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.

$$\lim_{z\to i}  \frac{\arctan(1+z^2)^2}{\sin^2(1+z^2)}=\lim_{w\to 0}  \frac{\dfrac{\arctan w^2}{w^2}}{\dfrac{\sin^2 w}{w^2}}=1.$$
