Finding $\lim_{z\to i} \frac{\arctan(1+z^2)^2}{\sin^2(1+z^2)}$ without using Taylor expansions I'm stuck trying to compute this limit: 
$$\lim_{z\to i}  \frac{\arctan(1+z^2)^2}{\sin^2(1+z^2)}$$
I tried to use the logarithm form of $\arctan$ and the exponential form of $\sin$ but the formula got too complicated.

Hint
$$\frac{\tan ^{-1}\left(z^2+1\right)^2}{\sin^2 \left(z^2+1\right)}=\frac{\tan ^{-1}\left(z^2+1\right)^2}{\left(z^2+1\right)^2}\times \left(\frac{{\left(z^2+1\right)}}{\sin \left(z^2+1\right)}\right)^2$$ should help.
