# Residue of Function - Cauchy's Theorem

I need to find the poles and residues of the following function:

$$f(z) = \frac{1}{az+\frac{1}{2}b(z^2+1)+\frac{c}{2i}(z^2-1))}$$

which can be rewritten (or at least I have) as:

$$f(z) = \frac{1}{z^2+\frac{2az}{b-ci}+\frac{b+ci}{b-ci}}$$

I find the poles by equating the polynomial to zero:

$$z_{+,-} = \frac{\left(-a\pm(a^2-b^2-c^2)^\frac{1}{2}\right)(b+ci)}{b^2+c^2}$$

And then I try to find the residues at the two simple poles (simple because the polynomial on the bottom has simple zeros at $z_{+,-}$) by using the residue limit formula that can be found here: http://en.wikipedia.org/wiki/Residue_%28complex_analysis%29#Simple_poles but I seem to get the wrong answer.

How can I find the residues using Cauchy's thereom and why does the limit formula above not work (I seem to get a real and imaginary part to the residue)?

• How did you get that quadratic in the numerator as the main coefficient is $\;\frac12b+\frac1{2i}c=\frac12\left(b-ci\right)\;$ ? Oh, I see...you divided through by the main coefficient....doesn't look like it made things easier... – DonAntonio Apr 25 '14 at 19:44
• I just used it so I could just use the quadratic formula on the denominator to find the poles. – msd27 Apr 25 '14 at 19:53

$$g(z):=\frac12\left(b-ci\right)z^2+az+\frac12\left(b+ci\right)=0$$

with discriminant

$$\Delta=a^2-b^2-c^2$$

and the quadratic's roots are different whenever $\;a\neq b^2+c^2\;$

The roots are

$$z_{1,2}=\begin{cases}\frac{-a-\sqrt\Delta}{b-ci}\\{}\\\frac{-a+\sqrt\Delta}{b-ci}\end{cases}$$

So we can write $\;g(z)=\frac12(b-ci)(z-z_1)(z-z_2)\;$, and assuming the roots are different (and thus the poles are simple), we have

$$\text{Res}(f)_{z_1}=\lim_{z\to z_1}(z-z_1)f(z)=\frac1{\frac12(b-ci)(z_1-z_2)}$$

and simmilarly for the residue at $\;z_2\;$

If the roots are equal we have then a double pole here, so the derivative must be used in the limit above and etc.

• Thanks I just forgot to factorise out the 0.5(b-ci) – msd27 Apr 25 '14 at 20:16