I am trying to refresh my memory about some complex analysis and was intrigued by an exercise in Ahlfors Complex Analys (Ex. 1 in section 5.1.2) which asks you to develop $\frac{1}{1+z^2}$ in powers of $z - a$, $a$ being a real number. Ahlfors has just introduced what amounts to the method of solving problems like this using formal power series and then deriving results about complex power series by checking convergence (he doesn't use formal power series terminology but instead talks about polynomial approximations to the formal power series). I derive the following linear recurrence for the coefficients: $$ \alpha_0 = \frac{1}{1+a^2} \quad \alpha_1 = \frac{-2a}{(1+a^2)^2} \quad \alpha_i = \frac{-2a\alpha_{i-1}-\alpha_{i-2}}{1+a^2}. $$ which you can solve using the usual method for solving linear recurrences based on the characteristic polynomial: $$ f(x) = x^2 + \frac{2a}{1+a^2}x + \frac{1}{1+a^2}. $$ although it looks pretty messy. He then asks you to simplify this for the special case $a = 1$, which is more easily done just by calculating the first few $\alpha_i$ and observing that up to a power of $2$ the solution is periodic with period $8$ (or see Write $\frac {1}{1+z^2}$ as a power series centered at $z_0=1$).

Now my question: why does Ahlfors say $a$ is real? As the discriminant of the characteristic polynomial is $\frac{-4}{1+a^2}$, it has two distinct roots for any complex $a \neq \pm i$. Does the assumption that $a$ is real lead to some neater solution to the linear recurrence? Or an alternative approach to the problem? Or what?

  • $\begingroup$ My Ahlfors doesn't have a section 5.1.2, but it looks like assuming $a$ real is just an excessive way of assuming $a\neq\pm i$, for which the series is different (one negative power) and therefore, the recurrence to be solved is also different (order $1$). $\endgroup$
    – user565560
    May 30 '18 at 13:26

I believe you do not need a reccurence relation to find the power series.

We have that $$\frac{1}{z^2+1}=\frac{1}{2i}(\frac{1}{z-i}-\frac{1}{z+i})$$

For $g(z)=\frac{1}{z-i}$ we have that $g^{(n)}(a)=(-1)^n \frac{1}{(a-i)^{n+1}}$

For $h(z)=\frac{1}{z+i}$ we have that $h^{(n)}(a)=(-1)^n \frac{1}{(a+i)^{n+1}}$

  • $\begingroup$ +1: good point: the method of partial fractions gives a neat solution that explains the periodicity when $a = 1$ nicely, but not it's the approach you'd expect to illustrate the topics preceding the exercise. $\endgroup$
    – Rob Arthan
    May 30 '18 at 15:39

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