Finding the residue of a pole of order 2 (complex analysis)

[mxmt]

The problem

Find Res(f,z1)

With: $f(z)=\frac{z}{(z^2+2aiz-1)^2}$

The attempt at a solution

The singularities are at $A=i(-a+\sqrt{a^2-1})$ and at $B=i(-a-\sqrt{a^2-1})$

With the normal equation (take limit z->A of $\frac{d}{dz}((z-A)^2 f(z))$ for finding the residue of a pole of order 2, my attempt fails.

I know the way the correct answer is constructed, but I do not understand it.

The solution is namely: take $\frac{d}{dz}$ of $\frac{z}{(z-B)^2}$
This is: $\frac{-(z+B)}{(z-B)^3}$
Then plug in z=A, which gives: $\frac{-(A+B)}{(A-B)^3}$

Now plugging in the values of A and B give the correct answer.

Any help is appreciated.

 

[mxmt]

Never mind, I figured it out.