# Laurent Series Expansion

Verify that the function $f(z) = \frac {z}{(z − 1)(z + 2)}$ is analytic in the annulus $4 < |z − 1 − 4i| < 5$ and find the Laurent series expansion of f centered at $1 + 4i$ for this annulus.

I understand that the poles are $z=1$ and $z=-2$. Because these values are not in the annulus, the function is analytic on said annulus.

I have a hard time with Laurent series expansion, though.

Hint:

Observe that

$$\frac z{(z-1)(z+2)}=\frac13\left(\frac1{z-1}+\frac2{z+2}\right)=\frac13\left(\frac1{4i+z-1-4i}+\frac2{3+4i+z-1-4i}\right)=$$

$$\frac13\left(\frac1{z-1-4i}\frac1{1+\frac{4i}{z-1-4i}}+\frac2{3+4i}\frac1{1+\frac{z-1-4i}{3+4i}}\right)$$

Finally (for me, not for you...), observe that we both have

$$\left|\frac{4i}{z-1-4i}\right|<1\;,\;\;\;\left|\frac{z-1-4i}{3+4i}\right|<1\;\;\;\;\text{(why?)}$$

Now use the development for geometric series with constant quotient $\;r\;,\;\;|r|<1\;$ ...

• I' not sure how you came to $\frac {1}{3} (\frac {1}{4i+z-1-4i}+\frac {2}{3+4i+z-1-4i})$. I am also not sure how you did your next step either – M Paul Apr 30 '17 at 21:18
• @MPaul I just added and substracted something to each of the two last denominators...For example, the first denominator is $$\color{red}{z-1}=4i+\color{red}{z-1}-4i$$and likewise for the second one...and then I factored something out. Try to recreate each algebraic step slowly and with paper and pencil... – DonAntonio Apr 30 '17 at 21:21
• Ok I see that now. Could I get a bit more guidance on the inequalities and the geometric series portion? – M Paul Apr 30 '17 at 21:48