Find Taylor Series of $\frac{1}{1+z^2}$ around $1$

Question

For $f(z)=\dfrac{1}{1+z^2}$ find the Taylor series centered at $1$.

While I know I could use partial fractions or perhaps maneuver this problem by adding constants, I would really like to use the following formula because I do not fully understand it. I have an exam in this tomorrow and this seems to conceptually give me a lot of problems. Any help is much appreciated. $$f^{(k)}(w)=\frac{k!}{2i\pi}\int\frac{f(z)}{(z-w)^{k+1}}dz$$

when I use partial fractions, I am confused on if $w$ becomes $\Bbb i$ (as opposed to $1$) and if $f(z)$ changes, such as for the Cauchy Integral Formula. Even then, how would I obtain the series representation? Thanks!

Answer 1

Your confusion makes sense: $w$ would become $\Bbb i$ only if you were required to compute the Taylor series around $\Bbb i$, which you will never be required to do because your $f$ is not analytic there.

Also, using Cauchy's formula is not easy in this example since it would lead to very ugly integrals. This formula is beautiful from a theoretical point of view, allowing one to conceive some proofs and gain mathematical insight, but in concrete problems it is wiser to use simpler techniques. Partial fraction expansion combined with geometric series, in this case, form one such simpler approach.

You have

$$\dfrac 1 {1+z^2} = \dfrac 1 2 \dfrac 1 {z- \Bbb i} + \dfrac 1 2 \dfrac 1 {z+ \Bbb i} .$$

You want to expand each fraction as a Taylor series around $1$, and this means coming up with some power series that exhibit the quantity $z-1$. Therefore, rewrite your fractions as

$$\dfrac 1 2 \dfrac 1 {(z-1) + (1 - \Bbb i)} + \dfrac 1 2 \dfrac 1 {(z-1) + (1 + \Bbb i)} .$$

Remember that $\dfrac 1 {1 - t} = \sum \limits _{t=0} ^\infty t^n$ (the geometric series), as long as $|t| < 1$, therefore let us try to produce formulae resembling $\dfrac 1 {1-t}$. We get that your function is

$$\dfrac 1 {2(1 - \Bbb i)} \dfrac 1 {\frac {z-1} {1 - \Bbb i} + 1} + \dfrac 1 {2(1 + \Bbb i)} \dfrac 1 {\frac {z-1} {1 + \Bbb i} + 1}$$

or, even clearer,

$$\dfrac 1 {2(1 - \Bbb i)} \dfrac 1 {1 - \left( -\frac {z-1} {1 - \Bbb i} \right)} + \dfrac 1 {2(1 + \Bbb i)} \dfrac 1 {1 - \left( -\frac {z-1} {1 + \Bbb i} \right)} ,$$

and now expand these two fractions as geometric series:

$$\dfrac 1 {2(1 - \Bbb i)} \sum \limits _{n=0} ^\infty \left( -\frac {z-1} {1 - \Bbb i} \right)^n + \dfrac 1 {2(1 + \Bbb i)} \sum \limits _{n=0} ^\infty \left( -\frac {z-1} {1 + \Bbb i} \right)^n .$$

Finally, put everything inside a single sum:

$$\sum \limits _{n=0} ^\infty \dfrac {(-1)^n} 2 \left( \dfrac 1 {(1 - \Bbb i)^{n+1}} + \dfrac 1 {(1 + \Bbb i)^{n+1}} \right) (z-1)^n .$$

It is possible to further slightly embelish this, but this is not relevant so let us stop here. This Taylor series exists only on that subset of $\Bbb C$ where the two intermediate geometric series exist, i.e. for

$$\{ z \in \Bbb C \mid \left| \dfrac {z-1} {1 - \Bbb i} \right| < 1 \} \cap \{ z \in \Bbb C \mid \left| \dfrac {z-1} {1 + \Bbb i} \right| < 1 \}$$

which is easily seen to be

$$\{ z \in \Bbb C \mid \left| {z-1} \right| < \sqrt 2 \} .$$

This is the standard approach for this type of exercise; attacking it with Cauchy's formula will only make you waste precious time during an examination.

Answer 2

You can take advantage of integrals: $$ \frac{1}{1+z^2}=\frac{1}{2}\frac{1}{z+i}+\frac{1}{2}\frac{1}{z-i} $$ An antiderivative of $1/(z+i)$ can be written $\log(z+i)$ (with an appropriate branch cut) and $$ \log(z\pm i)=\log(z-1+1\pm i)= \log(1\pm i)+\log\left(1+\frac{z-1}{1\pm i}\right) $$ (the actual term might differ by a constant, depending on what cut you choose); since $$ \log\left(1+\frac{z-1}{1\pm i}\right)= \sum_{n\ge0}\frac{(-1)^{n}}{n+1}\frac{(z-1)^{n+1}}{(1\pm i)^{n+1}} $$ you can conclude by differentiation that $$ \frac{1}{z\pm i}=\sum_{n\ge0}(-1)^n\frac{(z-1)^n}{(1\pm i)^{n+1}} $$