Vtgyjfy

I want to expand the function $ f(z) = frac {z-1}{z^2 -2z -3} $

in $ 0 < |z+1| < 4$

Then I want to use the result to evaluate this integral $ int_C frac {z-1}{(z+1)(z-3)} dz $

in $ C : |z+1|=2$

Now I approach this in 2 different ways, by extracting the residue from the expansion or by integrating the whole expansion

Laurent expansion:

$ f(z) = frac {z-1}{z^2 -2z -3} = frac {z-1}{(z+1)(z-3)} = (frac {z-1}{z+1} )(frac {1}{z-3}) = frac {z-1}{z+1} (frac {1}{(z+1)-4}) = frac {z-1}{z+1} (frac{-1}{4-(z+1)}) = frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n = frac {1-z}{4(z+1)} [1 + frac {z+1}{4} + frac {(z+1)^2}{16} + frac {(z+1)^3}{64} + ..... ] = frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... $

Now I use 2 approaches, first one is to take the residue and work out the integral using the residue theorem and the other is to just integrate the whole expansion

Residue theorem approach:

From the expansion, the residue for $ C : |z+1|=2$ is equal to $ frac{1}{2}$

Then, $int_C frac {z-1}{(z+1)(z-3)} dz = 2 pi i sum Res = 2 pi i ( frac {1}{2}) = pi i$

Now the second approach:

$int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} sum_{n=0}^{infty} (frac {z+1}{4})^n dz$

$ C: |z+1| = 2$ then, $ z+ 1 = 2 e^{i theta}$ , $ dz = 2 i z d theta$

$ int_C frac {z-1}{(z+1)(z-3)} dz = int_C frac {1-z}{4(z+1)} + frac {1-z}{16} + frac{(1-z)(z+1)}{64} + frac {(1-z)(z+1)^2}{256} + ..... dz = 2i int_0^{2 pi} frac{1-e^{i theta}}{4} + frac {(1-e^{i theta})(e^{i theta})}{8} + frac {(1-e^{i theta})(e^{2i theta})}{16} + frac {(1-e^{i theta})(e^{3i theta})}{32} + .... d theta$

= $2i [ frac {1}{4} theta - frac {e^{i theta}}{4i} + frac {e^{i theta}}{8i} - frac {e^{2i theta}}{16i} + frac {e^{2i theta}}{32i} - frac {e^{3i theta}}{48i} + frac {e^{3i theta}}{98i} - frac {e^{4i theta}}{128i} + ..... ]$ as $ theta $ goes from $0$ to $ 2 pi$

Finally $ = 2i [ frac { pi}{2} - frac {1}{8i} - frac {1}{32i} - ... ] = pi i - frac {1}{4} - frac {1}{16} - ....$

Now the problem is that with the residue theorem I got the answer $ pi i $ , but by integrating the whole expansion, keeping in mind that the radius is $2$ yields $ pi i $ minus a bunch of terms, is there a reason why?

$$f(z)=frac{z-1}{(z+1)(z-3)}=frac{(z+1)-2}{(z+1)(z+1-4)}$$Let $u=z+1$.

$$f(u)=frac1u(u-2)frac1{u-4}$$For $|u|in(0,4)$, we can expand the third term

$$begin{align}f(u)&=frac1u(2-u)frac14sum_{n=0}^infty left(frac u4right)^n\&=frac14sum_{n=0}^inftyleft(frac{2u^{n-1}}{4^n}-frac{u^n}{4^n}right)\&=frac1{2u}+frac14sum_{n=0}^inftyleft(-frac{2u^n}{4^{n+1}}right)end{align}$$

This means that the Laurent series for $f(z) $ is $$f(z)=frac1{2(z+1)}+frac14sum_{n=0}^inftyleft(-frac{2(z+1)^n}{4^{n+1}}right)=frac{1}{2(z+1)}-frac18-frac{(z+1)}{32}-frac{(z+1)^2}{512}-cdots$$

Now if you integrate the whole expansion, you should get $pi i$ again.