a) $Res(3i)$ for $R(z) = \frac{z^2-9}{(z^2+9)^2}$
So I solved this by using the formula is:
$\lim_{z \to 3i} \frac{1}{1!} \frac{d}{dz} [(z^2+9)^2 \times \frac{z^2-9}{(z^2+9)^2}]$
$\lim_{z \to 3i} \frac{d}{dz} [z^2-9]$
$\lim_{z \to 3i} 2z$ which will give $6i$ but my book says the answer is $0$.
b) $Res(-1)$ for $R(z) = \frac{z^3+4z+9}{(2z+2)(z-3)^5}$
So I solved this by using the formula as well:
$\lim_{z \to -1} \frac{1}{1!} \frac{d}{dz} [(2z+2) \times \frac{z^3+4z+9}{(2z+2)(z-3)^5}]$
$\lim_{z \to -1} [\frac{z^3+4z+9}{(z-3)^5}]$ which I get $\frac{-1}{256}$ but the answer is wrong the correct answer is $\frac{-1}{512}$.
I just don't know where I am wrong?