Pole-zeros of a real-valued causal FIR system Could someone please help me with the following question?

Below is the magnitude response of a real-valued causal linear phase FIR system of order N = 6. Determine the location of poles and zeros.



I know that for FIR systems all the poles are located at the origin, so we have a pole of order six at the origin. Also from the given diagram, I can say that we have a zero at 0.3pi and one at 0.8pi (both on the unit circle). Now since the system is real-valued, location of poles and zeros should be symmetric w.r.t. the real axis. But I don't know about the two other zeros?

Also, what about the pick in the diagram? Does it mean we have another pole?
Note the difference between the zeros at $0.3 \pi$ and at $0.8 \pi$.

The first one is clearly a zero crossing, much like $abs(x)$ at $x=0$.

At $\theta = 0.8 \pi$, however, the curve is tangent to the horizontal axis, much like $x^2$ at $x=0$. So you have a doulbe zero here.

So your zeros are:


2 zeros at $z = e^{\pm j 0.3 \pi}$
2 double zeros at $z = e^{\pm j 0.8 \pi}$