f(z) =z 4<br />

�<br />

(1 − z 1/2 ), where z 1/2 denotes the principal<br />

branch of the square root function. Discuss and justify your answer: Does f<br />

have a pole at z = 1? If so, find Res(f(z), 1).<br />

38. Residues can be used to find coefficients in partial fraction decompositions of<br />

rational functions. Suppose that p(z) is a polynomial of degree ≤ 2 and that<br />

z1, z2, and z3 are distinct complex numbers that are not zeros of p(z). Then<br />

f(z) =<br />

(a) Use (1) to show<br />

p(z)<br />

A<br />

= +<br />

(z − z1)(z − z2)(z − z3) z − z1<br />

B<br />

+<br />

z − z2<br />

C<br />

.<br />

z − z3<br />

p(z1)<br />

A = Res(f(z),z1) =<br />

(z1 − z2)(z1 − z3)<br />

p(z2)<br />

B = Res(f(z),z2) =<br />

(z2 − z1)(z2 − z3)<br />

p(z3)<br />

C = Res(f(z),z3) =<br />

(z3 − z1)(z3 − z2) .

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