For (i) and (ii), consider the function $f(z) = z/2+1/2$ on the unit disc, certainly $|f(z)|\le |z|/2+1/2 \le 1$, but $g(z)=1/2+1/(2z)$ has a pole at the origin and so cannot be analytic and $|g(z)|$ is not bounded.
For (iii) Sugata Adhya provides a counter example.
So (iv) has to be correct, and it follows form lee's suggestion:
recall the Cauchy integral formula for the derivatives: $ f^{'}(0) = \frac{n!}{2\pi i }\int_{\partial B(0,r)} \frac{f(z)}{z^2}dz. $ where $r<1$ so the ball $B(0,r)$ lies in the unit circle.
Using the standard estimation, we have $ |f'(0)|\le \frac{1}{2\pi}(2\pi r) \max_{|z|=r}\{\frac{|f(z)|}{|z|^2}\} \le \frac{1}{r} $ since $|f(z)|\le 1$.
Because this is true for every $r < 1$, passing to the limit shows $|f'(0)|\le 1$.