- #1

Gulli

- 96

- 0

I have to find the Laurent series for the following cases:

[tex]f(z)=\frac{1}{(z-a)(z-b)}[/tex] for [tex]0 < |a| < |b|[/tex] around [tex]z=0[/tex] [tex]z=a[/tex] [tex]z=\infty[/tex] and on the annulus [tex]|a| < |z| < |b|[/tex]

I know I can do a partial fraction thingy: [tex]f(z)=-\frac{1}{a(b-a)(\frac{z}{a}-1)}+\frac{1}{b(b-a)(\frac{z}{b}-1)}[/tex] which can be expressed as [tex]\frac{1}{a(b-a)}\sum(\frac{z}{a})^k - \frac{1}{b(b-a)}\sum(\frac{z}{b})^k[/tex] for [tex]\frac{z}{a}, \frac{z}{b} < 1[/tex]

The condition [tex]\frac{z}{j} < 1[/tex] is met for z = 0, but not for the other cases. I know I must use the substitution [tex]w=\frac{1}{z}[/tex] for the case where z must be taken at infinity (and thus w at 0). But I need to know how to really solve all of these cases.

[tex]f(z)=\frac{1}{(z-a)(z-b)}[/tex] for [tex]0 < |a| < |b|[/tex] around [tex]z=0[/tex] [tex]z=a[/tex] [tex]z=\infty[/tex] and on the annulus [tex]|a| < |z| < |b|[/tex]

I know I can do a partial fraction thingy: [tex]f(z)=-\frac{1}{a(b-a)(\frac{z}{a}-1)}+\frac{1}{b(b-a)(\frac{z}{b}-1)}[/tex] which can be expressed as [tex]\frac{1}{a(b-a)}\sum(\frac{z}{a})^k - \frac{1}{b(b-a)}\sum(\frac{z}{b})^k[/tex] for [tex]\frac{z}{a}, \frac{z}{b} < 1[/tex]

The condition [tex]\frac{z}{j} < 1[/tex] is met for z = 0, but not for the other cases. I know I must use the substitution [tex]w=\frac{1}{z}[/tex] for the case where z must be taken at infinity (and thus w at 0). But I need to know how to really solve all of these cases.

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