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

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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.
 
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Never mind, I figured it out.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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