Determining the complexity of calculating n-th root of an integer, and performing modulo arithmetic?
You can multiply two n-bit numbers trivially in $O(n^2)$, but using FFT (fast fourier transform) you can get the time down to $O (n \log n \log \log n)$. Lets call this time T(n). 

For the modulo, you would calculate ab, then z = floor (ab / m), then r = ab - mz. If mz > ab then z was too large, so decrease z by 1 and subtract m from mz. 

To find ab / r, you can use Newton iteration. There is a clever way to arrange this without any divisions, just multiplications. This is an iterative method with about log n iterations, but only the last iteration needs the full precision, so the total time is O (T(n)). 

Things are even better if you want $z = a^{1/b}$ rounded to an integer, and to check if it is an integer. Again you use Newton iteration, but you only need a result with a bit over n / b bits precision. The result itself is a bit harder to calculate, requiring about log b multiplications, so the time to calculate this is $O (T (n / b) \log b)$. If you can't decide whether z is an integer or not due to precision, you can calculate $round(z)^b$ in T(n).