smc

First, assume that $z_0\in \mathbb{R}$, so $z_0=1$ or $-1$. $1$ does not work because $P(1)\ge 4$. Assume that $z_0=-1$. Then $0=P(-1)=4-a+b-c+d$, we have $4+b+d=a+c\le 4+b$, so $d=0$. Also, $a=4$ has to be true since $4+b=a+c\le a+b$. Now $4+b=4+c$ gives $b=c$, therefore the only possible choices for $(a,b,c,d)$ are $(4,t,t,0)$. In these cases, $P(-1)=4-4+t-t+0=0$. The sum of $P(1)$ over these cases is ${\sum }_{t=0}^4(4+4+t+t)=40+20=60$. Second, assume that $z_0\in \mathbb{C}\backslash \mathbb{R}$, so $z_0=x_0+i{y}_0$ for some real $x_0,y_0$, $|x_0|<1$. By conjugate roots theorem we have that $P(z_0)=P(z_0^{\ast })=0$, therefore $(z-z_0)(z-z_0^{\ast })=(z^2-2x_0\ast z+1)$ is a factor of $P(z)$, and we may assume that $$P(z)=(z^2-2x_{0}z+1)(4z^2+pz+d)$$ for some real $p$. Expanding this polynomial and comparing the coefficients, we have the following equations: $$p-8x_0=a$$ $$d+4-2p{x}_0=b$$ $$p-2d{x}_0=c$$ From the first and the third we may deduce that $2x_0=\frac{a-c}{d-4}$ and that $p=\frac{da-4c}{d-4}$, if $d\ne 4$ (we will consider $d=4$ by the end). Let $k=2p{x}_0=\frac{(a-c)(da-4c)}{(4-d{)}^2}$. From the second equation, we know that $k=d+4-b$ is non-negative. Consider the following cases: Case 1: $a=c$. Then $k=0$, $b=d+4$, so $a=b=c=4$, $d=0$. However, this has already been found (i.e. the form of $(4,t,t,0)$). Case 2: $a>c\ge 0$. Then since $k\ge 0$, we have $da-4c\ge 0$. However, $da\le 4c$, therefore $da-4c=0$. This is true only when $d=c$. Also, we get $k=0$ again. In this case, $b=d+4$, so $a=b=4$, $c=d=0$, $x_0=-1/2$. $P(z)$ has a root $z_0=e^{i2\pi /3}$. $P(1)=12$. Last case: $d=4$. We have $a=b=c=d=4$ and that $P(z)$ has a root $z_0=e^{i2\pi /5}$. $P(1)=20$. Therefore the desired sum is $60+12+20=\mathrm{92...}\boxed{92}$.