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Mathematical Methods for Physicists: A concise introduction - Site Map

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FUNCTIONS OF A COMPLEX VARIABLE<br />

Solution:<br />

We ®rst apply the method of partial fractions to f …z† and obtain<br />

f …z† ˆ 3<br />

z ‡ 1 ‡ 1 z ‡ 2<br />

z 2 :<br />

Now the center of the given annulus is z ˆ1, so the series we are seeking must<br />

be one involving powers of z ‡ 1. This means that we have to modify the second<br />

and third terms in the partial fraction representation of f …z†:<br />

f …z† ˆ 3<br />

z ‡ 1 ‡ 1<br />

…z ‡ 1†1 ‡ 2<br />

…z ‡ 1†3 ;<br />

but the series <strong>for</strong> ‰…z ‡ 1†3Š 1 converges only where jz ‡ 1j > 3, whereas we<br />

require an expansion valid <strong>for</strong> jz ‡ 1j < 3. Hence we rewrite the third term in<br />

the other order:<br />

f …z† ˆ 3<br />

z ‡ 1 ‡ 1<br />

…z ‡ 1†1 ‡ 2<br />

3 ‡…z ‡ 1†<br />

ˆ3…z ‡ 1† 1 ‡‰…z ‡ 1†1Š 1 ‡ 2‰3 ‡…z ‡ 1†Š 1<br />

ˆ‡…z ‡ 1† 2 2…z ‡ 1† 1 2 3 2 …z ‡ 1†<br />

9<br />

2<br />

27 …z ‡ 1†2 ; 1 < jz ‡ 1j < 3:<br />

Example 6.24<br />

Given the following two functions:<br />

…a† e 3z …z ‡ 1† 3 ; …b† …z ‡ 2† sin 1<br />

z ‡ 2 ;<br />

®nd Laurent series about the singularity <strong>for</strong> each of the functions, name the<br />

singularity, and give the region of convergence.<br />

Solution: (a) z ˆ1 is a triple pole (pole of order 3). Let z ‡ 1 ˆ u, then<br />

z ˆ u 1 and<br />

!<br />

e 3z<br />

…z ‡ 1† 3 ˆ e3…u1†<br />

u 3 ˆ e 3 e 3u<br />

u 3 ˆ e3<br />

u 3 1 ‡ 3u ‡ …3u†2 ‡ …3u†3 ‡ …3u†4 ‡<br />

2! 3! 4!<br />

ˆ e 3 1<br />

…z ‡ 1† 3 ‡ 3<br />

…z ‡ 1† 2 ‡ 9<br />

2…z ‡ 1† ‡ 9 2<br />

27…z ‡ 1†<br />

‡ ‡<br />

8<br />

The series converges <strong>for</strong> all values of z 6ˆ 1.<br />

(b) z ˆ2 is an essential singularity. Let z ‡ 2 ˆ u, then z ˆ u 2, and<br />

278<br />

!<br />

:

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