I'm trying to understand this great answer from Matt L. . It's said that "One advantage of IIR filters is that steep filters with high stopband attenuation can be realized with much fewer coefficients (and delays) than in the FIR case, i.e. they are computationally more efficient." First of all why this is true? Is this because of poles? Actually this comes from my previous question on removing 400Hz noise. There are two filters for that purpose. The first one is a FIR filter with following frequency response and pole-zero plot: And the second one is a IIR filter: Removing noise using the IIR filter was far better. Why this happens? According to the frequency responses, attenuation in the case of FIR filter is greater in the magnitude relative to IIR filter. Also I don't understand why zeros in the FIR filter are placed so that and how this positioning corresponds to that frequency response. I understand well the relation between pole-zero plot and frequency response for the IIR filter.
Your assumption why IIR filters can have steeper transitions from passbands to stopbands compared to FIR filters of the same order is correct: IIR filter have poles away from the origin of the complex plane, and poles inside the unit circle close to zeros on the unit circle cause the corresponding frequency response to change rapidly with frequency.
The FIR filter in your question has a much broader notch than the IIR filter. This is not only caused by the fundamental difference between FIR and IIR filters, but also because for some reason the FIR filter has two zeros close to the notch frequency instead of only one. This makes the filter more robust with respect to errors in the estimation of the noise frequency, but it may also attenuate desired frequency components. It's also much harder for an FIR filter to approximate a constant response away from the notch frequency because the frequency response of an FIR filter is a polynomial and not a rational function as is the case for IIR filters.
From the zero locations of the FIR filter you can see that the filter is a linear phase filter: the zeros are either on the unit circle or they are mirrored at the unit circle. Note that FIR filters don't need to have linear phase. The required filter order for a certain specification can often be somewhat reduced if we don't impose the linear phase constraint.
Is this because of poles?
Yes. A steep drop or rise in a filter's frequency-domain response can only be achieved with a filter that has a long memory. In a FIR filter, this long memory can only come from, well, being long. In IIR filter, this long memory can come from having poles that are close to the unit circle.