Find principal part of Laurent expansion of $f(z) = \frac{1}{(z^2+1)^2}$ about $z=i$.

My attempt at a solution: First, I noticed that if I plug in $z=i$, I get a zero in the denominator. This leads me to think that it is an isolated singularity. If I look at the classification of singularities, I believe it is a pole since $\lim_{z \to z_0} \vert f(z_0) \vert = \infty$ for $z_0 = i$. By recalling the definition of the principal part of $f$, I am looking for the series containing all negative powers of $(z-z_0)$ in the Laurent expansion $\sum_{k=-\infty}^{\infty} a_k(z-z_0)^k$.

Based on what I have seen, I need to find the partial fraction decomposition of $f$. If so, then I have $\frac{1}{(z^2+1)^2} = \frac{A}{z^2+1}+\frac{B}{(z^2+1)^2}$. However, I think I am doing something wrong. From here, I believe I am supposed to use a geometric series.

I am using the textbook Complex Analysis, Third Edition by Joseph Bak and Donald J. Newman.

Any assistance and clarification would be greatly appreciated.

  • 2
    $\begingroup$ I have no idea what you're talking about but you might want to use $z^2+1=(z-i)(z+i)$ $\endgroup$ – xavierm02 Apr 14 '13 at 20:34
  • $\begingroup$ Ok, after using $z^2+1 = (z-i)(z+i)$, I then have $\frac{1}{((z-i)(z+i))^2}$. Using partial fractions, I have $\frac{A}{z+1}+\frac{B}{z-i} = 1$ which I then found $A = \frac{-1}{2i}, B=\frac{1}{2i}$. $\endgroup$ – Jamil_V Apr 14 '13 at 20:48

Rewrite $f(z)$ as $$ f(z) = \dfrac{1}{(z^2+1)^2}=\dfrac{1}{(z-i)^2(z+i)^2}=\dfrac{-1}{(z-i)^2}\cdot\dfrac{d}{dz}\left(\dfrac{1}{z+i}\right). $$ Using a geometric series, $$ \dfrac{1}{z+i}=\dfrac{1}{z-i+2i}=\dfrac{1}{2i}\cdot\dfrac{1}{\dfrac{z-i}{2i}+1}=\\ =\dfrac{1}{2i}\sum\limits_{n=0}^{\infty}{(-1)^n\left(\dfrac{z-i}{2i}\right)^n} =\dfrac{1}{2i}\sum\limits_{n=0}^{\infty}{\dfrac{(-1)^n}{(2i)^n}({z-i})^n}. $$ Therefore, $$f(z)=\dfrac{-1}{(z-i)^2}\dfrac{d}{dz}\left(\dfrac{1}{z+i}\right)=\dfrac{-1}{(z-i)^2}\dfrac{d}{dz}\left(\dfrac{1}{2i}\sum\limits_{n=0}^{\infty}{\dfrac{(-1)^n}{(2i)^n}({z-i})^n}\right)=\\ =\dfrac{-1}{(z-i)^2}\cdot\dfrac{1}{2i}\cdot \sum\limits_{n=1}^{\infty}{\dfrac{(-1)^n n}{(2i)^n}({z-i})^{n-1}}=\dfrac{-1}{2i}\cdot \sum\limits_{n=1}^{\infty}{\dfrac{(-1)^n n}{(2i)^n}({z-i})^{n-3}}.$$ Then the principal part of Laurent expansion is (for $n=1,2$) $$\dfrac{-1}{2i}\left({\dfrac{-1}{2i}({z-i})^{-2}}+{\dfrac{2}{(2i)^2}({z-i})^{-1}}\right)$$

  • $\begingroup$ I see how $f(z)$ is being rewritten, but what I don't understand is how do you get $\frac{1}{\frac{z-1}{2i}+1}$ to be equal to the series following it? $\endgroup$ – Jamil_V Apr 14 '13 at 20:58
  • 1
    $\begingroup$ Denote $q=\dfrac{z-i}{2i}.$ Then by using the geometric series expansion we get $$\dfrac{1}{\dfrac{z-i}{2i}+1}=\dfrac{1}{1+q}=\dfrac{1}{1-(-q)}=\sum\limits_{n=0}^{\infty}{(-q)^n}=\sum\limits_{n=0}^{\infty}{(-1)^n q^n}=\sum\limits_{n=0}^{\infty}{(-1)^n\left(\dfrac{z-i}{2i}\right)^n}.$$ $\endgroup$ – M. Strochyk Apr 14 '13 at 21:07
  • $\begingroup$ I missed the multiplier $\dfrac{1}{2i} $ in my answer. Corrected. $\endgroup$ – M. Strochyk Apr 14 '13 at 21:20
  • $\begingroup$ thank you very much for your assistance. I believe it is making more sense now. However, now that the geometric series is known, how can I apply that to the principal part of the Laurent expansion if the sum above has from $n=0$ to $\infty$? and the principal part requires the negative powers? $\endgroup$ – Jamil_V Apr 14 '13 at 21:23
  • $\begingroup$ I added that to the answer. $\endgroup$ – M. Strochyk Apr 14 '13 at 21:44

For all rational function problems, we can use long division. Instead of expanding in powers of $z-i$, let's expand in powers of $w$, where $w=z-i$. $$ f(z) = f(w+i) = \frac{1}{((w+i)^2+1)^2} = \frac{1}{w^2(2i+w)^2} = \frac{-1}{4w^2}\left(\frac{1}{1+\frac{w}{2i}}\right)^{2} \\ =\frac{-1}{4w^2}\left(1+iw+\frac{3}{4}\,w^2+\dots\right) \\ =\frac{-1}{4}\;w^{-2}+\frac{-i}{4}\,w^{-1}+\frac{3}{16}+\dots \\ =\frac{-1}{4}\;(z-i)^{-2}+\frac{-i}{4}\,(z-i)^{-1}+\frac{3}{16}+\dots $$


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