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.