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Complex Analysis - Maths KU

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386 Chapter 6 Series and Residues<br />

Projects<br />

26. In the application of the Laplace transform to problems involving partial differential<br />

equations, one often encounters an inverse such as<br />

f(x, t) =� −1<br />

�<br />

sinh xs<br />

(s2 �<br />

.<br />

+ 1) sinh s<br />

Investigate how (8) and (9) can be used to determine f (x, t).<br />

Answers to selected odd-numbered problems begin<br />

CHAPTER 6 REVIEW QUIZ<br />

on page ANS-21.<br />

In Problems 1–20, answer true or false. If the statement is false, justify your answer<br />

by either explaining why it is false or giving a counterexample; if the statement is<br />

true, justify your answer by either proving the statement or citing an appropriate<br />

result in this chapter.<br />

1. For the sequence {zn}, where zn = i n = xn + iyn, Re(zn) =xn = cos(nπ/2)<br />

and Im(zn) =yn = sin(nπ/2).<br />

2. The sequence { i n } converges.<br />

� �n 1+i<br />

3. limn→∞ √ =0.<br />

π<br />

4. limn→∞ zn = 0 if and only if limn→∞ |zn| =0.<br />

5. The power series �∞ convergence.<br />

k=1<br />

z k<br />

converges absolutely at every point on its circle of<br />

k2 6. There exists a power series centered at z0 =1+i that converges at z =25−4i and diverges at z =15+21i.<br />

7. A function f is analytic at a point z0 if f can be expanded in a convergent<br />

power series centered at z0.<br />

8. Suppose a function f has a Taylor series representation with circle of convergence<br />

|z − z0| = R, R>0. Then f is analytic everywhere on the circle of<br />

convergence.<br />

9. Suppose a function f has a Taylor series representation centered at z0.<br />

Then f is analytic everywhere inside the circle of convergence |z − z0| = R,<br />

R>0, and is not analytic everywhere outside |z − z0| = R.<br />

10. If the function f is entire, then the radius of convergence of a Taylor series<br />

expansion of f centered at z0 =1− i is necessarily R = ∞.<br />

11. Both power series<br />

1<br />

1+z =1− z + z2 − z 3 + ···<br />

and<br />

converge at z =0.86 − 0.52i.<br />

1 1 z − 1<br />

= −<br />

1+z 2 22 (z − 1)2<br />

+<br />

23 (z − 1)3<br />

−<br />

24 + ···

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