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Expand f(z)=1z(z3)f(z)=\frac{1}{z(z-3)} in a Laurent series valid for the indicated annular domain. 1<|z-4|<4

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By the geometric series of 11±z\frac{1}{1\pm z} in the domain z<1|z|<1, we have

11+z=1z+z2z3+\frac{1}{1+z}=1-z+z^2-z^3+\cdots

11z=1+z+z2+z3+\frac{1}{1-z}=1+z+z^2+z^3+\cdots

Consider the function f(z)=1z(z3)f(z)=\frac{1}{z(z-3)} in 1<z4<41<|z-4|<4. The domain 1<z4<41<|z-4|<4 equivalent to z44<1|\frac{z-4}{4}|<1 & 1z4<1\frac{1}{|z-4|}<1.

To expand f(z)f(z) in terms of 1z4\frac{1}{z-4} or z4z-4, rewrite the function f(z)f(z) as follows:

f(z)=1z1z3=14+z411+z4=14(z4)11+z4411+1z4\begin{align*} f(z)&=\frac{1}{z}\cdot \frac{1}{z-3}\\ &=\frac{1}{4+z-4}\cdot \frac{1}{1+z-4}\\ &=\frac{1}{4(z-4)}\frac{1}{1+\frac{z-4}{4}}\cdot \frac{1}{1+\frac{1}{z-4}}\\ \end{align*}

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