Residue Complex Function

Question

I've applied the formula $$\lim_{z \to i} \frac{d}{dz}{(z-z_0)^2.f(z)}$$ and I keep getting a value of 0. Could someone prove my sanity by starting this off?

Answer 1

Here a solution with less calculations.

• $f(z) = \frac{z^2-1}{(z^2+1)^2} = \frac{z^2+1-2}{(z^2+1)^2}= \frac{1}{z^2+1} - \frac{2}{(z^2+1)^2} $
• $Res_{z=i}\frac{1}{z^2+1} = \lim_{z\rightarrow i}\frac{z-i}{(z-i)(z+i)}=\frac{1}{2i}$
• For $Res_{z=i}\left(- \frac{2}{(z^2+1)^2}\right) = -2 Res_{z=i} \frac{1}{(z^2+1)^2}$ you need to calculate $$\frac{d}{dz}\frac{(z-i)^2}{(z-i)^2(z+i)^2} = \frac{d}{dz}\frac{1}{(z+i)^2}= \frac{-2}{(z+i)^3} \stackrel{z\rightarrow 0}{\longrightarrow}\frac{1}{4i}$$
• All together: $$Res_{z=i}f(z) = \frac{1}{2i} - 2\cdot\frac{1}{4i} = 0$$

p.s.: You can check your residues with wolfram alpha. Here the : output for above residue

Answer 2

Hint: Since $f(z)$ has a double pole at $z=i$, then we see that \begin{align} \operatorname{Res}_{z=i}f(z)=\lim_{z\rightarrow i}\frac{1}{2}\frac{d}{dz}\left((z-i)^2\frac{z^2-1}{(z^2+1)^2}\right). \end{align}

Edit: To put you at ease, observe the power series expansion at $z=i$ of the function is given by \begin{align} \frac{z^2-1}{(z^2+1)^2} =&\ \frac{1}{(z-i)^2}\frac{z^2-1}{(z+i)^2} =\frac{-1}{4(z-i)^2}\frac{(z-i)^2+2i(z-i)-2}{\left(1+\frac{z-i}{2i}\right)^2}\\ =&\ \frac{-1}{4(z-i)^2}\left((z-i)^2+2i(z-i)-2\right)\sum_{k=0}^\infty(-1)^k(k+1)\left(\frac{z-i}{2i} \right)^k\\ =&\ \left(\frac{-1}{4} -\frac{i}{2(z-i)}+\frac{1}{2(z-i)^2}\right)\sum_{k=0}^\infty(-1)^k(k+1)\left(\frac{z-i}{2i} \right)^k\\ =&\ \frac{1}{2(z-i)^2}-\frac{1}{8}-\frac{1}{8}i(z-i)+\text{higher order terms}. \end{align} Hence it is clear that the residue is zero.