If you want to calculate a number z, you may calculate it using the Taylor formula if you can express z as z = f (x), where f (x) and all its derivatives can all be calculated easily, and you usually gain a fixed number of digits at each iteration (unless x is close to the border of the circle where the Taylor formula converges).
You may find a function where z is the solution of f (z) = 0, and you can either use the Newton scheme, or some other approximation scheme if f' is hard to calculate, and double the number of digits at each step. Of course you have to find a function where f (z) can be calculated with the required precision.
Take sqrt (2). You can use the Taylor formula for sqrt (x) around x = 1.414^2, for example, which will converge at maybe three digits per round. Or you use the Newton formula for f (z) = z^2 - 2 and the number of digits doubles on each iteration.